Oswaal IIT JEE Main Solved Papers Physics Chapterwise and Topicwise 2019 and 2020 All shifts 32 Papers

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PHYSICS

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1ST edition, year 2021

ISBN SYLLABUS COVERED

“978-93-5423-007-3”

JEE (MAIN)

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PREFACE Logical thinkers, problem solvers, creators, engineers are definitely the most recognisable and valued people around the world. But many students preparing for JEE still ask; how good is engineering as a career option? Engineering has become the most lucrative profession now more than ever, with the demand for engineers exceeding their supply. Engineers bring ideas to life. In every field, engineers play a very important role in devising and executing new and better technologies. Thus, to take up engineering as a profession cracking the JEE (Main) exam is the first and the most important step. Every aspiring engineer in the country works diligently for this exam. This makes JEE a highly competitive exam. With the core objective of “Learning Made Simple” Oswaal Editorial Board has designed Oswaal JEE (Main) Solved Papers to help these aspiring candidates crack one of the most difficult examinations in the country. The book includes Previous Years’ Papers; to tell the students what to expect and familiarise them with the exam pattern that has been followed year on year along with the Scheme of Examination issued by the NTA on 16th Dec 2020.

Benefits of studying from this book are:

1. Based on the Scheme of Examination issued by the NTA on 16th Dec 2020 2. Fully Solved Papers of 2019 & 2020 JEE (Main) 3. Chapter-wise & Topic-wise presentation for systematic learning 4. Subjective (Integer Types) Questions for extensive practice 5. Concept Revision (Video Based) for hybrid learning 6. Commonly Made Errors to polish concepts 7. Mind Maps for better retention

8. Mnemonics for memorise the concepts.

This book aims to make students exam-ready, boost their confidence and help them achieve better results. We hope that OSWAAL JEE (Main) Solved Papers 2019 & 2020 will help you at every step as you move closer to your educational career goal. We wish you all great success ahead!

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TEAM OSWAAL

contents g g

Latest Syllabus for Academic Year 2021

7 - 8 9 - 9 10 - 10 11 - 16

10 Tips & Tricks to Crack JEE (Main) Trend Analysis JEE (Main) 2019 & 2020 g Mnemonics g

1. Physics and Measurement Topic 1. Dimensional Analysis

1 - 14

Topic 2. Least Count, Significant Figures and Error Analysis

2. Kinematics Topic 1. Motion in a Straight Line

15 - 31



Topic 2. Conservation of Linear Momentum

Topic 2. Rotational Motion

and Gravitational Potential Energy

Topic 2. Motion of Satellites and Escape Velocity

7. Properties of Solids and Liquids Topic 1. Elasticity

Topic 1. Electrostatic Force, Electric Field



Topic 2. Capacitors Topic 1. Electric Current, Ohm’s Law,



Topic 2. Kirchhoff ’s Laws and Electrical Instruments

113 - 135

Topic 2. Magnetism



Topic 1. Rectilinear Propagation of Light Topic 2. Wave Optics

8. Thermodynamics 136 - 153 Topic 1. Heat, Work and Internal Energy



Topic 2. Properties of Liquids and Calorimetry

Topic 2. Carnot Engine and P-V Diagrams

9. Kinetic Theory of Gases Topic 1. Kinetic Theory of Gases

154 - 169

Topic 2. Degrees of Freedom and Specific Heat Capacities of Gases

What some preparation tips ? There's an ocean of knowledge waiting for you. Dive in by scanning the QR code.

SCAN THE CODE

256 - 285

14. Electromagnetic Induction and Alternating Currents 286 - 305 Topic 1. Electromagnetic Induction Topic 2. RC, LR, LC and LCR Circuits 15. Electromagnetic Waves 306 - 318 16. Optics 319 - 347

17. Dual Nature of Matter and Radiation 18. Atoms and Nuclei Topic 1. Bohr’s Atomic Model



229 - 255





6. Gravitation 96 - 112 Topic 1. Gravitational Force, Gravitational Potential



13. Magnetic Effects of Current and Magnetism Topic 1. Magnetic Effects of Current

44 - 62

5. Rotational Motion 63 - 95 Topic 1. Centre of Mass and Moment of Inertia

194 - 228

Topic 2. Waves Motion

Electric Resistance

Topic 2. Frictional Forces

4. Work, Energy and Power Topic 1. Work, Energy and Power

11. Electrostatics



12. Current Electricity

32 - 43

Dynamics of Uniform Circular Motion



170 - 193

and Electrostatic Potential

Topic 2. Kinematics in Two Dimensions

3. Laws of Motion Topic 1. Equilibrium of Forces and

10. Oscillations and Waves Topic 1. Simple Harmonic Motion

348 - 362 363 - 379

Topic 2. Radioactivity

19. Electronic Devices 20. Communication Systems

380 - 393 394 - 399

• Appendix

400 - 408 qq

6 Tips to Prepare for JEE (Main) and Class 12th Board Exams Simultaneously

SCAN

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JEE (Main) Latest Syllabus for Academic Year (2021)

PHYSICS

The syllabus contains two sections - A and B. Section - A pertains to the Theory Part having 80% weightage, while Section - B contains Practical Component (Experimental Skills) having 20% weightage.

SECTION - A UNIT 1: PHYSICS AND MEASUREMENT Physics, technology and society. S I units. Fundamental and derived units. Least count, accuracy and precision of measuring instruments. Errors in measurement. Dimensions of Physical quantities, dimensional analysis and its applications. UNIT 2: KINEMATICS Frame of reference. Motion in a straight line: Position time graph, speed and velocity. Uniform and non-uniform motion, average speed and instantaneous velocity Uniformly accelerated motion, velocity-time. position-time graphs, relations for uniformly accelerated motion. Scalars and Vectors. Vector addition and Subtraction, Zero Vector, Scalar and Vector products. Unit vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion. UNIT 3: LAWS OF MOTION Force and Inertia, Newton’s First law of motion; Momentum, Newton’s Second Law of motion; Impulse; Newton’s Third Law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion; Centripetal force and its applications. UNIT 4: WORK, ENERGY AND POWER Work done by a constant force and a variable force; kinetic and potential energies, workenergy theorem, power. Potential energy of a spring, conservation of mechanical energy, conservative and nonconservative forces; Elastic and inelastic collisions in one and two dimensions. UNIT 5: ROTATIONAL MOTION Centre of mass of a two-particle system, Centre of mass of a rigid body: Basic concepts of rotational motion: moment of a force, torque, angular momentum, conservation of angular momentum and its applications; moment of inertia, radius of gyration. Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion. UNIT 6: GRAVITATION The universal law of gravitation. Acceleration due to gravity and its variation with attitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy; gravitational potential. Escape velocity, Orbital velocity of a satellite, Geo-stationary satellities. UNIT 7: PROPERTIES OF SOLIDS AND LIQUIDS Elastic behaviour, Stress-strain relationship, Hooke’s Law, Young’s modulus, bulk modulus, modulus of rigidity, Pressure due to a fluid column; Pascal’s law and its applications, Viscosity, Stokes’ Law, Reynolds number, Bernoulli’s principle and its applications. Surface energy and surface tension, angle of contact, application of surface tension- drops , bubbles and capillary rise. Heat, temperature, thermal expansion; specific Heat capacity, calorimetry; change of state, latent heat, Heat transferconduction, convection and radiation, Newton’s law of cooling.

UNIT 8: THERMODYNAMICS Thermal equilibrium, zeroth law of thermodynamics, concept of temperature. Heat, work and internal energy, First law of thermodynamics, Second law of thermodynamics: reversible and irreversible processes, Carnot engine and its efficiency. UNIT 9: KINETIC THEORY OF GASES Equation of state of a perfect gas, work doneon compressing a gas. kinetic theory of gases- assumptions concepts of pressure, kinetic energy and temperature: rms speed of gas molecules; Degrees of freedom, Law of equipartition of energy, applications to specific heat capacities of gases; Mean free path, Avogadro's number. UNIT 10: OSCILLATIONS AND WAVES Periodic motion- period, frequency, displacement as a function of time. Periodic functions, Simple harmonic motion (S.H.M) and its equation; phase; oscillations of a spring- restoring force and force constant; energy in S.H.M, - kinetic and potential energies; Simple pendulumderivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion, Longitudinal and transverse waves, speed of a wave,Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound. UNIT 11: ELECTROSTATICS Electric charges; Conservation of charge, Coulomb’s law- forces between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field: Electric field due to a point charge. Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its circulation for a point charge, electric dipole and system of charge; Equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, Dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, Energy stored in a capacitor. UNIT 12: CURRENT ELECTRICITY Electric current, Drift velocity, Ohm’s law. Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and nonohmic conductors, Electrical energy and power. Electrical resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. Electric Cell and its Internal resistance, potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff ’s laws and their applications. Wheatstone bridge, Metre bridge, Potentiometer- principle and its applications. UNIT 13: MAGNETIC EFFECTS OF CURRENT AND MAGNETISM Biot- Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields. Cyclotron.

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...CONTD. Force on a current- carrying conductor in a uniform magnetic field. Force between two parallel current carrying conductors- definition of ampere. Torque experienced by a current loop in uniform magnetic field; Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment, Bar magnet as an equivalent solenoid, magnetic field lines, Earth’s magnetic field and magnetic elements. Para-dia-and ferro-magnetic substances. Magnetic susceptibility and permeability, Hysteresis, Electromagnets and permanent magnets. UNIT 14: ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS Electromagnetic induction; Faraday’s law, induced emf and current; Lenz’s Law, Eddy currents, Self and mutual inductance. Alternating currents, peak and rms value of alternating current/ voltage; reactance and impedance; LCR series circuit, resonance; Quality factor power in AC circuits, wattless current AC generator and transformer. UNIT 15: ELECTROMAGNETIC WAVES Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnatic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, Xrays, gamma rays). Applications of e.m. waves. UNIT 16: OPTICS Reflection and refration of light at plane and spherical surfaces, mirror formula, Total internal reflaction and its applications, Deviation and Dispersion of light by a prism, Lens Formula, Magnification, Power of a Lens, Combination of thin lenses in contact, Microscope and Astronomical Telescope (reflecting and refracting) and their magnifying powers. Wave optics: wavefront and Huygen’s principle, Laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, Diffraction due to a single slit, width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, plane polarized light; Brewster’s law, uses of plane polarized light and Polaroids. UNIT 17: DUAL NATURE OF MATTER AND RADIATION Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; particle nature of light, Matter waves-wave nature of particle, de Broglie relation, Davisson- Germer experiment. UNIT 18: ATOMS AND NUCLEI Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars: isotones. Radioactivity-alpha, beta and gamma particles/rays and their properties; radioactive decay law. Mass- energy relation, mass defect; biding energy per nucleon and its variation with mass number, nuclear fission and fusion. UNIT 19: ELECTRONIC DEVICES Semiconductors, semiconductor diode: 1-V characteristics in forward and reverse bias; diode as a rectifier; I-V characteristics of LED, photodiode, solar cell and zener diode: zener diode as a voltage regulator. Junction transistor, transistor action. characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch.

UNIT 20: COMMUNICATION SYSTEMS Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and Frequency Modulation Bandwidth of signals, Bandwidth of Transmission medium, Basic Elements of a Communication System (Block Diagram only).

SECTION - B UNIT 21: EXPERIMENTAL SKILLS Familiarity with the basic approach and observations of the experiments and activities: 1. Vernier callipers- its use to measure internal and external diameter and depth of a vessel. 2. Screw gauge- its use to determine thickness/ diameter of thin sheet/wire. 3. Simple Pendulum- dissipation of energy by plotting a graph between square of amplitude and time. 4. Metre Scale- mass of a given object by principle of moments. 5. Young’s modulus of elasticity of the material of a metallic wire. 6. Surface tension of water by capillary rise and effect of detergents. 7. Co-efficient of Viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body. 8. Plotting a cooling curve for the relationship between the temperature of a hot body and time. 9. Speed of sound in air at room temperature using a resonance tube. 10. Specific heat capacity of a given (i) solid and (ii) liquid by method of mixtures. 11. Resistivity of the material of a given wire using metre bridge. 12. Resistance of a given wire using Ohm’s law. 13. Potentiometer : (i) Comparison of emf of two primary cells. (ii) Determination of internal resistance of a cell. 14. Resistance and figure of merit of a galvanometer by half deflection method. 15. Focal length of: (i) Convex mirror (ii) Concave mirror, and (iii) Convex lens using parallax method. 16. Plot of angle of deviation vs angle of incidence for a triangular prism. 17. Refractive index of a glass slab using a travelling microscope. 18. Characteristic curves of a p-n junction diode in forward and reverse bias. 19. Characteristic curves of a zener diode and finding reverse break down voltage. 20. Characteristic curves of a transistor and finding current gain and voltage gain. 21. Identification of Diode, LED, Transistor, IC, Resistor, Capacitor from mixed collection of such items. 22. Using multimeter to: (i) Identify base of a transistor. (ii) Distinguish between npn and pnp type transistor. (iii) See the unidirectional flow of current in case of a diode and an LED. (iv) Check the correctness or otherwise of a given electronic component (diode, transistor or IC). 

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10 TIPS & tricks TO CRACK JEE (main) EXAM IN THE FIRST ATTEMPT Joint Entrance Examination or JEE (Main) is conducted by NTA. After clearing JEE (Main), candidates will be eligible to apply for admission to IITs, NITs, CFTIs and almost all prestigious engineering colleges. Cracking the JEE (Main) Exam in the very first attempt; given the difficulty level, can be a tough task. But it is quite attainable if done diligently as well as smartly. Here we are giving you 10 tips that you must follow by heart to crack the exam in the very first attempt:

1. Start Studying from The Beginning

All the aspirants are aware of how vast, comprehensive and detailed the syllabus of the JEE (Main) exam is. To crack the exam in the first attempt you have to start preparing for the exam from the beginning of Class 12th It is only then that you will be able to complete the entire syllabus. Following this approach will also leave you with plenty of time to revise.

2. Prioritize & Plan Devise a study plan which is realistic rather than being idealistic. Try to cover a certain number of topics ever day.

3. Formulas Are Very Important

For JEE (Main) 2021, some of the questions will be based on direct formulas and thus students should remember them. A simple trick to remember all the formulas are that write all the formulas on a paper and paste them in front of study table, it will be helpful to revise anytime.

4. Get the Right Tools and Study Material

Collecting and preparing from the appropriate study material is something that you can’t ignore. You should refer to Oswaal JEE (Main) 15 Mock Test Papers to boost your preparation. The books you choose to study from should be on the lines of the current syllabus and the ones that could be trusted on before examination.

5. Practice Important Topics & Strengthen Your Weak Areas

You must analyse the detailed syllabus of all the sections and then start preparing first for the important topics, which are frequently questioned upon in the previous years’ exams. Try to cover those topics which are your weak areas and then cover the ones which are your strength.

6. Understand the Concepts

No one can crack the JEE (Main) exam just by mugging up all the concepts and topics. The syllabus of the exam is in-depth and to achieve good scores you need to understand every concept.

7. Revise Whenever You Get Time

Make sure you revise as much as possible. The revision will help you in keeping the concepts fresh in your mind until the day of the final examinations. You may refer to a few good Question Banks, Sample Papers and your self-made notes for this purpose.

8. Keep A Track on Time

While you are solving papers, make sure you keep a track on time i.e., how much time does it take to solve one sections and the type of questions which take minimum and maximum time.

9. Exam Day Strategy First & foremost, try to be in time at the exam centre, it will help you keep yourself calm. Scoring good marks is all about identifying the questions which you should be attempting first and the ones which are to be solved in the second round. Always try to attempt those questions first which seems familiar, less time consuming and easy.

10. Keep Yourself Motivated & Healthy

Don’t be Anxious, keep yourself Calm! Taking care of your thought process and keeping it positive is the first and the best course of action that one is required to take. This time is very important for you, so is everything you are eating or thinking. Eat healthy and easy to digest food and take proper sleep.

Always remember that to achieve good scores you will need consistent efforts and calm mind. Trust on your honest efforts, if in case at any point of time you feel stressed, don’t be hesitant in taking help from counsellors or family members. Your focus should only be on clearing the JEE exam and to give your best shot.

All the Best!!



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TREND ANALYSIS JEE (MAIN) 2019 & 2020 PHYSICS Total Papers Chap. No.

Chapter Name

8

8

6

10

Jan 2019

Apr 2019

Jan 2020

Sep 2020

Phase - 1

Phase - 2

Phase - 1

Phase - 2

1

Physics and Measurement

8

7

5

10

2

Kinematics

14

11

6

11

3

Laws of Motion

5

4

3

6

4

Work, Energy and Power

6

10

9

10

5

Rotational Motion

18

19

16

21

6

Gravitation

6

7

3

11

7

Properties of Solids and Liquids

6

13

7

9

8

Thermodynamics

18

13

11

16

9

Kinetic Theory of Gases

6

10

6

12

10

Oscillations and Waves

20

19

7

15

11

Electrostatics

23

20

12

21

12

Current Electricity

20

16

8

12

13

Magnetic Effects of Current

23

19

9

22

14

Electromagnetic Induction and Alternating Currents

9

11

9

14

15

Electromagnetic Waves

7

6

5

8

16

Optics

20

19

14

21

17

Dual Nature of Matter and Radiation

7

9

6

10

18

Atoms and Nuclei

9

11

6

11

19

Electronic Devices

8

9

8

8

20

Communication Systems

7

7

0

2

240

240

150

250

Total Questions



( 10 )

( 11 )

PHYSICS Class - 11, Unit-I

Class - 11, Unit-IV

Physical World

Work, Energy And Power

Good Workers work for Extended Session.

Fernandez d’souza ordered noodles, but was served pizza and pizza was a zest.

Strength wise arrangement of fundamental forces in ascending order : Gravitation < Weak Nuclear force < Electromagnetism < Strong Nuclear force

If force and Displacement are in opposite direction, then work done is negative. If force and Displacement are in same direction, then work done is positive. If force and Displacement are perpendicular to each other, then work done is zero.

Class - 11, Unit-II Motion In A Straight Line

Delhi to Vadodara via Tundla Agra.

Class - 11, Unit-V

Displacement/time = Velocity

Motion Of System Of Particles & Rigid Body

Velocity / time = acceleration

How rhino came swift? Since dino came slow.

Class - 11, Unit-III

Write 2MR2 under each figure and then divide by 2, 3, 4, 5 respectively.

1.(a) Newton's Laws of Motion

Newton, Newton don't kick cow She may move ahead little bit now* Newton hears her MAAA sound** Cow gives Newton a kick rebound***

Kelper's Laws of Planetary motion : Take Essential Foods Everyday 2/3 Times 1st Law: Planets move in elliptical orbits Sun is at one of foci of the orbit

3rd Law: Square of the Time-period of the planet is proportional to the cube of the semi major axes of the orbit. T2R3 nd

2 Law: A planet covers the equal area of space in equal interval of time no matter where it is in its orbit

∆D = Instantaneous velocity = dD/dT ∆Τ

Average Acceleration = ΔV/ΔT lim

Solid Cylinder

Class - 11, Unit-VI

Average Velocity = ΔD/ΔT

∆T → 0

Solid Sphere



A will be I, when 0 is close to T Replace the ''D'' simply with ''d'' lim

Solid Disk

Hollow Cylinder

1.(b) Motion In A Straight Line

∆T → 0

Hollow Shell

{

* Newton's 1st law. A body continues its state of rest or state of motion unless it is acted upon by an unbalanced force. ** Newton's 2nd law F = ma *** Newton's 3rd law : Every action has its equal and opposite reaction Interpretation : 1st two lines of the rhyme depicts the 1st law of motion 3rd line depicts the 2nd law of motion i.e. F = m × l Lat the depicts the 3rd law of motion

Hollow Ring

∆V = Instantaneous velocity = dV/dT ∆Τ

( 12 )

2. Thermal Properties of Matter

Interpretation: Letter E and F of Essential Food represents ''Elliptical'' and ''Foci". 1st Law : Planets move in elliptical orbits with Sun at one of the foci. Letter E of the word Everyday represents "Equal":

Fingers we have five Cats have nine lives. With 160 more Cat will help you sure!

2nd Law : A planet covers the equal area space in equal interval of time no matter where it is in its orbit.

Fingers we have five → 5F

2/3 and T of the last two words represents the "power of Time Period" and "power of semi-major axis:

Cat will help you sure! → 5F = 9C + 160

3rd Law : Square of the Time-period of the planet is proportional to the cube of the semi major axes of the orbit. T2 a R3.

Cats have nine lives. → 9C With 160 more → 9C + 160

Class - 11, Unit-VIII Thermodynamics

Temperature, Volume, Pressure No Heat is transferred Constant temperature → Isothermal process

Class - 11, Unit-VII

Constant volume → Isochoric process Constant pressure → Isobaric process

1. Mechanical Properties Of Solid

No heat transferred → Adiabatic process

Young Ravi bought a pen. (1) Relation between Y, B and σ : (write Y and B(1+ σ) with coefficients and an equal sign in between. 1Y = 3B (1 + σ) To find the coefficient of σ, refer the anti-clock circle, subtract the coefficients of B from coefficient of Y i.e. 1 – 3 = -2 So, the relation is 1Y = 3B (1 - 2σ) or, Y = 3B (1 - 2σ) (2) Relation between Y, η and σ : (write Y and η(1+ σ) with coefficients and an equal sign in between. 1Y = 2η (1 + σ) To find the coefficient of σ, subtract the coefficient of Y from coefficient of η i.e. 2 – 1 = 1 So, the relation is 1Y = 2η (1 + σ) or, Y = 2η (1 + σ)

Young Ravi bought a pen



Behaviour of Perfect Gas & Kinetic Theory

Degrees of freedom : Baa Baa Black Sheep Have you any wool? Yes sir, Mom has 3 bags full. Dadi needs 5 bags normally cool Papa keeps 6 bags normal rule. Papa, Dadi each needs 2 bags more High cold whenever, be very sure. Mom has 3 bags full → Degrees of freedom of Monoatomic gas is 3. Dadi needs 5 bags normally cool Degrees of freedom of diatomic gas at normal → (room) temperature is 5.

Poisson 's ratio

Bulk Modulus (3 for B) Rigidity Modulus (2 for ) Young's Modulus (1 for Y)

Class - 11, Unit-IX

Y(1)

B(3) n(2)

Papa keeps 6 bags normal rule → Degrees of freedom of Polyatomic gas at normal (room) temperature is 6. Papa, Dadi each needs 2 bags more → Degrees of freedom of Polyatomic gas at high temperature is 6+2=8. High cold whenever, be very sure → Degrees of freedom of Diatomic gas at high temperature is 5+2=7.

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Class - 11, Unit-X

Class - 12, Unit-III

Waves

Moving Charge And Magnetism

Teacher Punished Lazy Dog.

Fleming's left and right hand rule:

Particle oscillation in Transverse wave → Perpendicular to the direction of propagation of wave

Force Thumb First Finger Field Second Finger Current

Particle oscillation in Longitudinal wave → In the direction of propagation of wave

Class - 12, Unit-I

Thumb Motion

Electric Charge & Field

Equally divide cost per annum. To find electric field, divide the charge (enclosed) by the free space permittivity and area of the Gaussian

Left hand rule

Feel Free to Call Me

Class - 12, Unit-II First Finger Field

Resistor colour code :

Right hand rule

Thumb Motion Second Finger Current

Force Thumb In Fleming’s left hand rule, Thumb indicates FORCE. In Fleming’s left hand rule, Thumb indicates MOTION. In both rules, first finger indicates FIELD and second finger indicates CURRENT

Class - 12, Unit-IV

Interpretation :

Alternating Current

Colour codes of carbon resistors : Colour

Corresponding number

Black

0

Brown

1

Red

2

Orange

3

Yellow

4

Green

5

Berlin

6

Violet

7

Grey

8

White

9

Calcutta City Very Lovely and Very Congested For capacitive circuit → Current leads Voltage For inductive circuit → voltage leads current

Class - 12, Unit-V Electromagnetic Waves

Russian Magician showed an Interesting Very Unusual X-ray eye Game

( 14 )

Particle oscillation in Transverse wave → Perpendicular to the direction of propagation of wave

Electromagnetic waves with increasing frequency (decreasing wavelength) is in the order of: (a) Radio wave

Particle oscillation in Longitudinal wave → In the direction of propagation of wave

(b) Microwave (c) Infrared (d) Visible light

Class - 12, Unit-VII

(e) Ultraviolet

Einstein's equation of Photoelectric effect :

(f) X-Rays (g) Gamma Rays

W E Unite to form People

Class - 12, Unit-VI (a).

Photon energy

Ray Optics & Optical Instruments

Add Energy of electron emitted

M means MORE i.e Mirror Formula

Work Function Energy of emitted electron + Work function = Energy of incident Photon

M means MORE i.e+ 1 1 So, 1 –+– =– v u f Magnification will be of opposite sign : So, m =– v– u

(b).

Interpretation : E + ϕ = hf Or, E = hf = ϕ

Class - 12, Unit-VIII (a).

Atom : Hydrogen Spectra

Particle oscillation in Transverse wave → Perpendicular to the direction of propagation of wave

Papa brings Pastry for Babu and Lal

Particle oscillation in Longitudinal wave → In the direction of propagation of wave

When ni = 2, the series is Balmer

When ni = 1, the series is Lyman When ni = 3, the series is Paschen

Ray Optics & Optical Instruments

When ni = 4, the series is Brackett

L means MORE i.e Lens Formula

When ni = 5, the series is p-fund

(b).

Atom : Hydrogen Spectra

1 is Unimportant, 2 is Very important and rest are Important L means LESS i.e– 1 1 So, 1 –– –= – v u f Magnification will be of opposite sign : So, m =+ v– u

If ni = 1, i.e. Lyman series is in UV range. If ni = 2, i.e. Balmer series is in VISIBLE range. If ni = 3, 4 and 5, i.e. Paschen series, Brackett series and p-fund series are in IR range

( 15 )

(c).

Class - 12, Unit-IX

Isotope, Isobar, Isotone

ISO Tope Bar Tone

Electronic Devices

TOPE P i.e numbers of PROTONs are same and numbers of NEUTRONs different ISO BAR No P No N i.e. both PROTONs and NEUTRONs differ in number (Total remains same.) TONE N i.e. numbers of NEUTRONs are same and numbers of PROTONs different

Truth table of AND and OR gate

In isotopes, numbers of protons are same. Numbers of neutrons are different.

3V

total nucleons remain same.

5W

For AND gate, when both the switches are ON, then only the bulb is ON. i.e. When both the inputs are 1, then only output is 1. Otherwise the output is 0.

In isotones, numbers of neutrons are same. Numbers of protons are different. In isobars, numbers of neutrons are different. Numbers of protons are also different. But the

5W

5W 3V

5W

For OR gate, when both the switches are OFF, then only the bulb is OFF. i.e. When both the inputs are 0, then only output is 0. Otherwise the output is 1

( 16 )

PHYSICS AND MEASUREMENT

1

Chapter 1

Physics and Measurement

Syllabus Physics, technology and society, SI units, Fundamental and derived units. Least count, accuracy and precision of measuring instruments, Errors in measurement, Dimensions of Physical quantities, Dimensional analysis and its applications. Significant figures. LIST OF TOPICS : Topic-1 : Dimensional Analysis

Topic-1

.... P. 2

Topic-2 :  Least Count, Significant Figures

Dimensional Analysis

and Error Analysis

.... P. 7

Concept Revision (Video Based) Dimensions of Physical Quantities



Dimensional Analysis

Part - 1

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be : (b) FA2V–1 (a) FA2V–3 2 –2 (c) FA V (d) FA–1V0 [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. In momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is : (b) (P–1AT–1] (a) [P2 AT –2] 1/2 –1 (d) [PA–1T–2] (c) [PA T ] JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar constant. Dimension of solar constant is : (b) ML0T–3 (a) ML2T–2 2 0 –1 (c) M L T (d) MLT–2 [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.4. Dimensional formula for thermal conductivity is (here K denotes the temperature) : (b) MLT–3 K–1 (a) MLT–2 K –3 (d) MLT–2 K–2 (c) MLT K [JEE (Main) – 4th Sep. 2020 - Shift-1]

 1Fv 2  Q.5. A quantity x is given by  4  in terms of  WL  moment of inertia I, force F, velocity v, work W and length L. The dimensional formula for x is same as that of : (a) force constant (b) energy density (c) Planck’s constant (d) coefficient of viscosity [JEE (Main) – 4th Sep. 2020 - Shift-2]  E 1 ,y= Q.6. The quantities x = and z = CR B µ0 ∈0 are defined where C–capacitance, R–Resistance, l–length, E–Electric field, B–magnetic field and Œ0, m0, –free space permittivity and permeability respectively. Then : (a) Only y and z have the same dimension. (b) Only x and y have the same dimension. (c) Only x and z have the same dimension (d) x, y and z have the same dimension. [JEE (Main) – 5th Sep. 2020 - Shift-2]

3

PHYSICS AND MEASUREMENT

Q.7. The dimension of

B2 , where B is magnetic field 2m0

and m0 is the magnetic permeability of vacuum, is : (a) ML2T–2 (b) ML–1T–2 –2 (c) MLT (d) MLT–1 [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.8. The dimension of stopping potential V0 in photoelectric effect in units of Planck’s constant ‘h’, speed of light ‘c’ and Gravitational constant ‘G’ and ampere A is : (a) h0 c5 G–1A–1 (b) h2/3 c5/3 G1/3 A–1 2 3/2 1/3 –1 (c) h G c A (d) h1/3 G2/3 c1/3 A–1 [Modified - JEE (Main) – 8th Jan. 2020 - Shift-1] hc 5 where c is speed G of light, G universal gravitational constant and h is the Planck’s constant. Dimension of f is that of : (a) area (b) volume (c) momentum (d) energy [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.9. A quantity f is given by f =

Q.10. In SI units, the dimensions of

ε0 is : µ0

(a) [A–1 TML3]

(b) [AT2 M–1 L–1]

(c) [AT–3 ML3/2]

(d) [A2T3M–1L–2]

[JEE (Main) – 8th April 2019 - Shift-1] Q.11. If Surface tension (S), Moment of Inertia (I) and Planck’s constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be : (a) [S1/2 I3/2 h–1]

(b) [S1/2 I1/2 h–1]

(c) [S1/2 I1/2 h0]

(d) [S3/2 I1/2 h0]

[JEE (Main) – 8th April 2019 - Shift-2] Q.12. In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units? (a) [M–3 L–2 T8 A4] –2

0

–4

–2

(c) [M L T A ]

(b) [M–1 L–2 T4 A2] (d) [M–2 L–2 T6 A3]

[JEE (Main) – 10th April 2019 - Shift-2] Q.13. Which of the following combinations has the dimension of electrical resistance (e0 is the permittivity of vacuum and µ0 is the permeability of vacuum)? µ0 µ0 (a) (b) ε0 ε 0

(c)

ε0 µ0

(d)

ε0 µ0

[JEE (Main) – 12th April 2019 - Shift-1] Q.14. Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :

(a)

hc 5 G

GH c5 (c)

(b)

c3 Gh

(d)

GH c3

[JEE (Main) – 9th Jan. 2019 - Shift-2]

Q.15. The density of a material in SI units is 128 kg. m–3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is : (a) 40 (b) 16 (c) 640 (d) 410 [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.16. The force of interaction between two atoms is given  x2  by F = ab exp  −  ; where x is the distance, k is  α kt  the Boltzmann constant and T is temperature and a and b are two constants. The dimension of b is : (a) [MLT–2] (b) M2 L2 T–2] 0 2 –4 (c) [M L T ] (d) M2 LT–4] [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.17. If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be : (a) [V–2 A2 F–2] (b) [V–2 A2 F2] –4 –2 (c) [V A F] (d) [V–4 A2 F] [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.18. Let l, r, c and v represent inductance, resistance, capacitance and voltage respectively. The dimension l of in SI units will be : rcv (a) [LA–2] (b) [A–1] (c) [LTA] (d) [LA2] [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (d) 5. (b) 9. (d) 13. (a) 17. (d)

2. (c) 6. (d) 10. (d) 14. (c) 18. (b)

3. (b) 7. (b) 11. (c) 15. (a)

4. (b) 8. (a) 12. (a) 16. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. Given : The fundamental units are, speed V, area A and force F. To find : The dimensions of Young’s modulus (Y) in the new system of fundamental units. Dimensions of Young’s modulus : Y = [ ML−1T −2 ]

...(i)

Dimensions for force : [ F] = [ MLT −2 ]

...(ii)

Dimensions for area : [ A] = [ L2 ]

...(iii)

4 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Dimensions for velocity : −1

[ V] = [ LT ]

...(iv)

Let the dimensions for Young’s modulus in the new system of fundamental units be : [ Y ] = [ F]a [ A]b [ V]c

...(v)

Ans. 4. Option (b) is correct. Given : The fundamental units are, M, L and T. To find : The dimensions of thermal conductivity, k. Q  t  Dx k =   A DT

Using equations (i), (ii), (iii), (iv) and (v) : [k ] =

[ ML−1T −2 ] = [ MLT −2 ]a [ L2 ]b [ LT −1 ]c Compare the exponents. a = 1

[ ML2 T −2 ][ L]

Given : x =

−2 a − c =  −2

[ Y ] = [ F]1[ A]−1[ V]0 Ans. 2. Option (c) is correct. Given : The fundamental units are, momentum P, area A and time T. To find : The dimensions of energy (E) in the new system of fundamental units. Dimensions of Energy : ...(i)

Dimensions for force : [ P] = [ MLT −1 ]

...(ii)

Dimensions for area : [ A] = [ L2 ]

...(iii)

Dimensions for velocity : [ T] = [ T]

...(iv)

Let the dimensions for energy in the new system of fundamental units be : c

[ E] = [ P] [ A] [ T]

...(v)

Using equations (i), (ii), (iii), (iv) and (v) : [ ML2 T −2 ] = [ MLT −1 ]a [ L2 ]b [ T]c Compare the exponents. a = 1 a + 2b = 2 − a + c =  −2 On solving the above set of equations we get : 1 a = 1,b = , c = −1 2 That gives, [ E] = [ P]1[ A]1 / 2 [ T]−1 Ans. 3. Option (b) is correct. Given : The fundamental units are, M, L and T. To find : The dimensions of solar energy received on Earth’s surface per unit area per unit time, S. [ Energy ] [ ML2 T −2 ] [S] = = = [ ML0 T −3] 2 [ Area][ Time] [ L ][ T]

WL4

 IFv 2  [x] =  4  WL 

That gives,

[ E] = [ ML2 T −2 ]

IFv 2

To find : Dimensions of x.

On solving the above set of equations we get : a = 1, b = −1, c = 0

b

= [ M1L1T −3 K −1 ]

[ T][ L2 ][ K ]

Ans. 5. Option (b) is correct.

a + 2b + c =  −1

a

PHYSICS

=

[ ML2 ][ MLT −2 ][ L2 T −2 ] [ ML2 T −2 ][ L4 ]

= [ ML−1T −2 ] So, x has dimension of energy density. Ans. 6. Option (d) is correct. Given : The quantities E l 1 x = ,y = , z = . B CR µ0 ∈0 To find : Compare the dimensions of given quantities.  1  −1 [x] =   = [ c] = [ LT ]  µ0 ∈0  E [ y] =   = [ c] = [ LT −1 ] B  l  l  −1 [ z] =   =   = [ LT ]  CR   t  From the above equations it can be seen that all the given quantities x, y and z have dimensions of velocity. Ans. 7. Option (b) is correct. Given : B is magnetic field and m0 is magnetic permeability of vacuum. To find : Dimension of

B2 . 2m0

F  Dimensions of magnetic field  B =  : Il   [ B] = [ MLT −2 ][ A]−1[ L]−1 = [ M1T −2 A −1 ] ...(i)  1  : Dimensions of  m0 = 2 c ∈0   [ m0 ] = [ LT −1 ]−2 [ M −1L−3 T 4 A 2 ]−1 = [ L−2 T 2 ][ ML3 T −4 A −2 ] = [ MLT −2 A −2 ]

...(ii)

5

PHYSICS AND MEASUREMENT

From equations (i) and (ii), dimensions of

B2 : 2m0

[ MT −2 A −1 ]2 [ MLT −2 A −2 ]−1 = [ M 2 T −4 A −2 ][ M −1L−1T 2 A 2 ] = [ ML T ] Ans. 8. Option (a) is correct. Given : V0 is stopping potential, h is Planck’s constant, c is speed of light, G is gravitational constant and I is current. To find : Dimensions of V0 in terms of h, c, G and I. Let the required dimensions of V0 in terms of h, c, G and I be : z

1

1  [ ML2 T −1 ][ LT −1 ]5  2 2 4 −4 2 −2   = [ M L T ] 2 = [ ML T ] ...( i ) −1 3 −2 M L T [ ]  

From equation (i), we can see that f has dimensions of energy.

−1 −2

[ V0 ] = [ h]w [ c]x [G]y [ A ]

Dimensions of f :

...(i)

 F Fl  Dimensions of V0  V0 = El = l =  : q It  

Ans. 10. Option (d) is correct.

ε0 µ0 To find : Dimension of Given :

Unit of permittivity of free space, ε 0 =

(by Coulomb’s law), Reduce the unit, to get dimensional formula of

Also,

ε0 = µ0

ε 02 = c ε 0 ...(ii) ε 0 µ0  1  as, c =  ∈ 0 µ0 

Dimensions of h : ...(iii)

Dimensions of c :

From (i) and (ii) the dimensions of

−1

[ c] = [ LT ]

...(iv)

Dimensions of G : [G] = [ M −1L3 T −2 ]

...(v)

Dimensions of I : [ I] = [ A]

...(vi)

Put values from equations (ii), (iii), (iv), (v) and (vi) in equation (i), [ V0 ] = [ h]w [ c]x [G]y [ A]z [ ML2 T −3 A −1 ] = [ ML2 T −1 ]w [ LT −1 ]x [ M −1L3 T −2 ]y [ A]z Compare the exponents. 1 = w−y 2 = 2w + x + 3y −3 = − w − x − 2 y −1 = z On solving, we get : w = 0 ,x = 5,y = −1,z = −1. Put the values in equation (i), −1

[ V0 ] = [ h]0 [ c]5 [G]

[ A ]−1

So, none of the given options are correct. Ans. 9. Option (d) is correct. hc 5 , h is Planck’s constant, c is speed G of light, G is gravitational constant. To find : Dimensions of f. Given : f =

c2 ; m2 N

[e0] = [M–1L–3T4A2] ...(i)

[ V0 ] = [ MLT −2 ][ L][ A]−1[ T]−1 = [ ML2 T −3 A −1 ] ...(ii) [ h] = [ ML2 T −1 ]

ε0 in SI units. µ0

dimensions of c times dimensions of e0 :

   

ε0 will be µ0

ε0 = [c] [e0] µ0

= [L1T–1] [M–1L–3T4A2] = [M–1L–2T3A2] Ans. 11. Option (c) is correct. Given : Surface tension (S), moment of inertia (I) and Planck’s constant (h) are fundamental units. To find : Dimensional formula for linear momentum. Using the above formula, surface tension Force , = Length Since dimensions of surface tension : [S] = [M1 T –2] ...(i) Unit of moment of inertia = Mass × (distance)2, That gives dimensions of moment of inertia : [I] = [M1 L2] ...(ii) Unit of Planck’s constant = Joule × sec, Since dimensional formula of Planck’s constant : [h] = [M1 L2 T –1] ...(iii) Since of linear momentum = mass × velocity, Dimensional formula of linear momentum : [p] = [M1 L1 T –1] ...(iv) Expressing dimension of [p] in terms of [S], [I] and [h], let : p = (S)x (I)y (h)z ...(v) Putting values from equations (i), (ii), (iii) and (iv) in equation (v) of S, I, h and p : [M1 L1 T–1] = [M1 T–2 ]x [M1 L2]y [M1 L2 T–1]z p µ (S)x (I)y (h)z Equating the exponents : x + y + z = 1

6 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) 2y + 2z = 1 –2x – z = –1 Solving the above set of equations for x, y and z, we get : 1 1 x = , y = and z = 0. ...(vi) 2 2 Putting values of x, y and z in eqn. (v): 1

Ans. 14. Option (c) is correct. Given : Gravitational constant (G), Planck’s constant (h) and speed of light (c) are fundamental units. To find : The dimensions of time period. Dimensional formula of G : [G] = [M–1L3 T–2] ...(i)

1

0 p µ (S) 2 (I) 2 ( h ) . Ans. 12. Option (a) is correct. Given : X = 5YZ2, X has Dimensional formula

m1m2    F = G 2  Dimensional formula of h : r   [h] = [M1L2 T–1] ...(ii) of

 h  λ =  Dimensional formula of c : p  [c] = [L1 T–1] ...(iii) Let the dimensions of time period in terms of G, h and c be : t = K(G)x (h)y (L)z T = (G)x (h)y (c)z ...(iv) Putting values of G, h and L from equations (i), (ii) and (iii) in equation (iv) : [T] = [M–1 L3 T–2]x [M1 L2 T–1]y [L1 T–1]z Comparing the exponents : –x + y = 0 3x + 2y + z = 0 –2x – y – z = 1 Solving above set of linear equations, we get : 1 −5 1 . x = , y = and z = 2 2 2 Putting these values in equation (iv),

capacitance

ε0A   C = d  :   [X] = [M–1 L–2 T4 A2]. ...(i) Z has dimensional formula of magnetic field F  B = l  : I   [Z] = [M–1 T–2 A–1]. ...(ii) To find : Dimensions of X . ...(iii) Y µ 5Z 2 Putting dimensions of X and Y from (i) and (ii) in equation (iii), [Y] = [M–1 L–2 T4 A2] [M1 T–2 A–1]–2 [Y] = M–3 L–2 T8 A4 Ans. 13. Option (a) is correct. To find : The combination of e0 and mo which has dimensional formula of electrical resistance. Dimensional formula of resistance : [R] = [M–1 L2 T–3 A2] ...(i) Dimensions formula of e0 : e0 = [M–1L–3T4A2] ...(ii) (by Coulomb’s law) Dimensional formula of m0 : m0 = [M1L1T –2A–2] ...(iii) (B = m0 H) Let the required combination be : R = [e0]x [m0}y ...(iv) Putting dimensions or, R, e0 & m0 from equations (i), (ii) and (iii) in equation (iv), [M1L2T –3A–2] = [M–1L–3T4A2]x [M1L1T –2A–2]y Comparing the exponents : –x + y = 1 –3x + y = 2 4x – 2y = –3 2x – 2y = –2 Solving above set of linear equations, we get : 1 1 x= − ,y= . 2 2 Putting these values in equation (iv), we get the required expression : R =

µ0 . ε0

PHYSICS

1 −5 2

1



t = G 2 h 2 c

or

t µ

Gh c5

Ans. 15. Option (a) is correct. Given : Density of material in SI units = 128 kg/m3. To find : Density of material in new unit system, where unit of length = 25 cm and unit of mass = 50 g. Let the new density be r. That gives,

128

1000 g 3

(100 cm )

= ρ

r =



50 g ( 25 cm )3

;

128 ×1000 × 25 × 25 × 25 = 40  100 × 100 × 100 × 50

Ans. 16. Option (d) is correct. Given : Force of interaction between two atoms,

 x2   . F = αβ exp  −  α kT 

To find : Dimensions of b. As, exponential term will always be dimensionless, dimension of a  :

 x2  [ L]2 = [ M −1 T 2 ]. [a] =   = 1 2 −2  T k   [ M L T ]

7

PHYSICS AND MEASUREMENT

Now, dimensions of b, will be :

Comparing the exponents : z = 1; x + y + z = –1; –x – 2 y – 2z = –2 Solving above set of linear equations, we get : x = –4, y = 2 and z = 1. Putting these values in equation (v), [Y] = [V–4 A2 F]

[ F] = [M L T ] = [M2 L1T −4 ] [α ] [M−1T 2 ] 1 1 −2

[b] =

Ans. 17. Option (d) is correct. Given : Speed (V), acceleration (A) and force (F) are fundamental units. To find : Dimensions of Young’s modulus. Dimensions of speed : [V] = [L1T –1] ...(i) Dimensions of acceleration :

Ans. 18. Option (b) is correct. To find : Dimensions of l/rcv, where l is inductance, r is resistance, c is capacitance and v is voltage.

Dimensions of force :

[l] = [v1T1A –1]. Using Ohm’s law, Dimensional formula of resistance, will be : [r] = [v1A –1]. Using q =cv, Dimensional formula of capacitance, will be : [c] = [v–1T1A1]

1 1 –2

[F] = [M L T  ] ...(iii)

Dimensions of Young’s modulus :

[Y] = [M1L–1T–2] ...(iv)

Let the dimensions of Young’s modulus in terms of [V], [A] and [F] be : (Y) µ (V)x (A)x (F)x

di dimensions of inductance, will be : dt

Using v = l

[A] = [L1T  –2] ...(ii)

Y = K(V)x (A)x (F)x ...(v)

Now, reducing

Putting values of V, A, F & Y from equations (i), (ii), (iii) and (iv) in equation (v).



[M1L–1T –2] = [L1T –1]x [L1T –2]y [M1L1T –2]z

l to fundamental units : rcv

[ v1 T1 A −1 ]  l  = [ A −1 ]. =  rcv  ([ v1 A −1 ])([ v −1 T1 A1 ])([ v])  

Topic-2

Least Count, Significant Figures and Error Analysis Concept Revision (Video Based)    

Error Analysis     

Significant Figures

Reading Vernier Scale

Part -1

    

Part -2

Screw Gauge Principal & Description

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1 The least count of the main scale of a vernier callipers is 1 mm . Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of cylinder

the zero of the vernier scale between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division) (a) 2.99 cm (c) 3.21 cm

(b) 3.07 cm (d) 3.2 cm [JEE (Main) – 2nd Sep. 2020 - Shift-1]

8 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.2. Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as : (a) 2.125 cm (b) 2.124 cm (c) 2.123 cm (d) 2.121 cm [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. A physical quantity z depends on four observables a 2 b2 / 3 . The percentage of error a, b, c and d, as z = cd 3 in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is : (a) 13.5% (b) 16.5% (c) 14.5% (d) 12.25% [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.4. A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved and the lest count of the screw gauge, are respectively : (a) Positive, 0.1 mm (b) Negative, 2 mm (c) Positive 10 mm (d) Positive, 0.1 mm [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.5. A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm. 5.34 mm; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as : (a) (5.54 ± 0.07) mm (b) (5.538 ± 0.074) mm (c) (5.5375 ± 0.0740) mm (d) (5.5375 ± 0.0739) mm [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.6. A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is : (a) 2.40% (b) 5.40% (c) 4.40% (d) 3.40% [JEE (Main) – 8th Jan. 2020 - Shift-2] Q.7. If the screw on a screw-gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is : (a) 0.001 cm (b) 0.01 cm (c) 0.02 mm (d) 0.001 mm [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.8. For the four sets of three measured physical quantities as given below. Which of the following options is correct? (i) A1 = 24.36, B1 = 0.0724, C1 = 256.2 (ii) A2 = 24.44, B2 = 16.082, C2 = 240.2 (iii) A3 = 25.2, B2 = 19.2812, C3= 236.183 (iv) A4 = 25, B4 = 236.191, C4 = 19.5

PHYSICS

(a) A1 + B1 + C1 < A2 + B2 + C2 = A3 + B3 + C3 < A4 + B4 + C4 (b) A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4 (c) A1 + B1 + C1 < A3 + B3 + C2 < A2 + B2 + C2 < A4 + B4 + C4 (d) A4 + B4 + C4 < A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 [Modified - JEE (Main) – 9th Jan. 2020 - Shift-2] Q.9. The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is : (a) 5.755 mm (b) 5.950 mm (c) 5.725 mm (d) 5.740 mm [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.10. The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 mm diameter of a wire is : (a) 50 (b) 200 (c) 100 (d) 500 [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.11. In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to : (a) 0.7 % (b) 0.2 % (c) 3.5% (d) 6.8 % [JEE (Main) – 8th April 2019 - Shift-1] Q.12. In the density measured of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01) m, respectively. The error in the measurement of density is : (a) 0.01 kg/m3 (b) 0.10 kg/m3 3 (c) 0.31 kg/m (d) 0.07 kg/m3 [JEE (Main) – 9th April 2019 - Shift-1] Q.13. The area of a square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is : (a) 37 cm2 (b) 37.030 cm2 2 (c) 37.03 cm (d) 37.0 cm2 [JEE (Main) – 9th April 2019 - Shift-2] Q.14. The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures? (a) 4264 ± 81 cm3 (b) 4264.4 ± 81.0 cm3 (c) 4260 ± 80 cm3 (d) 4300 ± 80 cm3 [JEE (Main) – 10th Jan. 2019 - Shift-2]

9

PHYSICS AND MEASUREMENT

ANSWERS – KEY 1. (b) 5. (a) 9. (a) 13. (c)

2. (b) 6. (c) 10. (b) 14. (c)

3. (c) 4. (c) 7. (a) 8. (a) 11. (d) 12. (c)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. Given : The least count of the main scale of the vernier calliper is 1 mm, 10 divisions on vernier scale coincide with 9 divisions on main scale, when the jaws are touching the 7th division on the vernier scale is the coinciding reading, reading for the length of a cylinder lies between 3.1 cm and 3.2 cm and for that measurement the 4th division on the vernier scale is the coinciding reading. To find : The length of the cylinder. Least count of the Vernier calliper Least count of the main scale = Number of divisions on the Vernier scale =

0.1 = cm 0.01cm  10

Zeroerror = 7 × 0.01cm = 0.07 cm Length of cylinder = ( 3.1 + 4 × 0.01 − 0.07 )cm = 3.07 cm Ans. 2. Option (b) is correct. Given : Pitch of a screw gauge is p = 0.1 cm, number of divisions on the circular scale of the screw gauge is n = 50. To find : The correct thickness measured from the screw gauge. Least count of the screw gauge : pitch Least count = number of divisions on circular scale Least count =

0.1 = cm 0.002 cm 50

The given screw gauge can measure the thickness of an object up to third decimal place, with the third place being an even number. So, option b is correct. Ans. 3. Option (c) is correct. z=

a2b 2 / 3

, the percentage error in cd 3 measurement of the quantities a, b, c and d is 2%, 1.5%, 4% and 2.5%, respectively. To find : The percentage error in z. ∆z ∆a 2 ∆b 1 ∆c ∆d = 2 + + +3 z a 3 b 2 c d Given :

2 2 1.5 1 4 2.5 = 2× + × + × + 3×  100 3 100 2 100 100 =

4 + 1 + 2 + 7.5 14.5 = 100 100

∆z = 14.5% z

Ans. 4. Option (c) is correct. Given : Number of divisions on the circular scale of a screw gauge is 50, at zero reading the circular scale is 4 units ahead of the pitch scale, one rotation on the circular scale corresponds to 0.5 mm displacement on the pitch scale. To find : The nature of zero error and the least count of the screw gauge. As the circular scale is ahead of the pitch scale at zero reading, the zero error is positive. Least count of screw gauge pitch = number of divisions on the circular scale Least count =

0.5 mm = 0.01mm = 10 µm 50

Ans. 5. Option (a) is correct. Given : Four readings for the cross section of a pencil are, 5.50 mm, 5.55 mm, 5.34 mm, 5.65 mm, the average of the four readings is 5.5375 mm, the standard deviation for the data is 0.07395 mm. To find : The average diameter of the pencil, d. As the data is reported up to 2 decimal places, the average diameter of the pencil should also be reported up to 2 decimal places. So, d = (5.54 ± 0.07) mm Ans. 6. Option (c) is correct. Given : Least count of stop watch is ΔT= 1 s, Mean value of time taken by pendulum for 40 oscillations is T= 50 s, Least count of scale is Δl = 0.1 cm, Length of pendulum is l = 25.0 cm. To find : Percentage error in determination of g. Acceleration due to gravity, g =

4p 2 l

. T2 Percentage error in measurement of g : ∆g  ∆l 2 ∆T  × 100 =  + × 100 g T   l 1   0.1 =  + 2  × 100 = 4.4% 50   25 Ans. 7. Option (a) is correct. Given : Distance travelled by the screw of a screw gauge in 6 rotations on the main scale is d = 3 mm, Number of divisions on the circular scale is n = 50. To find : The least count of the screw gauge. Pitch of the screw gauge is the distance travelled by the screw in one rotation. d 3 Pitch= = mm = 0.5 mm 6 6 Least count of the screw gauge : Pitch 0.5 Least count = = mm = 0.001cm n 50 Ans. 8. Option (a) is correct. Given : (i) A1 = 24.36, B1 = 0.0724, C1 = 256.2 (ii) A2 = 24.44, B2 = 16.082, C2 = 240.2

10 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (iii) A3 = 25.2, B3 = 19.2812, C3 = 236.183 (iv) A4 = 25, B4 = 236.191, C4 = 19.5 To find : Compare values of A1 + B1 + C1, A2 + B2 + C2, A3 + B3 + C3 and A4 + B4 + C4. A1 + B1 + C1 = 24.36 + 0.0724 + 256.2 = 280.6324 = 280.6 A2 + B2 + C2 = 24.44 + 16.082 + 240.2 = 280.722 = 280.7 A3 + B3 + C3 = 25.2 + 19.2812 + 236.183 = 280.6642 = 280.7 A4 + B4 + C4 = 25 + 236.191 + 19.5 = 280.691 = 281 So, from above equations, we get : A1 + B1 + C1 < A2 + B2 + C2 = A3 + B3 + C3 < A4 + B4 + C4 Ans. 9. Option (a) is correct. Given : For a screw gauge the pitch = 0.5 mm, number of divisions on the circular scale is n = 100 the zero of the circular scale lies 3 divisions below the mean line when the screw gauge is fully tight without any object, the reading of the main scale and the circular scale of screw gauge for a thin sheet is 5.5 mm and 48 respectively. To find : t the thickness of the sheet. Least count of the screw gauge : pitch 0.5 LC = = mm = 0.005 mm n 100



As the zero of the circular scale lays 3 divisions below the mean line when the screw gauge is fully tight without any object, the error in screw gauge will be : Error = 3 ¥ LC = 3 ¥ 0.005 = 0.015 mm Thickness of the paper : t = Main scale reading + circular scale  reading ¥ LC + Error t = 5.5 mm + 48 ¥ 0.005 mm + 0.015 mm = 5.755 mm

Ans. 10. Option (b) is correct. Given : Least count of the main scale of a screw gauge is 1 mm diameter of wire to be measured is d = 5 mm To find : n, the number of divisions on the circular scale of the screw gauge. To be able to measure diameter of wire, the least count of the screw gauge shall be : LC = d = 5mm



We know, least count of a screw gauge is : least count of main scale LC = number of divisionson

circular scale Put given values in the equation above. 5 ¥ 10 -6 = n =

PHYSICS

Ans. 11. Option (d) is correct. Given : Least count of stop watch = 1 s, Mean value of time taken by pendulum for 20 oscillations = 30 s, Least count of scale = 1 mm = 0.1 cm, Length of pendulum = 55.0 cm. To find : Percentage error in determination of g. Acceleration due to gravity, 4p 2 l

. T2 Percentage error in measurement of g :



g =

 ∆l 2 ∆T  ∆g × 100 × 100 =  + T   l g



1   0.1 + 2  × 100 =  55 30  



= 6.8%

Ans. 12. Option (c) is correct. Given : Mass of cube,

m = (10.00 ±0.10) kg.

Edge length of cube,

l = (0.10 ± 0.01) m

To find : Error in measurement of density of cube. Density of cube, m

. l3 Error in measurement of density :



r =

∆m 3∆l 0.10 0.01 31 ∆ρ + = +3 = = m l 10 0.1 100 ρ = 0.31.

Ans. 13. Option (c) is correct. Given : Area of a square,

A = 5.29 cm2.

To find : Area of 7 such squares. Area of 7 squares,

A¢ = 5.29 cm2 + 5.29 cm2 + 5.29 cm2 + 5.29 cm2 + 5.29 cm2 + 5.29 cm2 + 5.29 cm2= 37.03 cm2.

As the number of digits following the decimal point in 5.29 = 2, the answer will also carry two decimal places. Ans. 14. Option (c) is correct. Given : Diameter of cylinder,

1 ¥ 10 -3 n



1 ¥ 10 3 = 200 5



D = (12.6 ± 0.1) cm,

Height of cylinder, h = (34.2 ± 0.1) cm.

To find : The volume of the cylinder.

11

PHYSICS AND MEASUREMENT

Volume of cylinder :

p D2 h 2 V = p r h = 4

=

3.14 × 12.6 × 12.6 × 34.2 4

= 4262.23 cm3.

As the diameter and height of the cylinder mentioned in the question carry three significant figures, so volume of the cylinder also need to be mentioned using up to three significant figures. Volume of cylinder in appropriate number of significant figures (3) :

V = 4260 cm3.

Error in volume, in appropriate number of significant figures (1) :

DD Dh 0.1 0.1 DV + =2 + ; = 2 D h 12.6 34.2 V

0.1   0.1 +  V = 80 cm 3 . DV =  2  12.6 34.2  So, volume of cylinder :

V = (4260 ± 80) cm3.

Subjective Questions (Chapter Based) Q.1. The density of a solid metal sphere is diameter. The maximum error in the density of the sphere  x  is   %. If the relative errors in measuring  100  the mass and the diameter are 6.0% and 1.5% respectively, the value of x is : [JEE (Main) – 6th Sep. 2020 - Shift-1] Sol. Given : Maximum error in measuring density of x %, the relative error in measuring a sphere is 100 6 mass is % and the relative error in measuring 100 1.5 %. diameter is 100 To find : The value of x. Density of sphere : 6m m ρ = = 3 3 pd 4 d p 3  2  From the above equation : ∆ρ ∆m 3∆d = + m d ρ x = 6 + 3 × 1.5 100 x = 10.5 100 x = 1050 Q.2. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways

engineer uses dimensional analysis and assumes that the distance depends on the mass density r of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer find that d is proportional to S1/n. Obtain the value of n Sol. Given : Density r, intensity of light S and frequency f are fundamental units and distance, d µ S1/n.

To find : Value of n.



Dimensional formula of density :



[r] = [M1L–3]



Dimensional formula of intensity :







(Using, unit of intensity S is Watt/m2)



Dimensional formula of frequency :







xpressing distance in terms of above three E fundamental units :





d µ (r)x (S)y (f  )z;





d = K(r)x (S)y (f)z;







Comparing the exponents :



x + y = 0; –3 x = 1; –3y – z = 0



We get,



So,



Hence, n = 3.

[S] = [M1T–3]

[f] = [T]–1

[L] = [M1L–3]x [M1T–3]y [T–1]z

1 1 x = − , y = and z = –1. 3 3 [L] µ [S]1/3

Q.3. Math List I with List II and select the correct answer using the codes given below the lists : List I List II P. Boltzmann constant 1. [ML2 T–1] Q. Coefficient of viscosity 2. [ML–1 T–1] R. Planck’s constant 3. [MLT–3 K–1] S. Thermal conductivity 4. [ML2 T–2 K–1] Sol. To find : Dimensions of Boltzmann constant, coefficient of viscosity, Planck’s constant and thermal conductivity. Boltzmann constant (using E = kT ) :

E 1 2 −2 −1 1 2 −2 −1 [k] =   = [ M L T ][ K ] = [ M L T K ] T



dv   Coefficient of viscosity  Using, F = η A  : dx  



 −1 dx  [h] =  FA = [M1 L1 T–2] [L–2] [T1] dv   = [M1 L–1 T –1]  h Planck’s constant  Using, λ =  : p  [h] = [lp] = [L1] [M1 L1 T–1] = [M1 L2 T–1] QL Thermal conductivity. (Using, K = , where A DT Q is in watts) : [K] = [M1 L2 T–3] [L] [L–2] [K–1] = [M1 L1 T–3 K–1] So, the correct match will be : P-4, Q-2, R-1, S-3.

12 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q. 4. The energy of a system as a function of time t is given as E(t) = A2 exp(–at), where a = 0.2 S–1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is Sol. Given : E(t) = A2 exp(-at), where a = 0.2 S–1 ...(i) Error in measurement of A, dA = 1.25 % A Error in measurement of time, dt = 1.50% t To find : The percentage error in measurement of E(t) at t = 5 s. Differentiating equation (i) :



2 dE = 2 AdA exp( −α t ) + A ( −α )exp( −α t )dt ;





dE = 2



dA 2 ( A exp( −α t )) − α ( A 2 exp( −α t ))dt ; A

dE dA − α dt ...(ii) = 2 E A Percentage error in measurement of E(t) : dA dt dE × 100 ± α t × 100. ...(iii) × 100 = ± 2 A t E Percentage error in measurement of E(t) at t = 5s : 1.25 1.50 dE × 100 ± ( 0.2 )( 5) × 100 = ±4% × 100 = ± 2 100 100 E

Q.5. Express the dimensions, length L and mass M, using Planck’s constant h, speed of light c and gravitational constant G as fundamental units. Sol. Given : Gravitational constant (G), Planck’s constant (h) and speed of light (c) are fundamental units. To find : The dimensions of mass [M] and length [L]. Dimensional formula of G : m1m2   [G] = [M–1 L3 T –2]  F = G 2  ...(i) r   Dimensional formula of h :  h   λ =  ...(ii) [h] = [M1 L2 T–1] p  Dimensional formula of c : [c] = [L1 T–1] ...(iii) Let L µ (G)x (h)x (c)z Let L = K(G)x (h)x (c)z ...(iv) Substituting the values of G, h & c from equations (i), (ii) and (iii) in equation (iv). [L] = [M–1 L3 T–2]x [M1 L2 T–1]y [L1 T –1]z Comparing the exponents : –x + y = 0; 3x + 2y + z = 1; –2x – y – z = 0 −3 1 1 We get, x = , y = and z = 2 2 2 That implies : L = (G)1/2 (h)1/2 (c)–3/2

1



1

PHYSICS

−3

L µ (G) 2 ( h ) 2 ( c ) 2



Doing similar exercise for [M], Let M µ (G)x (h)x (c)x Let M = K(G)x (h)x (c)x ...(v) Substituting values of G, h & c from equations (i), (ii) and (iii) in equation (v). [M] = [M–1 L3 T–2]x [M1 L2 T–1]y [L1 T –1]z Comparing the exponents : –x + y = 1; 3x + 2y + z = 0; –2x – y – z = 0 1 1 1 We get, x = − , y = and z = . 2 2 2 That implies : –1/2 1/2 1/2 [M] = [G] [h] [c] . Q.6. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of 7( R − r ) . The values a periodic motion is T = 2p 5g of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of the time period is 0.01 s. Give the error in measurement of r, T and g. Sol. Given : Formula used for time period of a periodic motion is,



7( R − r ) , ...(i) 5g R = (60 ± 1) mm, r = (10 ± 1) mm Time period measured in five successive measurements : 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. Least count of watch used = 0.01 s. To find : error in measurement of r, R, T and g. Mean time period : 0.52 + 0.56 + 0.57 + 0.54 + 0.59 = 0.556. ...(ii) T = 5 As time is measured up to two significant figures, T = 0.56 s ...(iii) Mean error in measurement of time, 0.52 − 0.56 + 0.56 − 0.56 + 0.57 − 0.56 T = 2p



d T =

+ 0.54 − 0.56 + 0.59 − 0.56



5 = 0.02. ...(iv) So, percentage error in measurement of T : 0.02 δT × 100 = 3.57% ...(v) × 100 = 0.56 T Percentage error in r, using equation (ii) :







Percentage error in R, using equation (ii) :







 4π 2 7( R − r )  , Percentage error in g,  g = 5 T 2  



1 δr × 100 = 10% × 100 = 10 r 1 δR × 100 = 1.67% × 100 = 60 R

13

PHYSICS AND MEASUREMENT



Using equations (i), (ii) and (v) : δg δ (R − r ) δT × 100 = × 100 + 2 × 100 g R−r T [d(R – r)) = 2, as errors always add.] 2 δg × 100 + 2 × 3.57% = 4% + . 7 14% × 100 = 50 g = 11%. Q.7. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line ? Sol. Given : For a screw gauge the pitch = 0.5 mm, number of divisions on the circular scale is n = 50, the zero of the circular scale lies 50 – 45 = 5 divisions below the mean line when the screw gauge is fully tight without any object, the reading of the main scale and the circular scale of screw gauge for a thin sheet is 0.5 mm and 25 respectively. To find : t the thickness of the sheet. Least count of the screw gauge : pitch 0.5 LC = = mm = 0.01mm n 50 As the zero of the circular scale lays 5 divisions below the mean line when the screw gauge is fully tight without any object, the error in screw gauge will be : Error = 5 ¥ LC = 5 ¥ 0.01 = 0.05 mm Thickness of the paper: t = Main scale reading + circular  scale reading ¥ LC + Error t = 0.5 mm + 25 ¥ 0.01mm + 0.05 mm = 0.8 mm Q.8. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. What are the measured values (in cm) by calipers C1 and C1? 2

4

3

C1 5

0 2

10 4

3

C2



For Vernier 2 : 10 divisions of Vernier scale coincide with 11 divisions of main scale. To find : The reading of the two Vernier scales as shown in the figure above. As 10 divisions of Vernier scale coincide with 9 divisions of main scale for Vernier 1, value of 1 division of Vernier scale : 9 mm = 0.9 mm 10 Least count of Vernier 1 : 1 main scale division – 1 Vernier scale division C1 = 1mm - 0.9 mm = 0.1mm





C1 = 28 mm + 7 ¥ 0.1mm = 28.7 mm = 2.87 cm



From the given figure, reading of C1 : main scale reading+coinciding scale × LC1 C1 = 28 mm + 7 ¥ 0.1mm = 28.7 mm = 2.87 cm



Similarly, for Vernier 2 : As 10 divisions of Vernier scale coincide with 11 divisions of main scale for Vernier 2, value of 1 division of Vernier scale : 11 mm = 1.1mm 10 The second Vernier seems odd. So, we match the readings from both the scale. Distance measured from main scale = distance measured from Vernier scale 28 mm + 8 ¥ 1mm = ( 28 mm + x ) + (7 ¥ 1.1mm )





x = 0.3 mm

So,

Q.9. A student reports length of a pencil to be 5.56 cm, describe the instrument used by her. Sol. Given : The length of the pencil is l = 5.56 cm. To find : The scale used for the measurement. As the measurement is reported up to the second decimal place, the least count of instrument used is LC = 0.01 cm = 0.1 mm The instrument has to be a Vernier caliper with value of one main scale division = 1 mm. So, first decimal place of l is measured from the main scale. The accuracy up to second decimal place comes from Vernier scale of the caliper whose 10 divisions match with 9 divisions of the main scale. So, value of 1 division of Vernier scale = 9 ¥ value of one main scale division 10 9 ¥ 1mm 10 = 0.9 mm =

0

5

10

Sol. Given : For both Vernier 1 and 2 the least count of main scale is 1 mm, For Vernier 1 : 10 divisions of Vernier scale coincide with 9 divisions of main scale,

C 2 = 28 mm + 0.3 mm = 28.3 mm = 2.83 cm



Least count of Vernier caliper used = value of 1 main scale division – value of 1 Vernier scale division LC = 1mm - 0.9 mm = 0.1mm

14 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.10. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. What is the diameter of the cylinder ? Sol. Given : During measurement of the diameter of a cylinder the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale, the 24th division of Vernier scale exactly coincides with one of the main scale divisions, total number of divisions in the Vernier scale is n = 50 which coincide with 2.45 cm of main scale. To find : d, the exact diameter of the cylinder. Value of one main scale division : 1MSD = 5.15 - 5.10 = 0.05 cm

Value of one Vernier scale division : 2.45 1VSD = cm = 0.049 cm 50



Sol. Given : Error in measured mass of a ball is Dm = 2%, diameter of the ball is measured on a m screw gauge, pitch of the screw gauge is pitch = 0.5 mm, number of divisions on the circular scale of screw gauge are n = 50, the main scale reading of the screw gauge for the measurement of diameter of the ball is MSR = 2.5 mm, reading on circular scale for the measurement of diameter of the ball is n¢ = 20 divisions. To find : The relative percentage error in measurement of density of the ball. Least count of the screw gauge : pitch 0.5 LC = = mm = 0.01mm n 50

Least count of Vernier scale : LC = 1MSD - 1VSD = 0.05 - 0.049 = 0.001cm



Diameter of cylinder : main scale reading + Vernier scale reading × LC d = 5.10 cm + 24 ¥ 0.001cm = 5.124 cm



Q.11. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, what will be the relative percentage error in the density?

PHYSICS



Diameter of ball as measured by the screw gauge : d = MSR + n¢ ¥ LC d = 2.5 mm + 20 ¥ 0.01mm = 2.70 mm Density of the ball : mass m ρ = = Volume 4 Ê d ˆ 3 p 3 ËÁ 2 ¯˜ Relative percentage error in measurement of density of the ball : Dρ Dm Dd ¥ 100 = ¥ 100 + 3 ¥ ¥ 100 ρ m d =

2 0.01 ¥ 100 + 3 ¥ ¥ 100 100 2.70

Dρ ¥ 100 = 2 + 1.11 = 3.11 ρ Dρ = 3.11% ρ







x = v0t + at

(ii)

A

B

Kinematics Part-1

t

vavg



Rate of change of position of an object w.r.t time in given direction .It in vector quantity Av er ag e



x t

Total displacement  Total time taken

y cit lo e V

In opposite direction, it will be sum and in same direction, it will be difference for the same frame of reference

(ii)

   vAB  vn  vB  A  vBn  vB  vA

a n-1) 2

a

Rate of change of velocity w.r.t. time,  v Types

orm

  x vinst  lim t  0 t   dx dt





Av e

v 1t 1  v2 t2  .....  vn tn t 1  t 2  .....  t n

When object traversed different speeds in different time of intervals.

S1 + S2 + S3+....Sn t1 + t2 + t3+....tn

When object traversed different distances with different time.

Distance traversed/time taken . Scalar quantity

The shotest distance traversed by any object in called displacement  It in a vector quantity. 

Total length of the path traversed by any object is called its distance  It is scalar quantity. 

When the magnitude or the direction of velocity changes w.r.t. time. o ti o n

n

ment Displace

e tanc Dis

Non - uni form m

if Un

ti o Mo

Equal distances are traversed in equal amount of time.

  v dv a inst lim  t  0 t dt

eed Sp e g ra

(i)

(iv) xnth = v0+

2 2 (iii) v = v0 + 2ax

v = v+ at 0

(i)

Free Fall

ns of quatio tic e a n motion nem cceleratio Ki a ly rm i fo n u

Earth’s gravity (g= 9.8m/s2) on neglecting air resistance. It is a case of motion with uniform acceleration. e.g Apple falling from a tree.

An object falling because of

Total time interval  v  t

Total change in velocity

Acceleration

aavg 

Ve locity



KINEMATICS

15



2 Horizontal range,R = u sin2θ g

Total time of flight,Tf = 2u sinθ g Maximum height,Hmax = u2sin2θ 2g

Vectors having common starting point.

If, A = – B



Vectors having same magnitude but opposite direction   eg. A is a negative of B

Vectors having same direction and magnitude

gnitu It has magnitude as one or unity  A ^ A =  A

Equ

Pla ne

tile jec n o tio Pr mo

Mo tio ni na

Kinematics Part-2

x=(ucosθ)t;

y=(u sinθ) t - 1 gt2 2

Motion of an object that is in flight after being thrown or projected.

r

Vec to

→ P

tile ation of path of projec

g y=x tanθ– 2 2 x2 2u cos

→ P

Law of Parallelogram Law of Triangle → → → R=P+Q → → → → R R Q Q

Addition of Vectors

It has zero magnitude and orbitary direction

at any instant po ax = 0,ay = g C o m

→ → |V & AB|=| VBA|

→ → VAB = –VBA

→→ → → V BA = VB –VA

→ → → V AB =VA –VB

ac=v2/r = r2 = 4π2 r2

A body in a circular motion acted upon by an acceleration directed towards centre of the circular motion.

Angular velocity ,ω= θ/t Angular acceleration,d=∆ω/∆t eg- merry go around.

When an object follows a circular path at a constant speed , the motion of the object is called uniform circular motion.

→ a = axi + ayj; ax=dvx/dt & ay= dvy/dt → |a|=√ax2+ay2

→ v =vx i + vy j , → magnitude|v | = √vx2+vy2

→ i + y^ j Position vector , r = x^ → Displacement vector ,  r =  x^ j i +y^

→ → λ A=λA

u= ucosθ,u= u sinθ x y

Motion of Body under two dimensional frame.

→→ → → A–B=A+(–B)

ele r a ti o n

f acc n ts o

Centripeta l

ne

16 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Acce lera tio n

Chapter 2

Kinematics

Syllabus Frame of reference; Motion in a straight line : Position-time graph speed and velocity; Uniform and nonuniform motion, average speed and instantaneous velocity; Uniformly accelerated motion, velocitytime and position-time graphs, relations for uniformly accelerated motion; Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector; Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.

Topic-1

LIST OF TOPICS : Topic-1 : Motion in straight line

Motion in Straight Line

.... P. 17

Topic-2 :  Kinematics in Two Dimensions .... P. 23

Concept Revision (Video Based) Speed, Velocity and Acceleration



Velocity-time and Position-time Graphs

Kinematics

Horizontal Motion  

Vertical Motion

  

  

Part-1     Part-2

Relative Velocity

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. Train A and train B are running on parallel tracks in opposite directions with speeds 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/hour. Speed (in ms–1) of this person as observed from train B will be close to : (take the distance between the tracks as negligible) (a) 30.5 m s–1 (b) 29.5 m s–1 –1 (c) 31.5 m s (d) 28.5 m s–1 [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. A tennis ball is released from a height h and after falling freely on a wooden floor it rebounds and h reaches height . The velocity versus height of 2 the ball during its motion may be represented graphically by :



(graphs are drawn schematically and on not to scale) (a)

(b)

v h h/2

(c)

v h/2

h(v) h

(d)

v h/2 h

h(v)

h(v)

v h h/2

h(v)

[JEE (Main) – 4th Sep. 2020 - Shift-1] Q.3. A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packed is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to (g is the acceleration due to gravity) :

18 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a) t =

 2h  (b) t =    3g 

2 h   3 g

h h (c) t = 3.4   (d) t = 1.8   g g   [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.4. A balloon is moving up in air vertically above a point A on the ground. When it is a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45° with respect to the vertical. When the balloon climbs up a further height h2, it is seen at an angle 60° with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2 is (given tan30° = 0.5774) :

h2

h1 A

60°

45° d

B

2.464 d

C

(a) 0.732 d (c) 1.464 d

(b) 0.464 d (d) d [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.5. The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is : v (m/s) 4 2 0 2

PHYSICS

Q.8. The bob of a simple pendulum has mass 2 g and a charge of 5.0 mC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take g =10 m/s2) (a) tan–1 (2.0) (b) tan–1 (0.2) –1 (c) tan (5.0) (d) tan–1 (0.5) [JEE (Main) – 8th April 2019 - Shift-1] Q.9. Ship A is sailing towards north-east with velocity  i + 50 ^ j km/hr where ^ i points east and ^ j, v = 30 ^ north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in : (a) 4.2 hrs. (b) 2.6 hrs. (c) 3.2 hrs. (d) 2.2 hrs. [JEE (Main) – 8th April 2019 - Shift - 1] Q.10. A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time) (A)

(B)

v

a O

O

t

(C)

t

(D)

A B

x

D 1 2 3 4 5 6 C S

x

t (in s)

(a) 12 m

49 m (b) 4

(c) 11 m

(d)

37 m 3

[JEE (Main) – 5th Sep. 2020 - Shift-2] Q.6. When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + b)v, this angle changes to 45°. The value of b is close to : (a) 0.41 (b) 0.50 (c) 0.73 (d) 0.37 [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.7. A particle moves such that its position vector  r (t ) = cos ω ti + sin ω t j where w is a constant and t is time. Then which of the following statements is   true for the velocity v(t ) and acceleration a(t ) of the particle  :  r (a) v and a both are parallel  to  (b) v is perpendicular to r and a is directed away from the   origin  r (c) v and a both are perpendicular to   (d) v is perpendicular to r and a is directed towards the origin [JEE (Main) – 8th Jan. 2020 - Shift-2]

O

O

t

t

[JEE (Main) – 8th April 2019 - Shift - 2] (a) A (b) A, B, C (c) B, C (d) A, B, D Q.11. The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight? (a) 90° (b) 150° (c) 120° (d) 60° [JEE (Main) – 9th April 2019 - Shift-1] Q.12. A ball is thrown upward with an initial velocity v0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mg v2 (where m is mass of the ball, v is its instantaneous velocity and g is a constant). Time taken by the ball to rise to its zenith is : (a)

(c)

 γ  1 tan −1  v  g 0  γg  

(b)

 γ  1 sin −1  v  g 0  γg  

 2γ  1  γ  tan −1  v ln  1 + v0  (d)  g 0    2γ g g  γg    th [JEE (Main) – 10­ April 2019 - Shift-1] 1

19

KINEMATICS

Q.13. A bullet of mass 20 g has an initial speed of 1 m s–1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5×10–2 N, the speed of the bullet after emerging from the other side of the wall is close to : (a) 0.1 m s–1 (b) 0.7 m s–1 –1 (c) 0.3 m s (d) 0.4 m s–1 [JEE (Main) – 10th April 2019 - Shift-2] Q.14. A particle is moving with speed v = b x along positive x-axis. Calculate the speed of the particle at time t = t (assume that the particle is at origin at t = 0). (a)

b 2τ 4

(b)

(c) b2t

(d)

b 2τ 2

a1a2 t

(d)

a1 + a2 t 2

v (m/s) 3 2 1 0

11 5

(d)

5 2

[JEE (Main) – 12th Jan. 2019 - Shift-1] Q.19. A person standing on an open ground hears the sound of a jet aeroplane coming from north at an angle 60° with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is : (a)

3 v 2

(b)

2v 3

v (d) 2

(c) v

2

[JEE (Main) – 9th Jan. 2019 - Shift-2] Q.16. A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5 s?

[JEE (Main) – 12th Jan. 2019 - Shift-1]



ANSWER – KEY

1. (b) 5. (d) 9. (b) 13. (b) 17. (b)

2. (d) 6. (c) 10. (d) 14. (b) 18. (c)

3. (c) 7. (d) 11. (c) 15. (c) 19. (d)

4. (d) 8. (d) 12. (a) 16. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. km Given : Speed of train A, v A = 36 km/hr, , speed of km hr are moving , train B, vB = 72 km/hr, both the trains hr in opposite direction, speed of a person walking km ,vp vp is parallel to vB. inside train A, vp = 1.8 km/hr, hr To find : The speed of the person as observed from m train B in m/s,,vp′ . s Let train A be moving in –x-direction and train B be moving in +x-direction. So, velocity of train A as observed from train B is : vA′ = ( −72 − 36 )) km /hr = 108 km /hr = 30 m /s

1 2 3 4 5 6 7 8 9 10 t(s)

(a) 10 m (b) 6 m (c) 3 m (d) 9 m [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.17. A particle moves from the point (2.0 ^ i + 4.0 ^ j) m, at t = 0, with an initial velocity (5.0 ^ i + 4.0 ^ j) m s–1. It is acted upon by a constant force which produces a constant acceleration (4.0 ^ i + 4.0 ^ j)m –2 s . What is the distance of the particle from the origin at time 2 s? (a) 15 m

(c)

b 2τ

[JEE (Main) – 12th April 2019 - Shift-2] Q.15. In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2, respectively. Then ‘v’ is equal to : 2a a (a) 1 2 t (b) 2 a1a2 t a1 + a2 (c)

i. They are moving in the same direction and ii. In the opposite direction is 25 3 (a) (b) 11 2

(b) 20 2 m (c) 5 m (d) 10 2 m [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.18. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of time taken by the passenger train to completely cross the freight train when :

Velocity of person (the person is moving in +x direction) : vp = +1.8 km /hr = 0.5 m / s Now, velocity of the person as observed by train B : vp′ = [ −30 − ( 0.5)] m /s = −29.5 m /s  Hence, the magnitude of velocity is vp′ = 29.5 m /s Ans. 2. Option (d) is correct. Given : A tennis ball is released from height h, it falls on a wooden floor and bounces back to a h height of . 2 To find : Choose the correct velocity versus height graph for the motion of the ball.

20 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

(a) The motion of the ball falling from height h to the wooden floor (h = 0) will be given by equation : v 2 = u2 − 2 gh

h2

(where u = 0, since the ball is dropped and 'h' is the distance by which the ball falls down) We see the equation is that of a parabola and the velocity is in –ve direction. It reaches its maximum value when the ball hits the floor. (b) The motion of the ball when it rebounds from the wooden floor is given by equation : v 2 = u2 − 2 gh (h is the instantaneous height to which the ball rises) Now the velocity has reversed its direction but the equation is still that of a parabola form. The velocity h reaches zero at height . 2 Both points (a) and (b) are satisfied by graph (d). Ans. 3. Option (c) is correct. Given : The initial velocity of the helicopter is u = 0, constant upward acceleration of the helicopter is g, height at which the helicopter releases a food packet is h. To find : t, the time taken by the food packet to reach the ground. Velocity of the helicopter when it drops the food packet : 2

2

v =

2 gh

h1 A

Distance travelled by the food packet before it hits the ground : 1 − h = 2 gh t − gt 2 2 1 2 gt − 2 gh t − h = 0 2 t =

t = t =

2 gh ± 4 gh g h ( 2 + 2) g

t = 3.4

h g

Ans. 4. Option (d) is correct. Given : A balloon released from point A is travelling vertically up in air, the height of the balloon is h1 when observed from B and h1 + h2 when observed from C.

C

h1 = d Also,

tan 30° =

h1 + h2 1 = d + 2.464 d 3

d + h2 1 = 3.464 d 3 h2 =

3.464 d 3

−d ≈d

Ans. 5. Option (d) is correct. Given : The velocity versus time graph for a body moving in straight line is shown below. v (m/s) 4 2 0 2

A B a b c S D e 1 2 3 4 d5 6 C

t (in s)

Point S is at t = 4.333 s To find : x, the distance travelled by the body in 6 seconds. Distance travelled by the body : x = area under the v − t graph x = area of a + area ofb + area ofc + area ofd + area ofe x=

1 2 gh ± 2 gh + 4 × gh 2 1 2× g 2

2.464 d

B

d

To find : The value of h2. From the graph above : h tan 45° = 1 = 1 d

v = u + 2 gh

Initial velocity of the food packet : u′ = 2 gh

60°

45°

1 1 1 × 2 × 4 + 1 × 4 + × 1.333 × 4 + × 0.667 2 2 2 1 × 2 + ×1× 2 2

x = 4 + 4 + 2.666 + 0.667 + 1 = 12.334 =

37 m 3

Ans. 6. Option (c) is correct. Given : The velocity of car is in +x-direction, the velocity of a rain drop for a stationary observer is in –y-direction, the velocity of rain drop for an observer in car moving with velocity v makes 60° angle with the horizontal, velocity of rain drop for an observer in car moving with velocity (1+b)v makes 45° angle with the horizontal. To find : The value of b. Let velocity of rain drop be vr. From the situation discussed above, v tan 60° = r ; vr = 3v v

21

KINEMATICS

Also, tan 45° =

Velocity of ship B  vB = –10i km/h Distance between two ships   = r – r = 80i + 150j.

vr ; 3v = (1 + β )v (1 + β )v

1+ β = 3

B

β = 3 − 1 = 0.73 Ans. 7. Option (d) is correct. Given : Position vector of particle as function of time is  r (t ) = cos ωti + sin ωt j , ...(i) w is a constant.  To find : The direction of velocity vector v(t ) and  acceleration vector a(t ).  From equation (i), expression for v(t ) :  dr  = −ω sin ωti + ω cos ωt j v t) = ( ...(ii) dt  From equation (ii), expression for a(t ) :  dv   ( = −ω 2 r (t ) ...(iii) a t) = dt The negative sign in equation (iii) shows that the   direction of a(t ) is opposite to that of r (t ), that is,  a(t ) is directed towards the origin. From equations (i) and (ii) :  v.r = −ω sin ωt cos ωt + ω sin ωt cos ωt = 0  So, direction of v is perpendicular to the direction  of r. Ans. 8. Option (d) is correct. Given : Mass of bob of simple pendulum = 2g = 2 × 10–3 kg, Charge on bob = 5.0 mC = 5.0 × 10–6 C, Intensity of applied electric field = 2000 V/m. To find : Angle the pendulum makes with the vertical, q. y 

x

T

T cos  

E qE

T sin  mg

Balancing the forces in equilibrium position of bob : T cos q = mg; T sin q = qE That gives, qE tan q = mg

 qE  5.0 × 10 −6 × 2000 q = tan −1  = 2 × 10 −3 × 10  mg 

= tan–1 (0.5). Ans. 9. Option (b) is correct. Given : Velocity of ship A, v = (30i + 50j) km/h, A

A

To find : The time at which the distance between the two ships is minimum.  Let r A = 0i + 0j; that gives :  r B = 80i + 150j After time t :  r = 30ti + 50tj; A

As ship B is moving towards west :  r = (80 – 10t)i + 150j B

Distance between two ships after time t :   d = r – r = (80 – 10t – 30t)i + (150 – 50t)j B

A

d2 = (80 – 40t)2 + (150 – 50t)2 For t to be the time when distance between the two ships is minimum : d (d2) = 0 dt d [(80 – 40t)2 + (150 – 50t)2] = 0 dt 2(80 – 40t)(– 40) + 2(150 – 50t)(–50) = 0 –3200 + 1600t – 7500 + 2500t = 0; 107 ; t = 41 t = 2.6 Ans. 10. Option (d) is correct. Given : At t = 0, (initial position) x = 0, (initial velocity) u = 0 For t > 0, a = constant. To find : The graph(s) among given which represent the motion qualitatively. The graph A shows : At t = 0, A = 0, that gives u = 0. t > 0, a = constant. So, graph A is correct. The graph B shows : At t = 0, u = 0. At t > 0, (velocity) v is linear function of time, that implies dv a = = constant. dt So, graph B is correct. The graph C shows : At t = 0, x ¹ 0. So, graph C is incorrect. The graph D shows : At t = 0, x = 0, At t > 0, the graph looks like a parabola, starting from origin : 1 1 x = ut + at 2 = at 2 2 2 So, graph D is correct as it correctly represents displacement-time graph for a particle moving with constant acceleration.

22 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 11. Option (c) is correct. Given : The speed of river stream, vr = 2 km/h, The speed of swimmer, vs = k km/h. To find : The angle between two velocities a, such that the swimmer crosses the river straight, with velocity vsr.  vr vs

For the above diagram : vr 2 1 = = ; θ = 30 o. 1. sin θ = vs 4 2 q is the angle between the direction the swimmer swims vs and the direction the swimmer actually flows across the river due to push from the river stream vsr. 2. a = 90° + q = 120°. Ans. 12. Option (a) is correct. Given : Initial velocity of ball = v0, Drag force acting on the ball, Fd = mg  v2. To find : Time taken by the ball to reach its maximum height (v = 0). Forces acting on the ball : F = ma = mg + mg v2; a = g + g v2 dv = − . dt

The negative sign in above equation indicates that the velocity and acceleration of the ball are in opposite directions. 0 dv  dv 1 v0  = − ∫ dt ; ∫v  = ∫ dt  2  2 0 γ ∫0  g +γv  g   v2 +   γ  Using identity : 1  dx  −1  x  ∫  2 = tan   2 a a x +a 



Resistance force offered by the wall, F = 2.5 × 10–2 N. To find : Final speed of the bullet. Deceleration of the bullet :

    v 1 γ tan −1  0  t =  g  γ g    γ 

 γ v0 1 tan −1  =  g γg  Ans. 13. Option (b) is correct. Given : Mass of bullet, m = 20 g = 20 × 10–3 kg, Initial speed of bullet, u = 1 m/s, Thickness of wall = 20 cm = 20 × 10–2 m,

   

a =

F −2.5 × 10 −2 m/s = = −1.25 m / s2,2 , m 20 × 10 −3

Final speed of bullet : v2 = 2as + u2 = –2 × 1.25 × 20 ×10–2 + 12; 1 = 0.7 m/s 2 Ans. 14. Option (b) is correct. Given : Velocity of particle, dx = b x, v = dt Direction of particle’s velocity = along +x axis, At t = 0, position of particle is at the origin (x = 0). To find : Speed of particle at t = t. Acceleration of particle :

 vsr

PHYSICS

v = 

b2 dv 1 b dx 1 b = =  b x = . a = 2 dt 2 x dt 2 x As acceleration is constant, velocity at t = t, will be v(t) = u + at. Here, u is initial velocity of particle at t = 0 :

x b= 0 0. (given, at t = 0 , x = 0) u = b= That gives

v(t) = at =

b2 τ. 2

Ans. 15. Option (c) is correct. Given : Initial velocity of both the cars : uA = uB = 0. At t > 0, acceleration of car A, aA = a1 and acceleration of car B, aB = a2, Car B finishes the race in time T, Car A finishes the race in time T – t, Final velocity of car B at time T = vB (T) = uB + aBT = a2T, Final velocity of car A at time (T – t) : vA(T – t) = uA + aA(T – t) = a1(T – t). To find : Value of v = vA – vB v = a1(T – t) – a2T; v = (a1 – a2)T – a1t ...(i) Also, distance travelled by car A in time (T – t) and car B in time T is same : 1 1 2 2 So, dA = dB ; a1 ( T − t ) = a2 T ...(ii) 2 2 From (ii) we get :

T =

a1 a1 − a2

t. ...(iii)

Put value of T from equation (iii) in equation (i).

v = ( a1 − a2 )

a1 a1 − a2

t − a1t =  a1a2 t.

23

KINEMATICS

Ans. 16. Option (d) is correct. Given : The graph of velocity versus time for a particle. To find : Position of particle at t = 5 s.

l1 = 60 m, Speed of passenger train, v1 = 80 km/h, Length of freight train, l2 = 120 m, Speed of freight train v2 = 30 km/h. To find : The ratio, t1/t2, of times taken by the two trains to cross each other when : (i) They are moving in same direction (t1). (ii) They are moving in opposite direction (t2). Relative velocity of trains, when they are moving in opposite direction = v1 + v2. Relative velocity of trains, when they are moving in same direction = v1 – v2. (l1 + l2 ) v − v2 v1 + v2 110 11 t1 = = . =  = 1 So, (l1 + l2 ) v1 − v2 50 5 t2 v1 + v2

v (m/s) D A 3 E G 2 1 F B C 0 O 1 2 3 4 5 6 7 8 9 10 t (s)

Displacement of particle at (t = 5 s) = area under the graph, “velocity versus time” from t = 0 s to t = 5 s. So displacement = area D OEF + area of square EFBG + area of rectangle DBCA 1  =  × 2 × 2  + ( 2 × 2 ) + ( 3 × 1) 2  = 2 + 4 + 3 = 9 m Ans. 17. Option (b) is correct. Given : Position of particle at (t = 0),  r 0 = (2.0i + 4.0j) m, Velocity of particle at (t = 0),  u = (5.0i + 4.0j) m/s, Acceleration of particle,  a = (4.0i + 4.0j) m/s2, To find : r = distance of particle from origin at (t = 2 s).   1  2  r = ro + ut + at 2  j + ( 5.0 i + 4.0 ) j t + 1 ( 4.0 i + 4.0 ) j t 2  m = ( 2.0 i + 4.0 )  2   Putting t = 2 s in above expression, we get :  r = (20 i + 20 j)m  r = 20  2 m

Ans. 19. Option (d) is correct. Given : Speed of sound = v, Position 1 (P1) of plane makes an angle of 60° with the ground, Position 2 (P2) of plane makes an angle of 90° with the ground. To find : Velocity of the plane, vp.

From the diagram : vp 1 v ; vp . = = cos 60° = v 2 2

Ans. 18. Option (c) is correct. Given : Length of passenger train

Topic-2

Kinematics in Two Dimensions Concept Revision (Video Based) Vectors

Part -1       Part -2

Projectile Motion

  Part -1               Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. Standing from the origin at time t = 0, with initial velocity 5 j ms–1, a particle moves in the x-y plane with a constant acceleration of (10i + 4 j ) ms–2. At time t, its coordinates are (20m, y0 m). The values

of t and y0 are, respectively : (a) 4s and 52 m (b) 2s and 24 m (c) 2s and 18 m (d) 5s and 25 m [JEE (Main) – 4th Sep. 2020 - Shift-1]

24 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.2. A particle moving in the xy-plane experiences a  velocity dependent force F = k(υ y i + υ x j ), where ux and uy are the x and y components of its velocity   υ . If a is the acceleration of the particle, then which of the following statements is true for the particle?   (a) quantity υ .a is constant in time   (b) quantity υ ×a is constant in time  (c) F arises due to a magnetic field (d) kinetic energy of particle is constant in time [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.3. A particle starts from the origin at t = 0 with an initial velocity of 3.0 i m/s and moves in the x-y plane with a constant acceleration (6.0i + 4.0 j ) m/s2. The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is : (a) 60 (b) 50 (c) 32 (d) 40 [JEE (Main) – 8th Jan. 2020 - Shift-2]

    A1 3= , A 2 5 and A1 + A 2 = 5. The value of Q.4. Let=     ( 2 A1 + 3A 2 ) • ( 3A1 − 2 A 2 ) is : (a) –106.5 (b) –99.5 (c) –112.5 (d) –118.5 [JEE (Main) – 8th April 2019 - Shift-2] Q.5. The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

(a) a +

b2 4c

(b) a +

b2 3c

(c) a +

b2 c

(d) a +

b2 2c

[JEE (Main) – 9th April 2019 - Shift-2] Q.6. The position vector of a particle changes with  time according to the relation r ( t ) = 15t2 ^ i +

(a) 20 cm (c) 26 cm

 u  

(b) 18 cm (b) 14 cm [JEE (Main) – 10th April 2019 - Shift-2]

Q.8. A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If y, and t, are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is : (a)

R 4g

(b)

R g

(c)

R 2g

(d)

2R g

[JEE (Main) – 12th April 2019 - Shift-1] Q.9. The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle q0 with speed v0 then (g =10 ms–2) : 5 −1  1  −1 (a) θ 0 = sin   and v0 = ms 3 5  

3  2  −1 (b) θ 0 = cos−1   and v0 = ms 5  5 5 −1  1  −1 (c) θ 0 = cos   and v0 = ms 3  5 3 −1  2  −1 (d) θ 0 = sin   and v0 = ms 5  5 [JEE (Main) – 12th April 2019 - Shift-1]

Q.10. Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct? (a) R2 = 4h1h2 (b) R2 = 16h1h2 2 (c) R = 2h1h2 (d) R2 = h1h2 [JEE (Main) – 12th April 2019 - Shift-2]  Q.11. A particle is moving with a velocity v = K(yi^+ xj^), where K is a constant. The general equation for its path is : (a) y = x2 + constant (b) y2 = x + constant (c) y2 = x2 + constant (d) xy = constant [JEE (Main) – 9th Jan. 2019 - Shift-1]

(4 – 20t2)j^. What is the magnitude of the acceleration at t = 1? (a) 40 (b) 25 (c) 100 (d) 50 [JEE (Main) – 9th April 2019 - Shift-2] Q.7. A plane is inclined at an angle a = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms–1, from the base of the plane, making an angle q = 15° with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to : (Take g = 10 ms–2)

PHYSICS

Q.12. The position co-ordinates of a particle moving in a 3-D coordinate system is given by

x = acos wt y = asin wt and  z = awt The speed of the particle is : (a)

2 aω

(c)

3 aω

(b) aw (d) 2aw [JEE (Main) – 9th Jan. 2019 - Shift-II]

Q.13. In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be :

25

Kinematics

z B

1 x = ux t + ax t 2 2

E

20 = 0 +

a A

H G

F

O

t = 2s

y

The y coordinate of the distance covered by the particle in time t : 1 y = uy t + a y t 2 2

a x D

a

(a)

1 ^ ^ a (k – i ) 2

(b)

1 ^ ^ a (i – k ) 2

(c)

1 ^ ^ a (j – i ) 2

(d)

1 ^ ^ a (j – k ) 2

yo = 5 × 2 +

  −1 n − 1 (c) sin  2   n + 1 

(d) sin

−1  n − 1 

n + 1  

th

[JEE (Main) – 10 Jan. 2019 - Shift-2] Q.16. A body is projected at t = 0 with a velocity 10 m s–1 at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1 s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 m s–2, the value of R is : (a) 2.5 m (b) 2.8 m (c) 10.3 m (d) 5.1 m [JEE (Main) – 11th Jan. 2019 - Shift-1]

ANSWER – KEY

1. (c) 5. (b) 9. (c) 13. (c)

2. (b) 6. (d) 10. (b) 14. (a)

3. (a) 7. (a) 11. (c) 15. (a)

1 × 4 × 22 2

y o = 18m

[JEE (Main) – 10th Jan. 2019 - Shift-1] Q.14. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is : (a) 1 : 16 (b) 1 : 2 (c) 1 : 4 (d) 1 : 8 [JEE (Main) – 10th Jan. 2019 - Shift-1]   Q.15. Two vectors A and B have equal magnitudes. The   magnitude of ( A + B ) is ’n’ times the magnitude     of ( A – B ). The angle between A and B is :  2  −1  n − 1  −1 n − 1 (b) cos  (a) cos  2   n + 1 n + 1   2

1 × 10t 2 2

4. (d) 8. (d) 12. (a) 16. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : Velocity of a particle moving in xy plane  / s, acceleration of the particle at t = 0 is u = 5 j mm/s,  2 , is a = (10i + 4 j )m/ m ss–2 , coordinates of the particle at time t are (20 m, yom). To find : Value of t and y0. The x coordinate of the distance covered by the particle in time t :

Ans. 2. Option (b) is correct. Given : Force acting on a particle moving in xy plane  is F = k ( v i + v j ), where v and v are components y

x

x

y

of velocity of the particle and acceleration of the  particle is a.   To find : Relation between v and a. From given expression of force acting on the particle  :   F || v

Also, by Newton’s second law of motion :     F = ma ;F || a From above set of equations :   v || a That gives :     v × a = 0 or v × a = constant Ans. 3. Option (a) is correct.  Given : Position of particle at t = 0 is r = 0, velocity  m/s, of particle at t = 0 is vox = 3i m / s,particle moves in an xy plane with a constant acceleration  2 a = ( 6.0i + 4.0 j ) m m/s/ 2s. . To find : x coordinate of the particle when its y coordinate is y = 32 m, y coordinate of the particle at some time t : 1  y = voy t + a yt 2 2 ⇒

32 = 0 +

⇒

t = 4, s

1 × 4 × t2 2

it is the time elapsed after t = 0 when y coordinate of particle is t = 32 m. x coordinate of the particle at time t = 4s : 1  D = vox t + ax t 2 2 D = 3×4 +

1 × 6 × 4 2 = 60 m. 2

Ans. 4. Option (d) is correct. Given :

    A1 = 3, A 2 = 5, A1 + A 2 = 5. ...(i)

26 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)     To find : ( 2 A1 + 3 A 2 ).( 3 A1 − 2 A 2 )   2 2 A1 + A 2 = A1 + A 2 + 2 A1 A 2 cos θ ;

using the given values (equation 1) in above equation, we get : 3 cos q = − . 10      2  2   ( 2 A1 + 3 A 2 ).( 3 A1 − 2 A 2 ) = 6 A1 − 6 A 2 + 5 A1 .A 2

Initial velocity of projectile, u = 2 m/s. g = 10 m/s2. To find : The distance from the base at which the particle hits the plane, d. u sin=uy y

Acceleration of particle, dv(t ) = 2b − 6 ct a(t) = dt Putting a(t) = 0 in above equation; we get : b t = . 3c Velocity at, t =

b 3c

= a + 2bt – 3ct2 2

b2  b   b  = a + 2b   − 3c   = a + 3c  3c   3c  Ans. 6. Option (d) is correct. Given : Position vector of particle as function of time :  r (t ) = 15t2 ^ i + (4 – 20t2 ) ^ j To find : Acceleration of particle at t = 1, a(t = 1). Velocity of particle :  dr  v(t ) = = 30t i^ – 40t j^ dt Acceleration of particle :    dv d 2 r a(t ) = = 2 = 30 i^ – 40 j^ dt dt Magnitude of acceleration at t = 1,  a = 30 2 + 40 2 = 50. Ans. 7. Option (a) is correct. Given : Angle of inclination of the plane with ground, a = 30°, Angle of inclination of the projectile with the plane, q = 15°,

u 

u cos=ux

g sin=ax

= 6 × 9 – 6 × 25 + 5(3)(5) cos q  3  = 54 – 150 + 75  −  = – 118.5  10  Ans. 5. Option (b) is correct. Given : Position of particle as function of time is x(t) = at + bt2 – ct3, where a, b and c are constants. To find : Velocity of particle when acceleration is zero. Velocity of particle, dx(t ) = a + 2bt − 3ct 2 v(t) = dt

PHYSICS



x

g cos=ay



g

Let t be the time the particle takes to reach the base : 2uy u sin θ = 2 t = ay g cosα Also, 1 2 d = ux t + ax t 2

  u sin θ   1  u sin θ  =  u cosθ  2    − ( g sin α )  2   cos g α 2    g cosα  

2

Putting the given values, we get : d = 20 cm. Ans. 8. Option (d) is correct. Given : Initial speed of shell = u, Distance between the point the shell is fired to the point it hits the ground = R. To find : The product t1t2, where t1 and t2 are two values of time taken by the shell to hit the target in two possible ways, such that the range for both the paths is same, R1 = R2. Let the two possible angles at which the shell is fired such that it maintains a range of R are q1 and q2 . For projectile motion time of flight, sin θ1 t1 = 2u g and range, u2 sin 2θ1 R1 =  g sin θ 2  Similarly, t2 = 2u g and range, u2 sin 2θ 2 R2 =  g As R1 = R2 ; u2 sin 2θ 2 u2 sin 2θ1 ; = g g sin 2q1 = sin 2q2 ; q1 + q2 = 90.



 sin θ1   sin θ 2    2u  t1t2 =  2u g  g  

27

KINEMATICS

= =

t1t2 =

= =

4 u2 g2 4 u2 g2 4 u2 g2 2u 2 g2

Maximum height attained by projectile is given as :

sin θ1 sin θ 2 sin θ1 cos ( 90 − θ1 ) sin θ1 cosθ1 sin 2θ1

launch angle = q0, initial speed = v0. To find : Value of q0 and v0. Given equation of trajectory of particle, y = 2x – 9x2 ...(i) Equation of trajectory in general for projectile, gx 2 y = tan θ o −  2 ...(ii) ( 2 v0 cos2 θ 0 )

Comparing (i) and (ii) : tan q0 = 2; g = 9 ( 2 v02 cos2 θ 0 )

...(iii) ...(iv)

From equation (iii) : 2 5

; cosθ o =

1 5

...(v)

Put value of cos q0 from equation (v) in equation (iv) : 5 v0 = m/s 3 Ans. 10. Option (b) is correct. Given : For two projectiles the initial speeds, u1 = u2 = u, The range of the two projectiles : R1 = R2 = R, The maximum heights reached by the two projectiles : h1 and h2. To find : Relation between R, h1 and h2. Let projectile 1 is projected at angle q1 and projectile 2 is projected at angle q2. As range and initial velocity of both projectiles is same q1 + q2 = 90°. Let q1 = 45° + q and q2 = 45° – q ...(i) Range for projectile motion is given as : R1 =

u2 sin 2 ( 45° + θ ) , 2g



h2 =

u2 sin 2 ( 45° − θ ) ...(ii) 2g

 u4 h1h2 =  2  4g

=

2  u2 sin 2θ1  2 R =   g g  g 

sin q0 =

h1 =



Ans. 9. Option (c) is correct. Given : Trajectory of particle near the surface of earth : y = 2x – 9x2, g = 10 m/s2,





u2 sin 2θ1 u2 sin 2θ 2 = R2 = =R ...(ii) g g

u4 16 g 2

  (sin ( 45° + θ ) sin ( 45° − θ ))2   ( 2 sin ( 45° + θ )sin ( 45° − θ ))2

Using identity 2, sin A sin B = cos (A – B) – cos (A + B)

h1h2 =

u4 16 g 2

(cos 2θ − cos 90 )2

 u2  u4 2 cos 2θ  =  2 cos 2θ =   16 g  4g 

2

2

 u2   u2  =  sin ( 90° + 2θ )  =  sin 2 ( 45° + θ )  4 4 g g     [using equations (i) and (ii)] : 2

2

2

 u2  R2 1  u2 sin 2θ1  sin 2θ1  =   = =     16  16 g  4g   So, relation between R, h1, h2 : R2 = 16h1h2 Ans. 11. Option (c) is correct. Given : Velocity of particle :  v = k(yi^ + xj^) ...(i) To find : Equation for path of the particle. Velocity :  dx ^ dy ^ v = i + j ...(ii) dt dt Comparing (i) and (ii) : dy dx = y and = x; dt dt x dy = ; = xdx y dx Integrating both sides we get : y2 = x2 + constant. Ans. 12. Option (a) is correct. Given : x = acos wt; y = asin wt; z = awt, are position coordinates of a particle. To find : Velocity of the particle. dz dy  dx v = ^ i+ ^ j+ ^ k dt dt dt

...(iii)

^ = –aw sin wt ^ i + aw cos wtj^+ awk v = (( a 2ω 2 sin 2 ωt + a 2ω 2 cos2 ωt + a 2ω 2 ))

= 2aω Ans. 13. Option (c) is correct. Given : Side of cube = a. To find : The vector GH.

28 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) By vector identity we know :  2 2 2 A + B = A + B + 2 AB cosθ

z B

E

= 2 A 2 + 2 A 2 cosθ [Using equation (i)]

a A

H G

F

O

y

a a x D Coordinates of point G : a xG = = ,yG 0= , zG 2 Coordinates of point H : a ,y H = , zH xH = 0= 2

a 2 a 2

2 uA ; g

So, area covered by bullets on the ground fired by gun A, 2 AA = π RA =

π uA4 2

g Similarly, maximum range of gun B, u2 RB = B ; g So, area covered by bullets on the ground fired by gun B, 2 AB = π R B =

π uB4 g2

The ratio,



π uA4 g2

= 2 A 2 (1 + cosθ ) Similarly,  2 2 2 A − B = A + B − 2 AB cosθ = 2 A 2 − 2 A 2 cosθ

So, the vector GH will be : (xH – xG) ^ i +(yH – yG) ^ j +(zH – zG) ^ k a ^ a ^ a ^ ^ = – i + j = (j – i ) 2 2 2 Ans. 14. Option (a) is correct. Given : Gun A fires bullets at speed, uA = 1 km/s, Gun B fires bullets at speed, uB = 2 km/s. To find : Ratio of maximum areas covered by the A bullets on the ground fired by the two guns, A . AB Maximum range of gun A, RA =

PHYSICS

4

4  uA  AA 1 1 = = =     = 16 . AB π uB4  uB   2  g2

Ans. 15. Option (a) is correct. Given :   A = B ...(i)     A + B = n A − B ...(ii) To find : The angle q between the two vectors A and B.

2 = 2 A (1 − cosθ ) Using equation (ii) 2 2 2 2 A (1 + cosθ ) = n 2 A (1 − cosθ ); 1 + cosθ . n2 = 1 − cosθ On simplification :

 2  −1 n − 1 θ ; = cos  2  2 ( n + 1)  n +1 Ans. 16. Option (b) is correct. Given : At t = 0, velocity of projectile, u = 10 m/s, Angle of projection, q = 60°, Acceleration due to gravity, g = 10 m/s2. To find : Radius of curvature of the projectile’s trajectory at t = 1 s, R. x component of velocity vector of a projectile at t = 1 s, m / s. vx = u cosθ + g x t = 10 × cos 60° = 5 m/s. cos q =

( n 2 − 1)

(as acceleration due to gravity is directed downwards, gx = 0) y component of velocity vector of a projectile at t = 1,

vy = u sin θ + g yt = 10 × sin 60° − 10 × 1

  3 − 10  m/s. =  10  2   angle the velocity vector makes with the horizontal at t = 1, vy =  3 − 2 ; φ = 15°. tan f = vx Radius of curvature of the projectile’s trajectory at t=1:

R =

vx2 + v y2 v2 = ≈ 2.8 m g cos φ 10 cos 15

Subjective Questions (Chapter Based) Q.1. The speed versus time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be ...... [JEE (Main) – 4th Sep. 2020 - Shift-2]

29

KINEMATICS

Q.3. The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is ____. [JEE (Main) – 9th Jan. 2020 - Shift-1] Sol. Given : Distance x travelled by a particle as function of time is

10 8

u (ms–1) 6 4 2 1

2 3 time (s)

4

5

x 2 = at 2 + 2bt + c ,

Sol. Given : A speed versus time graph for a particle

...(i)

acceleration of particle is A ∝ x −n

y

...(ii)

To find : Value of n. Differentiating equation (i) with respect to time: dx  dx  2x = 2 at + 2b  dt = v  dt  

8 m/s

vx = at + b

t 5s

To find : x, the distance travelled by the particle. Distance travelled by the particle : x = Area under the v − t graph 1 x = × 5 × 8 = 20m 2

x(t ) = 10 + 8t − 3t 2 ,

...(i)

position of particle-2 as function of time is 3

y( t ) = 5 − 8 t ,

Differentiating equation (iii) with respect to time : dx dv  dv  v + x = a = A  dt dt  dt  v 2 + Ax = a A =

Q.2. A particle is moving along the x-axis with its coordinate with time ‘t’ given by x(t) = 10 + 8t – 3t2. Another particle is moving along the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t3. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as v . Then v (in m/s) is _____________. [JEE (Main) – 8th Jan. 2020 - Shift-1] Sol. Given : Position of particle-1 as function of time is

...(ii)

speed of particle-2 as measured by particle-1 at t = 1 s is v21 = v m s ...(iii) To find : Value of v. From equation (i), velocity of particle-1 as function of time is   v1 = ( 8 − 6t )i ...(iv) From equation (ii), velocity of particle-2 as function of time is  v = −24t 2 j ...(v) 2

A =



a−v x a−

Comparing equations (iii) and (vii), we get : v = 580

at + b   v = x   

x2 x

=

ax 2 − a 2t 2 − b 2 − 2 abt x3 ( x 2 = at 2 + 2bt + c )

A = =

a 2t 2 + 2 abt + ac − a 2t 2 − b 2 − 2 abt x3 ac − b

2

...( iv )

x3

A ∝ x −3 n = 3

   Q.4. The sum of two forces P and Q is R such that   R = P . The angle q (in degrees) that the resultant    of 2P and Q will make with Q is, __________. [JEE (Main) – 7th Jan. 2020 - Shift-2] Sol. Given :    P + Q = R, ...(i)   |R| = P ...(ii)    To find : Angle q between 2 P + Q and Q. By construction : P 

Put t = 1 in equation (vi) : ...(vii)

2

( at + b )2

From equation (iv) and (v), speed of particle-2 as measured by particle-1 will be :    v21 = v2 − v1 = −24t 2 j − ( 8 − 6t )i ...(vi) v21 = −24 j − 2i = ( 24 )2 + 2 2 = 580 m s

...(iii)

 P

P+Q



 Q

We can see q = 90°.

2P+Q

30 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)  Q.5. The position vector r of a particle of mass m is given by the following equation  r (t ) = α t 3 i + β t 2 j , 10 m s, b = 5 m s and m = 0.1 kg. At 3 t = 1 s, find the velocity, angular momentum with respect to the origin, force and torque with respect to the origin for the given case in vector notation. Sol. Given : Position vector for particle of mass m :  r = α t 3 i + β t 2 j ,

A B

10 m s3 ,β = 5 m s2 , 3

That gives,   10  r =  t 3 i + 5t 2 j  m 3  

Sol. Given : Speed of plane A,





At t = 0 s, relative velocity of B with respect to A, vBA is along the line perpendicular to the line of motion of A,



At t = 0 s, distance between two planes is 500 m.



To find : The time t0, when the two planes just collide.



As relative motion of B is along the line perpendicular to the motion of A,







Plane B is approaching plane A along line perpendicular to the motion of A with velocity :



vB sin 30° = 200 ×

va = 100 3 m/s,

...(i)

Mass of particle, m = 0.1 kg. To find : Various quantities associated with the particle at t = 1 s. (a) Velocity of particle (from equation (i)) :  dr  v = = 10t 2 i + 10t j = (10i + 10 j )m s dt ...(ii) (b) Angular momentum with respect to origin :      L = r × p = r × mv At t = 1 s, position vector (from equation (i)) :   10  r =  i + 5 j  m ...( iii )  3  At t = 1 s, angular momentum (using equations (i) and (ii)) :   10  L =  i + 5 j  × ( 0.1)(10i + 10 j )  3  5 = − k N m s 3   (c) F = ma At t = 1 s, acceleration (using equation (ii)) :  dv  = 20ti + 10 j = 20i + 10 j a = dt At t = 1 s, Force :  F = ( 0.1)( 20i + 10 j ) = ( 2i + j )N

60°

30°

where α =

where  α =

PHYSICS

vA = vB cos 30° = 100 3 ; vB = 200 m/s



1 = 100 m/s. 2 Time t0 when the planes just escape the collision :





t0 =

500 = 5 . s 100

Q.7. A rocket is moving in a gravity free space with a constant acceleration of 2 ms–2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 ms–1 relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 ms–1­from its right end relative to the rocket. The time in seconds when the two balls hit each other is a = 2ms–2 0.3

...(iv)

(d) Torque with respect to origin (using equations (iii) and (iv)) : 20     10  τ = r × F =  i + 5 j  × ( 2i + j ) = − k Nm 3  3  Q.6. Air-planes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60° with respect to the horizontal respectively as shown in figure. The speed of A is 100 3 ms–1. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = t0, A just escapes being hit by B, calculate the value of t0

ms–1

0.2

ms–1

x

4m

Sol.

Given : Acceleration of rocket, a = +2 m/s2, Initial velocity of ball A, uA = 0.3 m/s, Initial velocity of ball B, uB = –0.2 m/s, Distance travelled by ball A in time t = SA, Distance travelled by ball B in time t = –SB, Length of rocket = 4 m. To find : the time of collision of two balls, t, when SA – (–SB) = 4.

31

KINEMATICS

a = 2ms–2 0.3 ms–1

0.2 ms–1

x

4m



1 2 1 2 SA = uA t + at = 0.3t + ( 2 )t 2 2







SA – (–SB) = SA + SB = 0.3t + 0.2t

1 2 1 2 – SB = uBt + at = −0.2t + ( 2 )t 2 2



Solving for a :









Displacement of boy for an observer on ground





1 2 = 1.15 + at = ut cosθ , 2

= 0.5t = 4; t =

2(ut cosθ − 1.15) t2

= 5 m/s2.

Q.9. A bullet hits a wooden block with an initial velocity of u. It loses half of its velocity within a distance of 5 cm inside the block. How much more the bullet will penetrate within the block before coming to rest? Sol. Given : Initial velocity of bullet = u,

4 = 8 s. 0.5 Q.8. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. Calculate the acceleration in m/s2. Sol. Given : Acceleration of train, a = constant, Initial speed of projectile, u = 10 m/s, Angle of projection, q = 60°, Displacement of boy inside the train to catch the ball at the initial height = 1.15 m. To find : Acceleration of the train. Time of flight of the projectile, 2u sin θ 2 × 10 × sin 60° = = 3 s t = g 10 Displacement of boy for an observer in train = 1.15 m

a =

u . 2 To find : s2, such that velocity of the bullet at s2 is v2, v2 = 0. At s1 = 5 cm, velocity of the bullet, v1 =



Applying









v12 = u2 – 2as1





u2 2 = u − 2 × 5 × a ; 4







Applying





v2 = u2 – 2as at s2





v22 = v12 – 2as2 











v2 = u2 – 2as at s = 5 cm

a =

0 =

3 2 u 40

3 u2 − 2 × u 2 s2 4 40

10 6 = 1.66 cm s2 =



Moti on of a

ca ro n

on est

Opposes impending relative motion FS = µsR

Kinetic Fric tion

A push or pull which changes or tends to change state of rest or of uniform motion of a body.

or µR < µk < µs

Oppose actual rolling motion FR < Fk 45°; sin q > cos q; F2 > F1. So, the block is stationary for q > 45° and frictional force acting towards Q. Similarly, the block is stationary for q < 45° and frictional force acting towards P. Q.4. A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is m. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. What should be the value of fN Sol. Given : Angle of inclination of plane, q = 45°, Coefficient of friction between the block and the incline = m, Force required to push the block up the incline = F1, Force required to just prevent the block from sliding down = F2 and F1 = 3F2, The normal force N = 10m. To find : Value of N.



From diagram (A) : F1 = mgsinq + mmgcosq ...(i) From diagram (B) : F2 = mgsinq – mmgcosq ...(ii) Given : F1 = 3F2; mgsinq + mmgcos q = 3(mgsin q – mmgcos q); 4 mmgcos q = 2mgsin q; 4mcos 45° = 2sin 45°;





m =





N = 10 m = 10 ¥

2 1 = . 4 2 1 =5 2

Q.5. Find the acceleration of mass m in both cases. The pulley arrangement for both figure a and b is identical and the mass of the rope is negligible. (a)

(b)

m

m 6m

F=6mg

Sol. Given : In figure (a) mass m is lifted by attaching another mass 6m at the other end of the rope, In figure (b) mass m is lifted by applying a downward force of 6mg at the other end of the rope. To find : The acceleration of mass m in both the cases. Case a : Total force acting on mass m, Fa = 6mg – mg = 5mg; Total mass to be pulled by this force : ma = 6m + m = 7m Acceleration of mass m : Fa 5mg 5 a = m = 7 m = 7 g . a

43

LAWS OF MOTION



Case b : Total force acting on mass m, Fb = 6mg – mg = 5mg; Total mass to be pulled by this force : mb = m. Acceleration of mass m : Fb 5mg = = 5 g. a = mb m Q.6. Find the angle q for the system of pulleys shown in the diagram below, where M = 3m. Neglect the mass of pulleys and strings and assume they are frictionless.

T

m

F=mg



Balancing the forces acting on mass M :

T

T q

q

q

M m

T

T m

q M

F=mg F=mg

F=mg

Sol. Given :

M = 3 m ...(i) the pulleys and strings are massless and frictionless. To find : Angle q. For the system to be in equilibrium : T = mg ...(ii)

F=mg



Tcos q + Tcos q = Mg; (Put value of M from equation (i) and value of T from equation (ii))



2mgcos q = 3 mg ;



cos q =





3 ; θ = 30∞. 2 q = 30°.





When force and displacement are perpendicular to each other. W = zero when θ = π 2

When force (F) and displacement (s) are in opposite direction. W will be (-ve) when π VQ > VP Q.7. The centres of two stars are separated by a distance of 6r. The masses of these stars are M and 4M and the respective radii are r and 2r. An object of mass m is fired from lighter star towards the heavier star. What is the minimum initial speed required by the object to reach the lighter star. Sol. Given : Distance between two stars = 6r, Mass of start 1 = M, mass of star 2 = 4M, radius of star 1 = r, radius of star 2 = 2r, mass of object fired from surface of star 1 = m.

112 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

To find : Minimum initial speed required by the object to reach star 2.

.

r 1

x1 P x2 6r

2r 2



PHYSICS

The potential energy of the object at point P :

GMm G( 2 M )m GMm G( 2 M )m = ...(v) 2r 4r x1 x2



For the object to cross point P, the minimum required initial kinetic energy will be :



At some distance x1 from star 1 the strength of gravitational field due to both the stars is zero. Let this point be P. P is at a distance x2 from centre of star 2. G( 4 M )m GMm ; x 2 = 2 x1 ...(i) 2 = x 22 x1 Also, x2 + x1 = 6r; x1 = 2r and x2 = 4r ...(ii) Let initial kinetic energy of the object fired from star 1 be : 1 2 Ki = mvi ...(iii) 2 The potential energy of the object at the surface of star 1 : G( 2 M )m GMm – ...(iv) 5r r

Kmin =

1 2 mv 2 min

Ê GMm G( 2 M )m ˆ Ê GMm G( 2 M )m ˆ - = ÁË 2r 4 r ˜¯ ÁË r 5r ˜¯ 2 GM GM GM 2GM vmin + + = 2r 2r r 5r 2

= Vmin =

2GM 5r 4GM 5r

The above equation gives the minimum required initial speed of the object to cross point P. Beyond



this the object will be attracted towards star 2.



Deforming force applied normal Fn  A Area

Force Changing Volume FV  A Area

pe

longitudinal stress F / A  longitudinal strain l / l

F l Mgl  Y   r 2l Al

Y 

Young’s modulus of elasticity

Stress

Property of material by virtue of which it regains its original shape & size after the removal of deforming force

s Modulus Young’

Restoring force per unit area i.e., stress= AF

s

(iii) Shearing area or tangential stress Tangential Force F   t A Area



(ii) Volumetric stress



(i) Longitudinal stress

9 1 3 9 B   or Y  Y B  3B

(iv)

E

3B  2  2  6B



(iii)

and Liquids Part-1

Properties of Solid

Elasticity

Elastic potential energy in a stretched wire(U) 1 = ×stress × strain × volume of the wire 2

Ty

B

P 1 ;  compressibility V B V

hydraulic stress B volume strain

Bulk modulus or volume modulus of elasticity

Bulk Modulus

Types of Modulus of Elasticity

Within the elastic limit, stress is directly proportional to strain. i.e., stress ∝ strain

Hooke’s law

Shearing strain=angular displacement of the plane perpendicular to the fixed surface = 

T

tangential stress shearing strain

s

F F   A  s  A



Rigidity or shear modulus of elasticity

Ratio of change in configuration to original configuration change in configuration Strain  original configuration Modulus of Rigid ity

Strain

(iii)

original volume

change in volume V Volumetric strain   V

change in length l  lo original length

(ii)



Longitudinal strain

(i)

Vo

Laternal strain () d / d  Longitudinal strain ( ) l / l

Value of σ lies between 0 and 0.5

Poision’s ratio(σ) =

es

rgy ne e l tia ten o cP sti a l

yp

Relation between Y, B,  & σ (i) Y=3B (1 − 2 σ) Y=2(1 +  (ii)

PROPERTIES OF SOLIDS AND LIQUIDS

113

2



of continuity

m=a1v11=a2v22 for an incompressible liquid, 1=2 then a1v1=a2v2 or av=constant  Pascal’s law : The pressure exerted at any point on an enclosed liquid is transmitted equally in all direction. Hydraulic brakes and hydraulic lifts are based on Pacal’s law.

 Equation

Fl

Ven turim eter

Applications

ow of

fl u

and Liquids Part-2

Properties of Solid

That can flow like liquids and gases =

solu

su te P res P)

re

(

G a

Total or actual pressure at a point. Absolute pressure= atmospheric pressure + gauge pressure=Pa+hρg

E A

Ab

y

Relative Density or specific gravity=

Difference between the absolute pressure at a point and the atmospheric pressure. ρg=absolute pressure(P) – atmospheric pressure(Pa)

density of substance density of water at 4oC

Density of water at 4°C i.e., maximum density of water=1.0×103 kg/m3

Volume(v)

Density(ρ)= Mass(m)

F tension S= l work done in increasing area W  Surface Energy=  increase in surfacearea A 2S cos  Capillary rise or fall, h= r g  Excess Pressure inside a drop (liquid) 2S Pexcess = R  Excess Pressure inside a bubble (soap) 4S Pexcess = R

Stroke’s law F=6 πηvr

Opposing force between different layers of fluid in relative motion Viscous drag F A dv dx η=coefficient of viscosity

Lift of an aircraft wing. Sprayer or atomizer Blowing off the roofs during windstorm.

 Surface

o

o

o

D e n sit y

(ρg) re u s es Pr e ug

Pressure (atm) exerted by the atmosphere. At sea level, 1 atm=pressure exerted by 0.76m of Hg=hρg=0.76×13.6×103×9.8=1.013×105 Nm-2 =101.3kPa

Pa) re( u ess Pr ir c he

Pressure(P)  thrust(F)  lim A 0

sit

S u rf a c e T e n s io n

V

dF area(A) dA Pressure exerted by a liquid column of height h, (p)=hρg

ids

o isc

For an incompressible, non-viscous, streamline,irrotational flow of fluid, 1 P   v 2   gh  constant 2

s Fluid s of w a L

Streamline : In liquid flow when the velocity is less than critical velocity, each particle of the liquid passing through a point travels along the same path and same velocity as the preceding particles. Turbulent : When velocity of liquid flow is greater than critical velocity and particles follow zig-zag path.

1

2hm g  a 2 – a 2

Bernoulli’s principle

Q  a1a2

Velocity of efflux of liquid through an orifice V  2 gh To ric ell’s Fluids

A t m os p

Law

Device used to measure the rate of flow of liquid. Volume of liquid flowing per second

114 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Properties of Solids and Liquids

Chapter 7 Syllabus

Elastic behaviour, Stress-strain relationship, Hooke’s Law; Young’s modulus, bulk modulus, modulus of rigidity; Pressure due to a fluid column; Pascal’s law and its applications; Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, Reynolds number; Bernoulli’s principle and its applications; Surface energy and surface tension, angle of contact, application of surface tension - drops, bubbles and capillary rise; Heat, temperature, thermal expansion; specific heat capacity, calorimetry; change of state, latent heat; Heat transfer-conduction, convection and radiation, Newton’s law of cooling.

Topic-1

LIST OF TOPICS : Topic-1 : Elasticity

Elasticity

.... P. 115

Topic-2 :  Properties of Liquids and Calorimetry

.... P. 120

Concept Revision (Video Based) Stress and Strain

Young’s Modulus

          Part -1       Part -2

 

         

       Part -1       Part -2

Bulk modulus

                 Part -1         Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. A cube of metal is subjected to a hydrostatic pressure 4 GPa. The percentage change in the length of the side of the cube is close to : (Given bulk modulus of metal, B = 8 × 1010 Pa) (a) 1.67 (b) 5 (c) 20 (d) 0.6 [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.2. Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms–1. If the Young’s modulus of wire is 16 × 1011 Nm–2, the extension of wire over its natural length is : (a) 0.03 mm (b) 0.04 mm (c) 0.02 mm (d) 0.01 mm [JEE (Main) – 7th Jan. 2020 - Shift-1]

Q.3. A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 104 N. When the tension is changed to T, the velocity v changed to . The value of T is close to : 2 (a) 10.2 × 102 N (b) 5.15 × 103 N 4 (c) 2.50 × 10 N (d) 30.5 × 104 N [JEE (Main) – 8th Jan. 2020 - Shift-2] Q.4. Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is : (a) 1 : 2 (c) 2 :1

(b) 2 : 1 (d) 1 : 2 [JEE (Main) – 9th Jan. 2020 - Shift-2]

116 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.5. A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1p ms–2, what will be the tensile stress that would be developed in the wire? (a) 6.2 × 106 N m–2 (b) 5.2 × 106 N m–2 6 –2 (c) 3.1 × 10 N m (c) 4.8 × 106 N m–2 [JEE (Main) – 8th April 2019 - Shift-1] Q.6. A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms–1. Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to : (a) 106 Nm–2 (b) 104 Nm–2 8 –2 (c) 10 Nm (d) 103 Nm–2 [JEE (Main) – 8th April 2019 - Shift-1] Q.7. Young’s moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to : (a) 1.5 mm (b) 1.9 mm (c) 1.7 mm (d) 1.3 mm [JEE (Main) – 8th April 2019 - Shift-2] Q.8. In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young’s Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2] (a) 1.2 × 106 N/m2 (b) 4.0 × 106 N/m2 6 2 (c) 1.8 × 10 N/m (d) 8 × 106 N/m2 [JEE (Main) – 10th April 2019 - Shift-2] Q.9. The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit? (a) 1.00 mm (b) 1.16 mm (c) 0.90 mm (d) 1.36 mm [JEE (Main) – 10th April 2019 - Shift-2] Q.10. At 40°C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40°C to 20°C it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young’s modulus of brass are 10–5/°C and 1011 N/m2, respectively; g = 10 ms–2) (a) 9 kg (b) 0.5 kg (c) 1.5 kg (d) 0.9 kg [JEE (Main) – 12th April 2019 - Shift-1]

PHYSICS

Q.11. A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to : (a) (c)

9F ( pr 2 YT) 3F 2

( pr YT)



(b)



(d)

6F ( pr 2 YT) F ( 3 pr 2 YT)

[JEE (Main) – 12th April 2019 - Shift-2]



Q.12. A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion a/°C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by DT K. Young’s modulus, Y, for this metal is : F A a( D T - 273)

(a)

F A pD T

(b)

(c)

F 2A aD T

2F (d) A aD T [JEE (Main) – 9th Jan. 2019 - Shift-1]



Q.13. A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8.

The new value of increase in length of the steel wire is : (a) 3.0 mm

(b) 4.0 mm

(c) 5.0 mm

(d) Zero [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY 1. (a) 5. (c)

2. (b) 6. (a)

3. (b) 7. (c)

4. (b) 8. (d)

9. (b)

10. (a)

11. (c)

12. (a)

13. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Given : Bulk modulus of the material of the cube is B = 8 × 1010 Pa, pressure on the cube is ΔP = 4 GPa. Dl × 100, the percentage change in the l length of the cube. To find :

117

PROPERTIES OF SOLIDS AND LIQUIDS

By definition of bulk modulus : ∆P B = ∆V V

v2 = 

v = v 2

Percentage change in length of cube : 1 ∆V ∆l × 100 = × × 100 = 1.67 l 3 V

T =

Ans. 2. Option (a) is correct. Given : Speed of transverse wave on a straight wire is v = 90 m/s, mass of wire is m = 6.0 g, length of wire is l = 60 cm, area of cross section of wire is A = 1.0 mm2, Young’s modulus for the wire is Y = 16 × 1011 N m 2 . To find : Dl, the extension of wire over its natural length. Velocity of transverse wave on a straight wire : T m

Ans. 4. Option (b) is correct. Given : Length of two steel wires is l1 = l2 = l, load on both wires is m1 = m2 = m, ratio of energy stored per unit volume for the two wires is E1 : E2 = 1 : 4.

...(i)

To find : d1 : d2, the ratio of diameters of the two wires. Energy stored per unit volume in wire 1 : 1 ( stress)2 1 F2 = Y 2 2 A12 Y

E1 =

1 ( stress)2 1 F2 = Y 2 2 A 22 Y

E2 =

...(ii)

...(iii)

2

T v 2m stress A v 2m Y = = = lA = ∆l strain ∆l A∆l l l

E1 E2

v2m ( 90 )2 × 0.006 = = 3 × 10 −5 m AY 1 × 10 −6 × 16 × 1011 = 0.03 mm

∆l =

Ans. 3. Option (b) is correct. Given : Case a : Velocity of wire on taut steel wire is v1 = v, the tension in wire is T1 = 2.06 × 104 N, Case b : Velocity of wire on the same taut steel wire v is v2 = , the tension in wire is T2 = T. 2 To find : Value of T. T m = ,m is mass per unit m l length of the wire which will be same for both cases a and b) : = (v Using relation

v1 =

2.06 × 10 4 = 5.15 × 10 3 N 4

As l1 = l2 = l and m1 = m2 = m, F and Y will be same for both wires. From equations (ii) and (iii) :

v 2m l

Young’s modulus :

T1 m

2.06 × 10 4 T

Energy stored per unit volume in wire 2 :

m (m = is mass per unit length of the wire and T is l the tension in the wire.) 2 v= m

T1 T2

v1 = v2

Percentage change in volume of cube : ∆V × 100 = 5 V

= T

...(ii)

From equations (i) and (ii) :

∆V ∆P 4 × 10 9 = = = 5 × 10 −2 V B 8 × 1010

v =

T2 m

...(i)

 d2  p 2   4  A2 d4 = 22 =   2 = 24 A1 d1  d2  p 1   4   

...(iv)

From equations (i) and (iv) : d24 d14

=

d1 : d2 =

1 4 2 :1

Ans. 5. Option (c) is correct. Given : Radius of steel wire is r = 2.0 mm = 0.002 m, load on steel wire is m = 4 kg, acceleration m due to gravity is g = 3.1 p 2 . s To find : Tensile stress in wire.

Tensile stress =

=

mg Force = 2 Area pr 4 ¥ 3.1p p ¥ 0.002 ¥ 0.002

= 3.1 × 106 N/m2

118 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 6. Option (a) is correct. Given : Length of the rubber cord of the catapult is L = 42 cm, mass of cord is negligible, diameter of cross section of cord is D = 6 mm = 0.006 m, mass of stone is m = 0.02 kg, The boy pulls the catapult by a length = DL = 20 cm, the initial velocity of the stone when it leaves the catapult is v = 20 m/s. To find : The Young’s modulus of the rubber cord. Potential energy stored by the stretched rubber cord : 1 ( DL) Ê D ˆ 1 Ê DL ˆ p Á ˜ ...(i) YÁ AL  = Y ˜ Ë 2¯ L 2 2 Ë L¯ 2

2



2

In equation (i), A is the area of cross section of the cord with diameter D. The potential energy stored in the stretched cord is given to the stone as kinetic energy. Kinetic energy of stone 1 2 = mv ...(ii) 2 As energy is always conserved, equate equations (i) and (ii), 2 1 2 1 ( DL) Ê D ˆ p Á ˜ = mv Y 2 Ë ¯ L 2 2



=

1 × 0.02 × 20 × 20 2

Y ª 106 N/m2 Ans. 7. Option (c) is correct. Given : Length of wire A is l1 = 2m, radius of wire A is r1 = R, Length of wire B is l2 = 1.5 m, radius of wire B is r2 = 2 mm, Young moduli of the two wires A and B are in ratio, Y1 7 = . Y2 4 To find : R if the two wires stretch by the same length (Dl1 = Dl2) ...(i) for a given load (F1 = F2 = F) ...(ii) Change in length of wire A with load, F1 : F1l1 Dl1 = ...(iii) p r12 Y1 Change in length of wire B with load F2 : F2 l2 Dl2 = ...(iv) p r22 Y2 From equations (i) and (ii), l2 l1 = 2 2 r2 Y2 r1 Y1 Put given values in the equation above.

=

1.5 ; 0.002 ¥ 0.002 ¥ Y2

R = 17 mm

Ans. 8. Option (d) is correct. Given : Length of brass wire = length of steel wire = l = 1 m, Area of cross section of brass wire = area of cross section of steel wire = A = 1 mm2, Young’s modulus for steel wire is YS = 120 × 109 N/m2, Young’s modulus for brass wire is YB = 60 × 109 N/m2. To find : Stress required to produce a net elongation of, Dl = 0.2 mm. Dl = DlS + DlB 1 ˆ Fl Ê 1 + ; = Á A Ë YS YB ˜¯

2

Put given values in the above equation. 1 0.2 ¥ 0.2 ¥ p ¥ 0.006 ¥ 0.006 Y 2 0.42 ¥ 4

2 R 2 Y1

PHYSICS

Dl F = Ê A Ê 1 1 Á l ÁË Y + Y Ë S



B

ˆˆ ˜¯ ˜¯

F = 8 × 106 N/m2. A

Ans. 9. Option (b) is correct. Given : Elastic limit of brass is s = 379 MPa. To find : Minimum diameter of brass rod, dmin for load, F = 400 N. The stress in brass rod is given as : F s = A =

F Êd ˆ p Á min ˜ Ë 2 ¯

2

=

4F 2 p dmin

Put given values in the above equation. 4 ¥ 400 ; 379 × 106 = 2 ( 3.14 ¥ dmin )

dmin = 1.16 mm

Ans. 10. Option (a) is correct. Given : Radius of brass wire is r = 1 mm, the wire when cooled from 40°C to 20°C regains its original length of, l = 0.2 m, Young’s modulus for brass is Y = 1011 N/m2, coefficient of linear expansion for brass is a = 10–5/°C, acceleration due to gravity is g = 10 m/s2 To find : Closest value of M, a small mass hung from the wire. Young’s modulus :

Y =

Mgl ADl

(A is the area of cross section of the brass wire)

119

PROPERTIES OF SOLIDS AND LIQUIDS



Y =

Mg Ê l ˆ  ...(i) A ÁË Dl ˜¯

Expansion in wire due to rise in its temperature from 20°C to 40°C will be : Dl = a D T l Here DT = 40° – 20° = 20°C That gives, Dl = 20a ...(ii) l Put value of

Dl from equation (ii) in equation (i), l



M =

Mg Ê 1 ˆ  ; A ÁË 20a˜¯



Y =

20aYA g

Put given values in the above equation :

M =

20 ¥ 10

-5

¥ 10

11

¥ p ¥ 0.001 ¥ 0.001 10

= 6.28 kg. The answer is closest to option a, 9 kg. Ans. 11. Option (c) is correct. Given : Length of cylindrical rod is L, radius of cylindrical rod is r, Young’s modulus of elasticity for the rod is Y, when the rod is heated by temperature T (DT = T) and simultaneously subjected to a net longitudinal compressional force F, change in its length is Dl = 0. To find : Coefficient of volume expansion of the rod. Strain in the rod due to compressional force F : F Dl Fl ˆ Ê = ...(i) ÁË as,Y = ADl ˜¯ YA l Thermal strain in rod : Dl = a D T = a T ...(ii) l (a is coefficient of linear expansion) As net elongation of the rod due to the simultaneous compressional force F and the thermal strain is 0, F F . = a T ; a = YTpr 2 YA Hence, coefficient of volume expansion of the rod will be : 3F g = 3a = YTpr 2 Ans. 12. Option (a) is correct. Given : Length of rod at room temperature is L, area of cross section of rod is A, coefficient of linear

expansion for rod is a/°C, when the temperature of rod is increased by DT and simultaneously it is also subjected to a net compressional force F, change in its length Dl = 0. To find : Young’s modulus for the rod. Strain in the rod due to compressional force F : F Dl Fl ˆ Ê = ...(i) ÁË as,Y = ADl ˜¯ YA l Thermal strain in rod : Dl = a DT ...(ii) l (a is coefficient of linear expansion) As net elongation of the rod due to the simultaneous compressional force F and the thermal strain is 0, F F = a DT ; Y = YA AaDT Ans. 13. Option (a) is correct. Given : Mass of the load is M kg, length of steel wire is l = 2 m, radius of steel wire is r = 1.0 mm, increase in length of wire due to load M is Dl = 4.0 mm, Relative density of a liquid is rl = 2, relative density of load is rM = 8. To find : Dll, the new elongation of steel wire when the load is immersed in the liquid. Increase in length of steel wire due to load M is

Dl =

Mgl Mgl = ...(i) AY p r 2 Y

Here, Y is Young’s modulus for the steel wire and g is acceleration due to gravity. When the load is fully immersed in the liquid :

Dll =

(Mg - FB ) l pr 2 Y

...(ii)

Here FB is buoyant force due to the liquid. FB = Mg

rl ...(iii) rM

Put value of buoyant force from equation (iii) in equation (ii)



Dll =

Ê r ˆ Mgl Á 1 - l ˜ rM ¯ Ë pr 2 Y

...(iv)

Divide equation (iv) by equation (i),

Ê rl ˆ Ê 2ˆ 3 Dll = Á 1 - ˜ = ; = Á 1 ˜ Ë 8¯ 4 rM ¯ Ë Dl Dll =

3 3 ¥ Dl = ¥ 4.0 mm = 3.0 mm 4 4

120 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Topic-2

Properties of Liquids and Calorimetry Concept Revision (Video Based) Pascal’s Law

Viscosity

         Part -1       Part -2

  

Bernoulli’s Principle

      Part -1          

 

Surface Tension

Part -2

 

                      Part -1    

Specific Heat

 

Part-1



Surface Tension by Capillary Rise Method



Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is w rad s–1. The difference in the height, h (in cm) of liquid at the centre of vessel and at the will be : 

h

10 cm

(a)

5ω 2 2g

(b)

2ω 2 25 g

(c)

25ω 2 2g

(d)

2ω 2 5g

[JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = 0.05 Nm–1, density = 667 kg m–3) which rises to height h in the tube. It is observed that the two tangents drawn from observed that the two tangents drawn from liquidglass interfaces (from opp. sides of the capillary) make an angle of 60° with one another. Then h is close to (g = 10 ms–2)

(b) 0.137 m (d) 0.049 m [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is : (a) 4 : 1 (b) 2 : 1 (c) 0.8 : 1 (d) 8 : 1 [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.4. A metallic sphere cools from 50°C to 40°C in 300 s. If atmospheric temperature around is 20°C, then the sphere’s temperature after the nest 5 minutes will be close to : (a) 33°C (b) 31°C (c) 35°C (d) 28°C [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.5. A air bubble of radius 1 cm in water has an upward acceleration of 9.8 cms–2. The density of water is 1 gm cm–3 and water offers negligible drag force on the bubble. The mass of the bubble is (g = 980 cm/s2). (a) 4.15 gm (b) 1.52 gm (c) 4.51 gm (d) 3.15 gm [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.6. Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is : (a) 0.087 m (c) 0.172 m

121

PROPERTIES OF SOLIDS AND LIQUIDS

(a)

3 gdS( x 2 − x1 )2 4

(b)

1 gdS( x 2 − x1 )2 4

(c) gdS( x 2 + x1 )2 (d) gdS( x 22 + x12 )2 [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.7. A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of 27 the shell material is with respect to water, the 8 value of r is : 1 (a) R (b) 4 R 3 9 8 2 (c) R (d) R 9 3 [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.8. In an experiment to verify Stokes law, a small spherical ball of radius r and density r falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to : (ignore viscosity of air) (a) r2 (b) r (c) r3 (d) r4 [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.9. A fluid is flowing through a horizontal pipe of varying cross-section, with v ms–1 at a point where the pressure is P Pascal. At another point where P pressure Pascal its speed is V ms–1. If the density 2 of the fluid is r kg-m–3 and the flow is streamline, then V is equal to : P P +υ2 +υ2 (a) (b) 2ρ ρ (c)

2P +υ2 ρ

(d)

P +υ ρ

[JEE (Main) – 6th Sep. 2020 - Shift-2] Q.10. An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is : 3 3 (a) (b) 4 2 9 81 (c) (d) 16 256 [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.11. Consider a solid sphere of radius R and mass  r2  density ρ ( r ) = ρ 0  1 − 2  , 0 < r £ R. The R   minimum density of a liquid in which it will float is : ρ0 2 ρ0 (a) (b) 5 5 ρ0 2 ρ0 (c) (d) 3 3 [JEE (Main) – 8th Jan. 2020 - Shift-1]

Q.12. A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to : (a) 1.03 (b) 1.01 (c) 1.26 (d) 1.04 [JEE (Main) – 8th Jan. 2020 - Shift-1] Q.13. M 5m N 5m O



Two liquids of densities r1 and r2 (r2 = 2r1) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing) : 2 1 (a) (b) 3 2 (c)

1 4

(d)

1 3

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.14. Water flows in a horizontal tube (see figure). The pressure of water changes by 700 Nm–2 between A and B where the area of cross section are 40 cm2 and 20 cm2, respectively. Find the rate of flow of water through the tube. (density of water = 1000 kg m–3)

. A

.

B

(a) 3020 cm3/s (c) 2720 cm3/s

(b) 2420 cm3/s (d) 1810 cm3/s [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.15. A small spherical droplet of density d is floating exactly half immersed in a liquid of density r and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet) : (a) r =

2T 3( d + ρ ) g

(b) r =

T (d + ρ ) g

(c) r =

T (d − ρ ) g

(d) r =

3T ( 2d − ρ ) g

[JEE (Main) – 9th Jan. 2020 - Shift-2]

122 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.16. Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of  : (density of water = 1000 kg/m3, coefficient of viscosity of water=1 mPa s) (a) 103 (b) 104 2 (c) 10 (d) 106 [JEE (Main) – 8th April 2019 - Shift-1] Q.17. Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8 × 102 kg/m3 and specific heat of 2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3 and specific heat of 4000 J kg–1 K–1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same) (a)

(b)

60°C T

A

60°C T

B

(c)

A

60°C T

A and B

t t [JEE (Main) – 8th April 2019 - Shift-1] Q.18. If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ’2r’ is : M (a) M (b) 2

(c) 4 M (d) 2 M [JEE (Main) – 9th April 2019 - Shift-1] Q.19. A wooden block floating in a bucket of water has 4 of its volume submerged. When certain amount 5 of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is : (a) 0.5 (b) 0.8 (c) 0.6 (d) 0.7 [JEE (Main) – 9th April 2019 - Shift-2] Q.20. The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius 12. The Êr ˆ ratio, Á 1 ˜ , is then close to : Ë r2 ¯

(c)

3 5

2 (b) 5 (d)

[Take, density of water = 103 kg/m3] (a) 46.3 kg

(b) 87.5 kg

(c) 64.5 kg

(d) 30.1 kg [JEE (Main) – 10th April 2019 - Shift-2]



Q.22. A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 – d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms –2) : (a) 300 m

(b) 400 m

(c) 600 m

B

4 (a) 5



2 3

[JEE (Main) – 10th April 2019 - Shift-1]

(d) 500 m [JEE (Main) – 10th April 2019 - Shift-2]

t

(d)

60°C T

Q.21. A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water?

B A

t

PHYSICS

Q.23. Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The crosssectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The crosssectional area of the stream, 0.15 m below the tap would be : (Take g = 10 ms–2) (a) 2 ×10–5 m2

(b) 5 × 10–5 m2

(c) 5 × 10–4 m2

(d) 1 × 10–5 m2

[JEE (Main) – 10th April 2019 - Shift-2] Q.24. When M, gram of ice at –10°C (specific heat = 0.5 cal g–1 °C–1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is : (a)

50 M 2 -5 M1

(b)

5 M1 - 50 M2

(c)

50 M 2 M1

(d)

5M 2 -5 M1

[JEE (Main) – 12th April 2019 - Shift-1]



Q.25. A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity h. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio Ê v1 ˆ ÁË v ˜¯ equals : 2

(a) 9 (c)

1 9

(b)

1 27

(d) 27 [JEE (Main) – 12th April 2019 - Shift-2]

123

PROPERTIES OF SOLIDS AND LIQUIDS

Q.26. One kg of water, at 20°C, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 W. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to : [Specific heat of water = 4200 J/(kg °C), Latent heat of water = 2260 kJ/kg] (a) 16 minutes (b) 22 minutes (c) 3 minutes (d) 10 minutes [JEE (Main) – 12th April 2019 - Shift-2] Q.27. The top of a water tank is open to air and its water lavel is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to : (a) 6.0 m (b) 4.8 m (c) 9.6 m (d) 2.9 m [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.28. Water flows into a large tank with flat bottom at the rate of 10–4 m3s–1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is : (a) 5.1 cm (b) 1.7 cm (c) 4 cm (d) 2.9 cm [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.29. A liquid of density r is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be : 3 2 (a) rv (b) rv2 4 (c)

1 2 rv 2

(d)

Q.32. A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by : (a)

(b) P

P

log(t)

1 t

(c)

(d) P

P t

1 3 t

[JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (c)

2. (a)

3. (d)

4. (a)



5. (a)



9. (b)

6. (b)

7. (d)

8. (d)

10. (d)

11. (a)

12. (b)



13. (c)

14. (c)

15. (d)

16. (b)



17. (b)

18. (d)

19. (c)

20. (b)



21. (b)

22. (a)

23. (b)

24. (a)



25. (a)

26. (b)

27. (b)

28. (a)



29. (a)

30. (c)

31. (a)

32. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : Radius of a cylindrical vessel is R = 5 cm, the angular speed of the vessel is w rad/s. 

1 2 rv 4

[JEE (Main) – 11th Jan. 2019 - Shift-1] Q.30. When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be : (a) 85°C (b) 60°C (c) 80°C (d) 70°C [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.31. A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be : (a) 2.0 (b) 0.1 (c) 0.4 (d) 1.2 [JEE (Main) – 12th Jan. 2019 - Shift-2]

 dh

h dr

10 cm

To find : h, the difference in height of the liquid at the periphery of the vessel and at the centre. Let r be the density of the liquid. Balancing the pressure on the liquid column due to weight and due to rotational motion : ( ρω 2 r )dr = ρ gdh R

∫0 (ω

2

h

r ) dr = g ∫ dh 0

h =

ω 2 R 2 25ω 2 = 2g 2g

124 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 2. Option (a) is correct. Given : Radius of the capillary tube is r = 0.15 mm, the capillary is dipped in to the methylene iodide kept in the beaker, the solution rises to height h within the capillary, the two tangents drawn from liquid-glass interface make 2q = 60° angle with one another, surface tension of methylene iodide is T = 0.05 N/m, density of methylene iodide is r = 667 kg/m3. To find : The value of h.

PHYSICS

Ans. 4. Option (a) is correct. Given : Initial temperature of metallic sphere is Ti = 50°C, temperature of the sphere after t¢ = 300 s = 5 minutes is T¢ = 40°C, temperature of surrounding is T = 20°C. To find : Tf, the temperature of the sphere after next t = 5 minutes. By Newton’s law of cooling : Ti − T′  T + T′  = K i − T t′  2  After putting the given values in the above equation :

30°

K =

60°

2 25

Also, The height of the liquid column within the capillary: 2T cosθ h = ρ gr From construction, q = 30°. 2 × 0.05 × cos 30° h = = 0.087 m 667 × 10 × 0.15 × 10 −3

4T R1

4T 0.01 = R1

0.02 =

...( i )

R13

=

V1 8 = V2 1

40 − Tf =

 2  40 + Tf − 20   25  2  Tf 5

Tf ≈ 33°C  Ans. 5. Option (a) is correct. Given : Radius of an air bubble in water is r = 1 cm, the upward acceleration of the bubble cm 2 , , density of water is in water is a = 9.8  cm/s s2 ρ = 1 g cm 3 ,g = 980 cm s2 . To find : m, the mass of the air bubble. Fb =

4 3 pr ρ g 3

Fb − mg = ma 4 3 pr ρ g = m( g + a ) 3

4T R2

4T R2

1 8

=

Balancing the forces acting on the bubble :

From equations (i) and (ii) : R 1 = 2 2 R1 R 32

40 − Tf

  T′ + Tf = K  − T    2

The buoyant force acting on the bubble :

Pressure inside soap bubble 2 : 4T P2 = Po + R2 1.02 = 1 +

t

5

Ans. 3. Option (d) is correct. Given : Pressure inside soap bubble 1 is P1 = 1.01 atm, pressure inside soap bubble 2 is P2 = 1.02 atm. To find : V1 : V2, the ratio of the volume of soap bubble 1 to the volume of soap bubble 2. Let R1 and R2 be the radii of soap bubble 1 and 2 respectively and Po be the atmospheric pressure. Pressure inside soap bubble 1 : 4T P1 = Po + R1 1.01 = 1 +

T′ − Tf

...(ii)

4 3 pr ρ g = 4.15 g m = 3 g+a Ans. 6. Option (b) is correct. Given : Density of liquid filled in two identical cylindrical vessels is d area of the base of both the vessels is S, height of liquid in vessel 1 is x1, height of liquid in vessel 2 is x2, both the cylinders are connected through a pipe at the bottom and a new equilibrium height is attained. To find : DU, the change in energy of the system in this process.

125

PROPERTIES OF SOLIDS AND LIQUIDS

∆U = dSx1 g ×

x1 x  x + x2  + dS x 2 g × 2 − 2 dS  1 g 2 2  2   x + x2  × 1   4 

  x2 x2 1 ∆U = dSg  1 + 2 − ( x12 + x 22 + 2 x1x 2 )    2 2 4   dSg = ( x1 − x 2 ) 2 4 Ans. 7. Option (d) is correct. Given : Outer radius of a hollow spherical shell is R, inner radius of the shell is r, the shell floats just submerged under water surface, specific gravity of the shell material is 27 . ρs = 8 To find : Value of r. Let r be the density of the shell material and rw be the density of water. Then, 27 ρ = . 8 ρw As the shell floats in equilibrium : mg = FB

P 1 2 1 + ρ v = ρ V2 2 2 2 V =

P + v2 ρ

Ans. 10. Option (d) is correct. Given : An ideal fluid maintains a laminar flow through a pipe, maximum diameter of the pipe is d1 = 6.4 cm, minimum diameter of the pipe is d2 = 4.8 cm. v To find : 1 , the ratio of minimum to maximum v2 velocity of fluid through the pipe. Using equation of continuity : (A1, A2 is area of cross section of pipe corresponding to d1, d2 respectively.) p d12 p d22 v1 = v2 4 4

 r 3  27 =1  1 − 3  × R  8 

v1 = v2

 r3  8  1 − 3  = 27 R   r = 0.889 R ≈

8 R 9

Ans. 8. Option (d) is correct. Given : Radius of ball is r, density of ball is r, the ball falls through a height h in air before entering a water tank. To find : The dependence of h on r, if the terminal velocity of the ball inside the water is the same as the velocity with which it hits the water surface. The velocity with which the ball hits the water surface after falling a height h in air : 2 gh

...(i)

2 r 2 ( ρ − σ )g 9 η

d22 ( 4.8 )2 9 = = d12 ( 6.4 )2 16

Ans. 11. Option (a) is correct. Given : Radius of solid sphere is R, mass density of sphere is  r2  ρ (r ) = ρo  1 − 2  ,  R   0 ≤ r ≤ R.

...( i )

To find : r¢, minimum density of liquid in which the sphere will float. For the sphere to float the force of buoyancy due to the liquid should balance the weight of the sphere. FB = mg

...( i )

m is mass of the sphere. If we consider a spherical ring of radius r and thickness dr, within the given solid sphere, its mass will be :

Let s be density of water. Terminal velocity of the ball inside water : v =

To find : V As the flow is stream-lined in nature : 1 P 1 P + ρ v2 = + ρ V2 2 2 2

A1v1 = A 2 v2

4 4 p ( R3 − r 3 )ρ g = p R3 ρw g 3 3

v =

flow rate at point A is v m/s, the pressure at point A is P pascal, the flow rate at point B is V m/s, the P pressure at point B is pascal. 2

...(ii)

From equations (i) and (ii) : h ∝ r4 Ans. 9. Option (b) is correct. Given : Density of fluid is r kg/m3, the flow through a horizontal pipeline is streamlined in nature, the

dm = ρ ( r )4p r 2 dr. That gives, total mass of the sphere to be : m =

R

∫0

ρ ( r )4p r 2 dr

From equation (i) : R 4 ρ ′ p R 3 g = ∫ ρ ( r )4p r 2 dr × g 0 3

126 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PN = ρ1 g × ( MN ) = ρ1 g × 5

 r2  ρ o  1 − 2  r 2 dr  R   1 R  R  = ρ o  ∫ r 2 dr − 2 ∫ r 2 r 4 dr  R 0  0  1 1   = ρoR3  −  3 5

ρ ′R 3 = 3

R

∫0

...(i)

So, net force due to liquid 1 on wall MN will be : 5 ρ1 g  P + PN  FMN =  M A = 2 A 2  

...(ii)

A is area of the wall MN. Pressure at point O due to liquid 1 and 2 is :

ρ′ 2 = ρo × 3 15 ρ′ =

PHYSICS

PO = 5 ρ1 g + ρ 2 g × ( NO) = 5 ρ1 g + 2 ρ1 g × 5 = 15 ρ1 g 

2 ρo 5

Ans. 12. Option (b) is correct. Given : Length of the cylinder is l = 1 m, the coefficient of expansion for cylinder is very low, the cylinder is floating on water. Case a : Temperature of water is T1 = 0°C, height of cylinder above water surface is h1 = 20 cm. Case b : temperature of water is T2 = 4°C, height of cylinder above water surface is h2 = 21 cm. ρ To find : 4 , density of water at T2 = 4°C relative ρo to density of water at T1 = 0°C. In both the cases the force of buoyancy due to water is balancing the weight of cylinder (mg) Case a : mg = ρ o A(l − h1 ) ...(i)

So, net force due to liquid 1 and 2 on wall NO will be : PN + PO 5 ρ g + 15 ρ1 g A= 1 A 2 2 20 ρ1 g = A 2

FNO =

...(iii)

From equations (ii) and (iii) : FMN = FNO

5 1 = 20 4

Ans. 14. Option (c) is correct. Given : A horizontal tube as shown below,

. A

.

B

Case b : mg = ρ 4 A(l − h2 )

...(ii)

In the above equations : r0 is density of water at 0°C, r4 is density of water at 4°C, A is area of cross section of cylinder and (l – h1) is the length of cylinder below water surface for case a. Equating equations (i) and (ii) : ρ o (l − h1 ) = ρ 4 (l − h2 )

ρ4 (l − h1 ) 80 = = = 1.01 ρo (l − h2 ) 79 Ans. 13. Option (c) is correct. Given : Edge length of square wall is a = 10 m, liquid 1 fills the MN part of the square wall, density of liquid 1 is r1, liquid 2 fills the NO part of the square wall, density of liquid 2 is r2 = 2r1, height MN = NO = 5 m. F To find : MN , the ratio of force due to liquids FNO exerted on MN to that on NO part of the wall.

Difference in pressure of water at A and B is N 2 PA − PB = 700 N/m , , area of cross section of tube at m2 A is AA = 40 cm2, area of cross section of tube at B kg 3 . . is AB = 20 cm2, density of water is ρ = 1000 kg/m m3 To find : The rate of flow of water through the tube. By equation of continuity : v A A A = vB A B (vA, vB are the velocities of water at point A and B of the tube) vA × 40 = vB × 20 vB = 2 vA By Bernoulli’s equation: PA +

Pressure at point M due to liquid 1 is PM = 0. Pressure at point N due to liquid 1 is :

...(i)

1 2 1 ρ vA = PB + ρ vB2 2 2

PA − PB =

1 ρ ( vB2 − vA2 ) 2

PA − PB =

1 ρ ( 4 vA2 − vA2 ) 2

PA − PB =

3 2 ρ vA 2

127

PROPERTIES OF SOLIDS AND LIQUIDS

vA =

 2 ( PA − PB )  2 × 700   = ρ 3 3 × 1000   7 m s 15

=

= ...(ii)

Rate of flow of water through the tube : 7 R = vA A A = × 40 × 100 cm 3 s 15 = 2733 cm 3 s Ans. 15. Option (d) is correct. Given : Density of spherical droplet is d density of liquid in which the spherical droplet is half immersed is r, surface tension of liquid is T. To find : r, the radius of the droplet. At equilibrium the forces acting on the droplet will balance as : weight = force due to buoyancy + the upward surface tension mg = FB + FT 4 14  d × p r 3 × g = ρ ×  p r 3  × g + T × ( 2p r ) 3 23  4 2 ρ r d −  g = 2T 3  2 r2 =

3T ρ  2d −  g 2 

r =

3T ( 2d − ρ ) g

Ans. 16. Option (b) is correct. Given : Radius of pipe is r = 5 cm, rate of flow of water coming out of the pipe is 100 litres litres V = 100 = , minute 60 second t density of water is kg , m3 coefficient of viscosity of water is h = 1 mPa s. To find : Order of magnitude of Reynolds number for the flow. Reynolds number for flow of a liquid : r vD Re = h

r = 1000

V ( v = t is velocity of flow, A is area of cross section A of pipe and D = 2r is diameter of pipe.) 2 rVr Re = tAh =

2 rVr tpr 2 h

=

2 rV r pt h 2 ¥ 1000 ¥ 100 ¥ 10 -3 0.05 ¥ 3.14 ¥ 60 ¥ 1 ¥ 10 -3

= 2 × 104

Ans. 17. Option (b) is correct. Given : Initial temperature of both the beakers A and B is T = 60°C Volume of liquids in both the beakers is same, V = VA = VB Density of liquid in beaker A is kg 3 rA = 8 ¥ 10 2 kg/m m3 Density of liquid in beaker B is kg 3 , , rB = 10 3 kg/m m3 Specific heat of liquid in beaker A is J SA = 2000 J kg–1, K–1, kg.K Specific heat of liquid in beaker B is J . SB = 4000 kg.K To find : Choose the correct cooling graph for both the liquids. Newton’s law of cooling : h dQ ( T - To ) = ms dt =

h ( T - To ) Vr s

dQ is rate of cooling, h is dt heat transfer coefficient, m = Vr is mass of liquid, s is specific heat of liquid, T is temperature of liquid, T0 is temperature of surrounding, V is volume and r is density of the liquid. As for both the liquids in beaker A and B, h, V, T and T0 are constants : dQ A 1 dQ B 1 ...(i) µ ; µ dt rA sA dt r B sB In the above equation :

Put given values in equation (i), dQ A 1 ; µ dt 800 ¥ 2000 dQ A µ 6.25 ¥ 10 -7 ...(ii) dt From equation (ii) and (iii) we can see : dQ dQ B  A > dt dt for the entire cooling process. So, graph of B will always lie above graph of A as shown in option b. Ans. 18. Option (d) is correct. Given : Radius of capillary 1 is r1 = r, radius of capillary 2 is r2 = 2r, mass of water that rises in capillary 1 is m1 = M.

128 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

To find : The mass of water that rises in capillary 2, m2. Height of liquid that rises in a capillary tube : 2T cosqc h = ...(i) r rg

For case b : Weight of block of volume V = weight of

(T is surface tension of liquid and qC is the contact angle) As we are using same liquid in capillary 1 and capillary 2 : 1 h µ ...(ii) r So, for capillary 1 : 1 h1 µ r1

(ro is density of oil) From equation (i), rw ro 4 r = + ; 5 w 2 2

and for capillary 2 : 1 h2 µ . r2 That gives h r1 2r h1 = h1 = 1 ...(iii) r2 r 2 Mass of water that has risen in capillary 1 : h2 =

m1 = M = r V1 = r p r12 h1 = r p r 2 h1 ...(iv) V1 is volume of water that has risen in capillary 1. Mass of water that has risen in capillary 2, using equation (iii), will be : m2 = r V2 = r p r22 h2

r p ( 2 r )2

h1 = 2 r p r 2 h1 ...(v) 2

Divide equation (v) by equation (iv). m2 = 2 ; m2 = 2 M M Ans. 19. Option (c) is correct. Given : Case a : fractional volume of block submerged in 4 water is V, 5 Case b : Fractional volume of block submerged in V and fractional volume of block submerged 2 V in water is . 2 To find : The density of oil relative to that of water. By Archimedes principle, Weight of the block = weight of the water displaced For case a : Weight of block of volume 4 V = Weight of water of volume V. 5 oil is

rb Vg = rw

4 Vg  5

(rb is density of block and rw is density of water) rb 4 ...(i) = rw 5

oil of volume

V V + weight of water of volume . 2 2

r b Vg = r w

V V g + r o g 2 2

ro 3 r = ; 10 w 2 3 ro = rw 5 ro = 0.6 rw Ans. 20. Option (b) is correct. Given : TM = 7.5, TW TM is surface tension of mercury, TW is surface tension of water, rM = 13.6 ,r M rW is density of mercury and rM is density of water, Contact angle of mercury with glass is qM = 135°, Contact angle of water with glass is qM = 0°, Depression of mercury in a capillary tube of radius r1 is h, Rise of water in a capillary tube of radius r2 is h. r To find : Ratio 1 . r2 Height of mercury that rises in a capillary tube : 2 TM cosqM ...(i) h = rM r1 g Height of water that rises in a capillary tube: 2 TW cosqW ...(ii) h = rW r2 g Equating equations (i) and (ii), 2TM cosqM 2 TW cosqW = ; rM r1 g rW r2 g Ê T ˆ Ê cosqM ˆ Ê rW ˆ r1 = Á M ˜ Á r2 Ë TW ¯ Ë cosqW ˜¯ ÁË rM ˜¯

= 7.5 ¥

cos 135∞ 1 2 ¥ ª  cos 0∞ 13.6 5

Ans. 21. Option (b) is correct. Given : Edge length of cubical block is a = 0.5 m, Fractional volume of block submerged in water = 0.3 V, Density of water is kg 3 rw = 10 3 kg/m . . m3

129

PROPERTIES OF SOLIDS AND LIQUIDS

To find : The maximum weight m, that can be put on the block without fully submerging it in water. By Archimedes principle, Weight of the block = weight of the water displaced rb Vg = rw (0.3V ) g  (rb is density of block) rb = 0.3 rw ...(i) After placing m on top of block, the block is almost completely submerged. So, rw Vg = rb Vg + mg m = ρ w V - 30 ρ w V = 0.7 rw a3 = 0.7 ¥ 10 3 ¥ ( 0.5)3

= 87.5 kg Ans. 22. Option (a) is correct. Given : Pressure on submarine at depth d1 is p1 = 5.05 ¥ 10 6 Pa, Pressure on submarine at depth d2 is p2 = 8.08 ¥ 10 6 Pa, Density of water is rw

kg 3 = 10 kg/m . . m3 3

To find : d2 – d1. Pressure on submarine located at a depth d1 below the sea surface : p1 = po + d1 rw g ...(i) Pressure on submarine located at a depth d2 below the sea surface : p2 = po + d2 rw g ...(ii) In equations (i) and (ii) p0 is the pressure on the surface of a sea. Subtract equation (ii) from equation (i) and put given values. p2 - p1 = ( d2 - d1 ) rw g ( 8.08 ¥ 10 6 - 5.05 ¥ 10 6 ) = ( d2 - d1 ) ¥ 10 3 ¥ 10 ( d2 - d1 ) ª 300 m Ans. 23. Option (b) is correct. Given : Initial speed of water emerging from tap is m v1 = 1.0 m/s, , s Cross sectional area of tap is A1 = 10 -4 m 2 ,

The flow of water is streamlined. To find : Cross sectional area A2, of stream at h = 0.15 m below the tap. Speed of water a distance 0.15 m below the tap : v2 =  v12 + 2 gh =

m 1 + 2 ¥ 10 ¥ 0.15 = 2 m/s s

( as v 2 - u2 = 2 gh )

As speed of water increases from v1 to v2, its cross sectional area decreases keeping the product of area and speed constant. v1 A1 = v2 A 2 ; A 2 =

v1 A1 v2

1 ¥ 10 -4 = 5 ¥ 10 -5 m 2  2 Ans. 24. Option (a) is correct. Given : Mass of ice is M1g, temperature of ice is T1 = –10°C, specific heat of ice is SI = 0.5 cal g–1 °C–1, Mass of water is M2g temperature of water is T2 = 50°C, Final temperature of ice-water mixture is T = 0°C. To find : Latent heat of ice, Lice. When ice and water are mixed together : Heat given by water = heat gained by ice ...(i) Heat given by water : ( DH )liquid = M 2 ¥ S w ¥ DT = M 2 ¥ 1 ¥ ( 50 - 0 ) = 50 M 2 ...(ii) In the equation above, S w = 1calg -1 ∞C -1 is the specific heat of water. Heat gained by ice : ( DH )solid = M1 ¥ S1 ¥ DT + M1 Lice

= M1 ¥ 0.5 ¥ ( 0 - ( -10 ))+ M1 Lice = 5 M1 + M1 Lice ...(iii)

Equating equations (ii) and (iii), 50 M 2 = 5 M1 + M1 L ice

Lice =

50 M 2 - 5 M1

Ans. 25. Option (a) is correct. Given : Case a : Radius of solid sphere is R, terminal velocity of sphere when falling through a viscous fluid of viscosity h is v1, Case b : The sphere in case a breaks into 27 identical solid spheres, the terminal velocity of each of the 27 spherical pieces falling through the same fluid is v2 v1 To find : The ratio  . v2 Terminal velocity of a sphere falling through a viscous fluid of viscosity h : v =

2 r2 ( ro - r f ) g ...(i) 9 h

In equation (i) : r is the radius of the sphere, ro is density of sphere and rf is density of fluid. As for both case a and case b the parameters h, ro, rf, g are same : v µ r 2 ...(ii)

130 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) For case a : 2

v1 µ R ...(iii) For case b : v2 µ r 2 (r is radius of each of the 27 spherical pieces) As the large sphere breaks into 27 identical spherical pieces : 4 p R3 R 4 3 3 p r ; r = ...(iv) = 3 3 27 So, 2

Ê Rˆ v2 µ Á ˜ ...(v) Ë 3¯ Divide equation (iii) by equation (v). v1 = 9 v2 Ans. 26. Option (b) is correct. Given : Weight of water is m = 1 kg, temperature of water is T = 20°C, specific heat of water is J S W = 4200 , kg°C latent heat of water is kJ Lw = 2260 , kg Mean resistance of heating element of the kettle is R = 20 W, rms voltage in the mains supply is Vrms = 200 V. To find : Time taken by the water to fully evaporate. Total heat required to first heat the water up to 100°C and then its complete evaporation : DQ = m ¥ S W ¥ DT + mLw = 1 ¥ 4200 ¥ 80 + 1 ¥ 2260 ¥ 10 3 = 2596kJ ...(i) Heat supplied by the kettle heating element in time t : V2 DQ = Pt = rms t R ( 200 )2 ...(ii) t 20 In equation (ii), P is power and t is the time. Equating equations (i) and (ii), =

( 200 )2 t = 2596 ¥ 10 3 20 t = 1298 s t ª 22 min Ans. 27. Option (b) is correct. Given : Rate of flow of water coming out of an opening is V 0.74 3 m s = 0.74 m 3 min. = t 60 (V is volume and t is time), radius of the circular opening is r = 2 cm the water level in the tank is maintained.

PHYSICS

To find : The depth of the centre of the opening from the surface of water in the tank. Velocity of flow of water : v =

V 0.74 m = = 9.8 m/s tA 60 ¥ p ( 0.02 )2 s

...(i)

(A = pr2 is the area of cross section of the circular opening) Also, from Torricelli’s law : v =  2 gh ...(ii) Equate equations (i) and (ii), 2 gh = 9.8 h =  4.8 m Ans. 28. Option (a) is correct. Given : Rate of flow of water into the tank is V m3 = 10 -4 t s (V is volume and t is time), area of cross section of the hole through which the water is leaking is A = 1 cm2, the height of water in the tank above the hole remains unchanged. To find : Height h, of water above the hole. As the height of water in the tank remains same: Rate of inflow of water = rate of outflow of water Rate of inflow of water V ...(i) = t Rate of outflow of water from the hole vA =  2 gh ¥ A ...(ii) In equation (ii) h is height of water surface above the hole. Equate equations (i) and (ii), V 2gh ¥ A =   t h =

2

1 Ê Vˆ  = 5.1cm 2 g ÁË tA ˜¯

Ans. 29. Option (a) is correct. Given : Density of liquid is r, radius of hose pipe is a, horizontal speed of liquid is v, amount of liquid passing through the mesh unaffected is 50%, amount of liquid losing all its momentum at the mesh is 25%, amount of liquid coming back with the same speed after hitting the mesh is 25%. To find : Resultant pressure on the mesh. Rate of flow of mass per unit time of the given liquid : dm = rAv ...(i) dt In the above equation, A is area of cross section of the hose pipe. For 25% of liquid the change in velocity is from v to 0 : Dv = v ...(ii)

131

PROPERTIES OF SOLIDS AND LIQUIDS

So, corresponding change in momentum will be : dp1 1 dm 1 = v = rAv 2 ...(iii) dt 4 dt 4 For 25% of liquid the change in velocity is from v to –v : Dv = 2 v ...(ii) So, corresponding change in momentum will be : dp2 1 1 dm = ( 2 v ) = rAv 2 ...(iv) dt 4 dt 2 For 50% of liquid the change in velocity is from v to Dv : Dv = 0 ...(ii) So, corresponding change in momentum will be : dp3 1 dm = ( 0 ) = 0 ...(v) dt 4 dt Net force on the mesh : dp1 dp2 dp3 3 + + = rAv 2 ...(vi) dt dt dt 4 Net pressure on the mesh : 3 rAv 2 F 4 3 p = = =  rv 2 A A 4 Ans. 30. Option (c) is correct. Given : Case a : Weight of liquid A is mA = 100 g, temperature of liquid A is TA = 100 °C, Weight of liquid B is mB = 50 g, temperature of liquid B is TB = 75 °C, Temperature of mixture of A and B is T = 90°C, Case b : Weight of liquid A is mA = 100 g, temperature of liquid A is TA = 100 °C, Weight of liquid B is mB = 50 g, temperature of liquid B is TB = 50 °C. To find : Temperature T’, of mixture of A and B in case b. In case a : heat lost by liquid A   = heat gained by liquid B mA ¥ S A ¥ ( TA - T ) = mB ¥ S B ¥ ( T - TB ) ...(i) SA and SB are specific heat capacity of liquids A and B respectively. Put given values in equation (i), 0.1 ¥ S A ¥ 10 = 0.05 ¥ S B ¥ 15 SA =

3 S B ...(ii) 4

In case b : heat lost by liquid A  =  heat gained by liquid B mA ¥ S A ¥ ( TA - T¢ ) = mB ¥ S B ¥ ( T¢ - TB ) 0.1 ¥ S A ¥ (100 - T¢ ) = 0.05 ¥ S B ¥ ( T¢ - 50 ) 3 0.1 ¥ S B ¥ (100 - T¢ ) = 0.05 ¥ S B ¥ ( T¢ - 50 ) 4 3(100 - T¢ ) = 2( T¢ - 50 ) 5T¢ = 400 ; T¢ = 80 ∞C

Ans. 31. Option (a) is correct. Given : Radius of cylindrical vessel is r = 5 cm, rotational speed of cylinder is w = 2 rps = ( 2 ¥ 2p ) rad s = 4p rad s , h Initial height of liquid inside the cylinder is , 2 where h is height of cylinder. To find : Difference in height of liquid at point O (centre of the cylinder) and the height at point P.

.

P

r y

O

Pressure at point O : pO = - rgy ...(i) Total pressure at point P : 1 pP = p + rr 2 w2 - rgy ...(ii) 2 The first term in equation (ii) is atmospheric pressure, second term is pressure due to rotation and the third term is the gauge pressure at depth y. Also, pressure at surface of a rotating fluid : pS = p ...(iii) Equating equations (ii) and (iii), r 2 w2 1 2 2 ...(iv) r w = gy ; y = 2 2g Put given values in equation (iv), y =

( 0.05)2 ( 4 p )2 ª 2 cm 2 ¥ 10

Ans. 32. Option (a) is correct. Given : Increase in volume of a soap bubble with time is constant dV = constant = c. ...(i) dt To find : Correct graph that depicts pressure inside the bubble as function of time. As the temperature of the bubble is constant : k pV = constant = k ; p = ...(ii) V (Boyle’s law) Differentiate equation II, with respect to time : dp k dV kc = - 2 = - 2 ...(iii) dt V dt V From equation (i), V = ct ...(iv) Put V = ct in equation (iii),

132 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) αV = αx + α y + αz

dp kc k = - 2 2 = - 2 ...(v) dt c t ct Integrating equation (v) : k -2 Údp = - c Út dt

C × 10 −6 = 5 × 10 −5 + 5 × 10 −6 + 5 × 10 −6 C × 10 −6 = 6 × 10 −5 C × 10 −6 = 60 × 10 −6

k p =   ct p µ

C = 60

1 ...(vi) t

From equation (vi) graph of pressure versus be a straight line as depicted in option a.

1 will t

Subjective Questions (Chapter Based) Q.1. When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to 0°, the surface tension of the liquid, in milliNewton m–1, is [r(liquid) = 900 kgm–3, g = 10 ms–2] (Give answer in closest integer) _________. [JEE (Main) – 3rd Sep. 2020 - Shift-1] Sol. Given : Radius of capillary tube is r = 0.015 cm, height of liquid column inside the capillary is h = 15 cm, contact angle between the liquid and glass is q = 0°, density of liquid is r = 900 kg/m3. To find : T, the surface tension of the liquid. Height of a liquid column inside a capillary tube : 2T cosθ h = ρ gr T =

h ρ gr 2

T =

1 × 15 × 10 −2 × 900 × 10 × 0.015 × 10 −2 2

T = 101 mN m Q.2. A non-isotropic solid metal cube has coefficients of linear expansion as : 5 × 10–5/°C along the x-axis and 5 × 10–6/°C along the y and the z-axis. If the coefficient of volume expansion of the solid is C × 10–6/°C then the value of C is _________. [JEE (Main) – 7th Jan. 2020 - Shift-1] Sol. Given : Coefficient of linear expansion along x axis for a solid metal cube is ax= 5 × 10–5 /°C, coefficient of linear expansion along the y and the z-axis for cube is ay = az = 5 × 10–6 /°C, coefficient of volume expansion for the cube is av = C × 10–6/°C. To find : Value of C. Volume of cube : V = xyz ∆V ∆x ∆y ∆z = + + V x y z 1 ∆V 1 ∆x 1 ∆y 1 ∆z = + + T V T x T y T z

PHYSICS

Q.3. M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g], the value of M is ________. [JEE (Main) – 7th Jan. 2020 - Shift-2] Sol. Given : Temperature of steam is T1 = 100°C, mass of steam is Mg, temperature of ice is T2 = 0°C, mass of ice is m = 200 g, both steam and ice are mixed to get water at T = 40°C heat of vaporisation of water is Lv = 540 cal/g, heat of fusion of ice is Lf = 80 cal/g, specific heat of water is Sw = 1 cal/g/°C. To find : Value of M. Heat absorbed by ice : mL f + mS w ( T − T2 )

...(i)

Heat released by steam : MLv + MS w ( T1 − T )

...(ii)

As heat absorbed will be same as heat released, equate equations (i) and (ii). mL f + mS w ( T − T2 ) = MLv + MS w ( T1 − T ) 200 × 80 + 200 × 1 × ( 40 − 0 ) = M × 540 + M × 1 × (100 − 40 ) 24000 = 600 M M = 40 Q.4. Three containers C1, C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)



C1

C2

C3

T

1l –

2l

--

60°C

2l

2l

30°C

2l

–

1l

60°C

1l

1l

1l

q

The value of q (in °C to the nearest integer) is ______.

[JEE (Main) – 8th Jan. 2020 - Shift-2] Sol. Given : C1, C2, C3 are three containers that contain water at different temperatures, T is the final temperature of the mixture of water from C1, C2, C3 the table below gives further details.

133

PROPERTIES OF SOLIDS AND LIQUIDS

C1

C2

C3

T

1l

2l

--

60°C

–

2l

2l

30°C

2l

–

1l

60°C

1l

1l

1l

q

To find: The value of q. Let the corresponding temperatures of the three containers be : q1, q2, q3. As all the three containers hold water, the specific heat capacity is same for all. So, from the table’s 1st entry : 1 × θ1 + 2 × θ 2 + 0 × θ 3 = (1 + 2 + 0 ) × 60

θ1 + 2θ 2 = 180

...(i)

From the table’s 2nd entry : 0 × θ1 + 1 × θ 2 + 2 × θ 3 = ( 0 + 1 + 2 ) × 30

θ 2 + 2θ 3 = 90

...(ii)

th

From the table’s 4 entry : 1 × θ1 + 1 × θ 2 + 1 × θ 3 = (1 + 1 + 1) × θ

θ1 + θ 2 + θ 3 = 3θ

...(iv)

Add equations (i), (ii) and (iii) : 3θ1 + 3θ 2 + 3θ 3 = 450

θ1 + θ 2 + θ 3 = 150





...(iii)

(v)

ω =

σA 4.8 × 107 × 10 −6 = = 4 rad / s ml 10 × 0.3

F = - hA

dv ...(i) dy

In equation (i), h is viscosity of the liquid and A is the area of plate. As h is very small :

Q.7. A drop of liquid of radius R = 10–2 m having 1 .0 surface tension S = Nm–1 divides itself into K 4p identical drops. In this process the total change in the surface energy DU = 10–3 J. If K = 10a then the value of a is Sol. Given : Radius of liquid drop is R = 10–2 m, surface tension of water is S =

θ = 50°C

(for rotational motion, v, lw)

The resistive force F :

hAuo u dv = o ; F = dy h h

On comparing equations (iv) and (v) : 3θ = 150 Q.5. A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s–1) with which it can be rotated about its other end in space station is (Breaking stress of wire = 4.8 × 107 Nm–2 and area of cross section of the wire = 10–2 cm2) [JEE (Main) – 9th Jan. 2020 - Shift-1] Sol. Given : Length of wire is l = 0.3 m, mass of the body attached to one end of the wire is m = 10 kg, area of cross section of wire is A = 10–2 cm2 = 10–6 m2, N 2 . . breaking stress of wire is σ = 4.8 × 107 N/m m2 To find : w, the maximum angular speed with which the wire can be rotated about its other end on a space station. Breaking stress : m l 2ω 2   2 T v2 m  l  = mω l σ = = = A A A A

uo

F



From the table’s 3rd entry : 2 × θ1 + 0 × θ 2 + 1 × θ 3 = ( 2 + 0 + 1) × 60 2θ1 + θ 3 = 180

Q.6. Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u0. If A is the area of the square plate then, what will be the resistive force of liquid on the plate? Sol. Given : A thin square plate floating on a liquid is pulled horizontally with constant velocity u0, height h of liquid tank is much less than its width. To find : Resistive force of liquid on the plate.







0.1 N , change in surface 4p m

energy when the drop divides itself in K identical drops is DU = 10–3 J. To find : a, for K = 10 a. ...(i) Mass of single drop of radius R =  mass of K drops of radius r. 4 4 R r p R 3 = Kr p r 3 ; r = 1 3 3 K3 (r is density of liquid) Change in surface energy when the drop divides itself in K identical drops : DU = SD A = S( K ¥ 4 p r 2 - 4 p R 2 ) Ê ˆ 4 p R2 2˜ Á = S K¥ - 4p R 2 Á ˜ ÁË ˜¯ 3 K Ê 1 ˆ = 4 pR 2 S Á K 3 - 1˜ ...(iii) ÁË ˜¯

DA is change in surface area of the drop when it divides itself in K identical drops.

134 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)



Put given values in equation (iii), 10 -3 = 4 ¥ p ¥ (10 -2 )2 ¥



1 0.1 Ê 3 ÁK

4 p ÁË

ˆ - 1˜ ˜¯

1



Comparing equations (i) and (iv),



a = 6.

Q.8. Consider two solid spheres P and Q each of density 8 gm cm–3 and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm–3 and viscosity h = 3 poiseuille. Sphere Q is dropped into a liquid of density 1.6 gm cm–3 and viscosity n = 2 poiseuille. The ratio of the terminal velocities of P and Q is Sol. Given : Density of solid sphere P = density of solid sphere Q = r = 8 g cm 3 ,



Diameter of P is DP = 1 cm, diameter of Q is DQ = 0.5 cm,

2

Ê 1 ˆ Ê 2 ˆ Ê 8 - 0.8 ˆ = Á ˜ Á ˜Á = 3 Ë 0.5 ¯ Ë 3 ¯ Ë 8 - 1.6 ˜¯ Q.9. A solid sphere of radius R and density r is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3r. The complete arrangement is placed in a liquid of density 2r and is allowed to reach equilibrium.

rP = 0.8 g cm 3 ,

viscosity of liquid in which P is dropped is hP = 3,



Density of liquid in which Q is dropped is g cm 3

Find net elongation of the spring?

Sol. Given : Radius of sphere-1 is r1 = R, density of sphere-1 is r1 = r,

Radius of sphere-2 is r2 = R, density of sphere-2 is r2 = 3r,



The two spheres are connected with a massless spring of force constant k,



Density of liquid in which the two sphere are placed is 2r.



To find : Net elongation of the spring.

Density of liquid in which P is dropped is

rQ = 1.6

That gives, D P2 h Q( r - rP ) vP =  2 vQ DQ h P( r - rQ )

K 3 ª 100 ; K = 10 6 ...(iv)



1

,



viscosity of liquid in which Q is dropped is hQ = 2.



To find : Ratio of terminal velocities of P and Q,

k 2

vP . vQ

Terminal velocity of P : vP =

2 rP2 ( r - rP ) g 9 hp



2 D2 =  P ( r - rP ) g 9 4 hp =

PHYSICS

D 2P ( r - rP ) g 18hp

...(iii)



Terminal velocity of Q : vQ =

2 2 rQ ( r - rQ ) g 9 hQ



2 2 DQ =  ( r - rQ ) g 9 4hQ

=

2 DQ

( r - rQ )g 18hQ



...(ii)



Balancing the forces acting on sphere-1 : 4 4 p R 3 rg + kx = p R 3 ( 2 r) g . ..(i) 3 3 Balancing the forces acting on sphere-2 : 4 4 p R 3 ( 3 r) g = p R 3 ( 2 r) g + kx ...(ii) 3 3 Equation (ii) minus equation (i) : 4 4 p R 3 ( 3 r) g - p R 3 rg 3 3 = 2 kx 4 p R 3 rg = kx 3 x =

4 p R 3 rg 3k

135

PROPERTIES OF SOLIDS AND LIQUIDS

Strain

Q.10. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure.

P

Strain

Q

Stress

P Q



Stress

Compare the tensile strength, ductility and the Young’s modulus of the two materials, P and Q. Sol. Given : Strain versus stress plot for two materials P and Q. To find : Compare the tensile strength, ductility and the Young’s modulus of the two materials. Young’s modulus : stress Y = , strain







So, for given stress, s : 1 ...(i) Y µ strain In the above plot, for stress = s, strain in P is higher than strain in Q. Therefore, from equation (i), YQ > YP ...(ii) From equation (ii), P is more ductile than Q. Also, from the graph breaking stress for P is more than that of Q, so P has more tensile strength.







T

1

or  m 

T

b



Wien ’s d isp lac em en t la w

Ki rc hh

Stefan’s Boltzmann law

Tempera

t

ure



l – l0 l  l0  l0(T – T0)

Increase of length of a solid on heating. Coefficient of linear expansion

TC  0 T  32 TK  273.15 –0  F  – =R 100  0 212  32 373.15  273.15 80 – 0

Relation among different temperature scales

Degree of hotness or coldness of a body or measuring device = Thermometer

(i) Conduction : heat transfer through molecular collisions without any actual motion of matter. (ii) Convection : heat transfer by actual motion of matter within the medium. Land breeze, sea breeze, trade winds based on natural convection are some examples. (iii) Radiation : method of heat transfer requiring no material medium.

4 4 E   T T0



e=emissivity For a perfectly black body, e=1

  5.672 Js 1 m2 k4 and

4

eT Here, E or E   eT 4

Energy per unit area (E) in given as

3 b(wine’s constant)  2.9 10 mK

m∝

al c

Types (In solids)

Increase in dimensions due to increase in temperature

Pri nci ple of

A – A0 A dA    A  T A dT A0(T–T0)

Increase in area of a solid on heating. Coefficient of superficial expansion

or α:β:

=2β=3α

Speci fic H ea t

T

Q s Q  n n T

V0 T

V0 dT

dV   V  

V0(T – T0)

V – V0

Increase in volume of a solid on heating. Coefficient of cubical expansion.

F  l   Y   , Y = Young’s modulus A  l 

Heat lost = Heat gained

Molar specific heat capacity, c =

Heat capacity s =

s Q Specific heat capacity C = m mT

calo rime try

Cubical or vo lum e ex pan Re lat sio ion n

t

ea

H

A form of energy, transferred between two systems by virtue of temperature difference.

t

5 For water, latent heat of fusion, L f  3.33 10 J / kg 5 Latent heat of vapourisation, Lv  22.6  10 J / kg

Heat required to change the state of unit mass substance without changing its temperature , L  Q m

Q. x K  A T2  T1  t

d on

Thermodynamics Part-1

T h er m

g lin oo c of law s ’ n to New

For small temperature difference between a body and its surroundings, the loss of dQ heat is given by   k  T2  T1  dt

Superficial or area Expansion

eλ aλ = Eλ = constant

al erm Th aws L

Therm Expan al sion

y v it

law uc ti

f ’s of La t e n tH ea

At any given temperature

136 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

,

,

V

– T

Work done w=

P

P2V2–P1V1 1–

A thermally insulated system neither gains nor loses heat

Thermodynamics Part-2

It is the statement of the law of conservation of Energy;

St a

lau

It is impossible for an engine working between a cyclic process to extract heat from a reservoir and convert completely into work.

la Kelvin-P

C

si tem us e nt

T

Measure of molecular disorder of a system.

No process in possible whose sole result in the transfer of heat from a colder object to a hotter object.

No process in possible whose sole result in the absorption of heat from a reservoir and the complete conversion of the heat into work.

nck Statement

THERMODYNAMICS

137

Chapter 8 Thermodynamics Syllabus Thermal equilibrium, zeroth law of thermodynamics, concept of temperature; Heat, work and internal energy; First law of thermodynamics; Second law of thermodynamics: reversible and irreversible processes; Carnot engine and its efficiency. Specific heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases; Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law; Wien’s displacement law, Stefan’s law.

Topic-1

Heat, Work and Internal Energy

list of Topics : Topic-1 : Heat, Work and Internal Energy

.... P. 138

Topic-2 :  Carnot Engine and P-V Diagrams .... P. 145

Concept Revision (Video Based)

     

Adiabatic and Isothermal Process

Heat Transfer

 

Thermodynamics

     

 

Latent Heat

   

 

        

Calculating Work Done

Blackbody Radiation

   Part -1       



 

Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. When the temperature of metal wire is increased from 0°C to 10°C, its length increases by 0.02%. The percentage change in its mass density will be closed to : (a) 0.8 (b) 0.008 (c) 2.3 (d) 0.06 [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of ‘m’ is close to (Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1°C–1)

(a) 2 (c) 3.2

(b) 4 (d) 2.6 [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.3. The specific heat of water = 4200 J-kg–1 K–1 and the latent heat of ice = 3.4 × 105 J-kg–1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (a) 61.7 (b) 69.3 (d) 63.8 (c) 64.6 [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.4. Match the thermodynamics processes taking place in a system with the correct conditions. In the table DQ is the heat supplied, DW is the work done and

139

THERMODYNAMICS

DU is change in internal energy of the system. Match the following (I) Adiabatic (A) DW = 0 (II) Isothermal (B) DQ = 0 (III) Isochoric (C) DU ¹ 0, DW ¹O DQ ¹ 0 (IV) Isobaric (D) DU = 0 (a) I Æ A   II Æ A   III Æ B   IV Æ C (b) I Æ B   II Æ D   III Æ A   IV Æ C (c) I Æ A   II Æ B   III Æ D   IV Æ D (d) I Æ B   II Æ A   III Æ D   IV Æ Cs [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.5. A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal (g –°C) (1 cal = 4.2 × 107 ergs) close to : (a) 38.4°C (b) 83.3°C (c) 87.5°C (d) 119.2°C [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.6. Two different wires having lengths L1 and L2 and respective temperature coefficient of linear expansion a1 and a2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is : α 2α 2 L2 L1 α L + α 2 L2 (a) 1 1 (b) α1 + α 2 (L2 + L1 )2 L1 + L2 (c) 2 α1α 2

(d)

α1 + α 2 2

[JEE (Main) – 5th Sep. 2020 - Shift-2] Q.7. The rods of identical cross-section and lengths are made of three different materials of thermal conductivity K1, K2 and K3, respectively. They are joined together at their ends to make a long rod (see figure). One end ofr the long rod is maintained at 100°C and the other at 0°C (see figure). If the joints of the rod are at 70°C and 20°C in steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1, K2 and K3 is : K1

K2

K3 0°C

100°C 70°C

20°C

(a) K1 < K2 < K3 (b) K1 : K3 = 2 : 3 (c) K1 : K2 = 5 : 2 (d) K1 > K2 > K3 [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.8. A litre of dry air at STP expands adiabatically to a volume of 3 litres. If g = 1.40, the work done by air is : (31.4 = 4.6555) [Take air to be an ideal gas] (a) 60.7 J (b) 100.8 J (c) 90.5 J (d) 48 J [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.9. A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to :

(Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 105 J kg–1) (a) 150 g (b) 20 g (c) 130 g (d) 35 g [JEE (Main) – 8th April 2019 - Shift-1] Q.10. Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘q2’ and ‘q1’ respectively, (q2 > q1). The temperature at the interface is :

d q2 3K

3d q1

K

q 2 + q1 q1 9q 2 (b) + 2 10 10 q 2q q 5q (c) 1 + 2 (d) 1 + 2 3 3 6 6 [JEE (Main) – 9th April 2019 - Shift-2] Q.11. A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is : [Given that R = 8.31 J mol–1 K–1] (a) 350 J (b) 374 J (c) 748 J (d) 700 J [JEE (Main) – 10th April 2019 - Shift-1] Q.12. n moles of an ideal gas with constant volume heat capacity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is : nR nR (a) (b) C V - nR C V + nR (a)

(c)

4nR C V - nR

(d)

7 Q 5

(d)

4nR C V + nR

[JEE (Main) – 10th April 2019 - Shift-1] Q.13. When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by DT. The heat required to produce the same change in temperature, at a constant pressure is : 5 2 (a) Q (b) Q 3 3 (c)

3 Q 2

[JEE (Main) – 10th April 2019 - Shift-2] Q.14. A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process ? (a) 25 J (b) 35 J (c) 30 J (d) 40 J [JEE (Main) – 12th April 2019 - Shift-2] Q.15. Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as

140 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) 3L , is connected across AB (See 2 figure). In steady state, temperature difference between P and Q will be close to : AB and length

Q.21. A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the

L — 4

water ?

A

B L — 2

P

L

Q

(a) 45°C (b) 75°C (c) 60°C (d) 35°C [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.16. A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 W K–1 m–1, the energy flux through it in the steady state is : (a) 90 W m–2 (b) 120 W m–2 –2 (c) 65 W m (d) 200 W m–2 [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.17. An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394 J kg–1 K–1) (a) 458 J kg–1 K–1 (b) 1232 J kg–1 K–1 (c) 916 J kg–1 K–1 (d) 654 J kg–1 K–1 [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.18. Ice at –20°C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/°C Specific heat of Ice = 2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g) (a) 50 g (b) 60 g (c) 40 g (d) 100 g [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.19. A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling x water, and 0 when in contact with ice. What is the 3 temperature of an object in °C, if this thermometer x in the contact with the object reads 0 ? 2 (a) 25 (c) 40

PHYSICS

(b) 60 (d) 35 [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.20. Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180°C and B upto T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is : (a) 230°C (b) 270°C (c) 200°C (d) 250°C [JEE (Main) – 11th Jan. 2019 - Shift-2]



[Specific Heat Capacities of water and metal are, respectively, 4200 J kg–1 K–1 and 400 J kg–1 K–1] (a) 15%

(b) 30%

(c) 25%

(d) 20% [JEE (Main) – 11th Jan. 2019 - Shift-2]

Q.22. A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is : (a)

K1 + K 2 2

(b) K1 + K2

(c)

2 K 1 + 3K 2 5

(d)

K 1 + 3K 2 4

[JEE (Main) – 12th Jan. 2019 - Shift-1]

ANSWER – KEY

1. (d)

2. (a)

3. (a)

4. (b)



5. (c)

6. (a)

7. (b)

8. (c)



9. (b)

10. (a)

11. (d)

12. (a)



13. (c)

14. (b)

15. (a)

16. (a)



17. (c)

18. (c)

19. (a)

20. (a)



21. (d)

22. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. Given : Change in temperature of a metal wire is DT = 10°C, the corresponding percentage change in its length is

∆l = 0.02%. l

To find :

Dm , the percentage change in the mass m

density of the wire. Coefficient of linear expansion of wire :

α =

Dl 0.02 = = 2 × 10 −5 lDT 100 × 10

Coefficient of volume expansion of wire :

γ = 3α = 6 × 10 −5

141

THERMODYNAMICS

The percentage change in the volume of the wire : DV = γ DT = 6 × 10 −2% V So, the percentage change in the mass density of the wire : Dm = m

Half of the kinetic energy lost by the bullet goes in heating the bullet. 1 1 2 × mv = mCDT 2 2

DV = 0.06% V

Ans. 2. Option (a) is correct. Given : Mass of water in a calorimeter is M = 200 g, initial temperature of water is T1 = 25 °C, mass of steam is m g, initial temperature of steam is T2 cal = 100 °C latent heat of water is Lw = 540 cal, g–1, cal –1 –1g specific heat of water is C w = 1 cal g, °C , final g.°C temperature of the mixture is T = 31°C. To find : m Heat gained by water is equal to the heat lost by the steam : MC w ( T − T1 ) = mLw + mC w ( T2 − T ) 200 × 1 × ( 31 − 25) = m × 540 + m × 1 × 69 m = 2g Ans. 3. Option (a) is correct.

5 −1 −1 Given : Specific heat of water is Li = 3.4 × 10 J kg K , 5 −1 latent heat of ice is Li = 3.4 × 10 J kg , mass of ice is m = 100 g, temperature of ice is T1 = 0°C, mass of water is M = 200 g, temperature of water is T2 = 25 °C.

To find : m¢, the amount of ice that will melt when the temperature of water reaches T = 0°C. Amount of heat lost by water, melts the m¢ amount of ice. MC w ( T2 − T ) = m′Li 200 × 4200 × 25 = m′ × 3.4 × 10 5 1000 m′ = 61.7 g Ans. 4. Option (b) is correct. Given : Adiabatic, isothermal, isochoric and isobaric thermodynamic processes. To find : The correct match for the thermodynamic processes. In adiabatic process : DQ = 0 In isothermal process : DU = 0 In isochoric process : DW = 0 Ans. 5. Option (c) is correct. Given : Mass of the bullet is m = 5 g, speed of the bullet is v = 210 m/s, the bullet strikes a wooden block and half of its kinetic energy is converted into heat, specific heat of material of bullet is C = 0.030 cal/(g◊°C) = 0.030 × 4.2 × 1000 J/(kg◊°C). To find : DT, the rise in temperature of the bullet.

DT =

v2 4C

DT =

210 × 210 4 × 0.030 × 4.2 × 1000

DT = 87.5 °C Ans. 6. Option (a) is correct. Given : Length of wire 1 is L1, length of wire 2 is L2, coefficient of linear expansion for wire 1 is a1, coefficient of linear expansion for wire 2 is a2, the two wires are joined end to end. To find : a, the coefficient of linear expansion of the new wire. The length of the new wire is L = L1 + L2 ...(i) If we increase the temperature of the new wire by DT, its length will change as : L′ = L + Lα DT = L(1 + α DT ) ...(ii) Also, L′ = L1 + L1α1 DT + L2 + L2α 2 DT = L1 (1 + α1 DT ) + L2 (1 + α 2 DT )

...(iii)

From equations (i), (ii) and (iii) : L(1 + α DT ) = L1 (1 + α1 DT ) + L2 (1 + α 2 DT ) Lα = L1α1 + L2α 2

α =

L1α1 + L2α 2 L1 + L2

Ans. 7. Option (b) is correct. Given : Three rods have same area of cross section and same lengths, the rods are made of different materials, the rods are joined together and the temperatures at various interfaces is shown below, there is no loss of energy from the surfaces of the rods. K1

K2

K3 0°C

100°C 70°C

20°C

To find : The correct relation between K1, K2, K3. From the diagram above : K 1 (100 − 70 ) = K 2 (70 − 20 ) = K 3 ( 20 − 0 ) 30 = K 1 50 = K 2 20 K 3 K1 : K 3 = 2 : 3 Ans. 8. Option (c) is correct. Given : Initial volume of gas is V1 = 1L, initial pressure of gas is p1 = 1 atm, the gas expands adiabatically to final volume V2 = 3l, g = 1.40, 31.4 = 4.6555. To find : W, the work done by the gas in the adiabatic expansion.

142 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) For an adiabatic process : p1V1γ

=

p2 V2γ

(p2 is pressure corresponding to volume V2) γ

1.4 V  1 p2 = p1  1  = 1   = 0.215 atm 3  V2  = 0.215 × 101.325 kPa = 21.785 kPa 3

= 21.785 × 10 Pa Work done during the adiabatic process : (1litre = 10 −3 m 3 ) W =

p1V1 − p2 V2 γ −1

1 × 101.325 × 10 3 × 1 × 10 −3 =

− 21.785 × 10 3 × 3 × 10 −3 ≈ 90 J 1.40 − 1

Ans. 9. Option (b) is correct. Given : Total mass of water in thermally insulated vessel is M = 150 g, temperature of water in vessel is T = 0°C, latent heat of vaporisation of water is LV = 2.10 ¥ 10 6 J kg -1 latent heat of fusion of water is LF = 3.36 ¥ 10 5 J kg -1 . To find : The mass of evaporated water when the vessel is pumped out adiabatically. Let mass of evaporated water be x g = x ¥ 10 -3 kg. So, heat lost by freezing water (amount of water that turns into ice) = heat gained by the evaporated water (amount of water that turns into steam). (150 – x) × 10–3 × 3.36 × 105 = x × 10–3 × 2.10 × 106 150 × 3.36 × 105 = (2.10 × 10–6 + 3.36 ×10–5)x x  20g Ans. 10. Option (a) is correct. Given : Thermal conductivity of material 1 is K, thermal conductivity of material 2 is 3K, thickness of material 1 is 3d, thickness of material 2 is d, the temperature at the outer surface of the material 1 is q1 and temperature at the outer surface of the material 2 is q2, q2 > q1. d q2 3K 2

3d K 1

q1

To find : The temperature T at the interface of the two materials. As q2 > q1, the heat will flow from material 2 to material 1.

PHYSICS

Rate of heat flow through material 2 : 3 K ¥ A ¥ ( θ2 - T )      ...(i) d Rate of heat flow through material 1 : K ¥ A ¥ ( T - θ1 )     3d

...(ii)

In equations (i) and (ii), A is area of cross section of two materials which is same. At steady state : K ¥ A ¥ ( T - θ1 ) 3 K ¥ A ¥ ( θ2 - T ) = 3d d 9(q2 – T) = (T – q1) 9q2 – 9T = T – q1 θ1 θ +9 2 T = 10 10 Ans. 11. Option (d) is correct. Given : Volume of He gas at STP is V = 67.2 L, gas constant is J R = 8.31 . mol ◊ K To find : Amount of heat DQ, needed to raise the temperature of the gas by DT = 20∞C． At STP volume of 1 mole of gas 22.4 L. So, total number of moles in 67.2 L of gas at STP will be : 67.2 n = =3 22.4 Amount of heat needed to raise the temperature of gas by : DT = 20∞C DQ = nC V DT (For monoatomic gas C V = DQ = 3 ¥

3 R) 2

3 ¥ 8.31 ¥ 20 ª 748 J 2

Ans. 12. Option (a) is correct. Given : Number of moles in an ideal gas is n, C V = constant. To find : The ratio of work done to heat supplied W ,when the gas undergoes an isobaric DQ expansion. For isobaric process, work done : W = nRDT ...(i) Heat supplied : DQ = nC p DT ...(ii) From relation C C C p - v = R ; C p = R + v ...(iii) n n Substitute in equation (ii) : C ˆ Ê DQ = n Á R + v ˜ DT ...(iv) Ë n ¯

143

THERMODYNAMICS

Take ratio of equation (i) and (iv), W nRDT nR = = Cv ˆ DQ nR + C v Ê nÁ R + DT Ë n ˜¯ Ans. 13. Option (c) is correct. Given : Amount of heat supplied to a diatomic gas at constant volume is Q, increase in temperature of gas is DT. To find : Q¢, heat required to increase the temperature of gas by DT at constant pressure. Amount of heat supplied to a diatomic gas at constant volume to raise its temperature by DT : Q = nC v DT ...(i) Amount of heat supplied to a diatomic gas at constant pressure to raise its temperature by DT : Q¢ = nC p DT ...(ii) Take ratio of above two equations. nC p DT C p Q¢ ...(iii) = = nC v DT C v Q As for diatomic gas : Cp 7 γ = = , Cv 5 equation (iii) becomes : 7 Q¢ = Q 5 Ans. 14. Option (b) is correct. Given : Work done by a diatomic gas in expanding at constant pressure is W = 10 J. To find : DQ, heat energy absorbed by gas in this process. Work done by the gas at constant pressure : W = PDV = nRDT = 10J ...(i) Heat absorbed by the gas at constant pressure : Ê7 ˆ DQ = nC p DT = n Á R˜ DT ...(ii) Ë2 ¯ 7 Ê ˆ ÁË C p = 2 R for a diatomic gas˜¯ Put value of nDT = equation (ii).

10 R

from equation (i) into

Ê 10 ˆ Ê 7 ˆ DQ = Á ˜ Á R˜ = 35 J Ë R¯Ë2 ¯ Ans. 15. Option (a) is correct. Given : Length of rod AB is 2L, temperature difference between two ends of rod AB is TAB = 3L 120°C, length of rod PQ is . 2 L — 4 A

B L — 2

P

L

Q

To find : TPQ, temperature difference between points P and Q in steady state. As every section of the arrangement of two rods has same thermal conductivity, the thermal resistance of different sections of the arrangement will be : R

S R/4 A R/2

R R/4 Q

P R

R/2

Net thermal resistance of network PQRS : 1 3 R PQRS = = R 1 1 5 + R R R +R+ 4 4

B

...( i )

Net thermal resistance between points A and B : R 3 R 8 R AB =  + R + = R ...( ii ) 2 5 2 5 Thermal current that flows between points A and B: DTAB 120 ¥ 5 I = = ...( iii ) R AB 8R So, net temperature difference between points P and Q : 120 ¥ 5 3R DTPQ = IR PQRS = ¥ = 45∞C  8R 5 Ans. 16. Option (a) is correct. Given : Temperature of heat source is T1 = 103K, temperature of the heat reservoir is T2 = 103K, thickness of copper slab connecting the source and the reservoir is d = 1 m, thermal conductivity of copper is W K = 0.1 . Km To find : Energy flux through the copper slab. Rate of flow of heat through copper slab : KA ( T1 - T2 ) dQ ...( i ) = dt d In equation (i), A is area of cross section of copper slab. Energy flux through copper slab : K( T1 - T2 ) 1 dQ = A dt d =

0.1 ¥ (1000 - 100 ) 1

= 90

W m2

Ans. 17. Option (c) is correct. Given : Mass of unknown metal is m = 192 g, initial temperature of metal is T1 = 100°C, mass of brass calorimeter is M = 128 g, mass of water inside the calorimeter is M¢ = 240 g, initial temperature of water is T2 = 8.4°C, the final temperature of water metal mixture is T = 21.5°C, specific heat of brass is

144 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) C B = 394

J J = 394 ¥ 10 -3 . kg ◊ K gK

To find : Specific heat CM of the unknown metal. Heat lost by unknown metal : DQ = mC M ( T1 - T ) = 192 ¥ C M ¥ (100 - 21.5) ...(i) Heat gained by the brass calorimeter: DQ1 = M ¥ C B ¥ ( T - T2 ) = 128 ¥ 0.394 ¥ ( 21.5 - 8.4 ) Heat gained by water : DQ2 = M¢ ¥ C W ¥ ( T - T2 ) = 240 ¥ 4.18 ¥ ( 21.5 - 8.4 ) In the equation above, C W = 4.18

...(ii)

...( iii )

J is specific gK

heat of water. Now, amount of heat lost by the metal will be equal to sum of amount of heat gained by the calorimeter and by the water. DQ = DQ1 + DQ2 192 ¥ C M ¥ (100 - 21.5) = 128 ¥ 0.394 ¥ ( 21.5 - 8.4 ) + 240 ¥ 4.18 ¥ ( 21.5 - 8.4 ) C M = 0.916

J J = 916 gK kg ◊ K

Ans. 18. Option (c) is correct. Given : Temperature of ice is T1 = –20°C, mass of water is m2 = 50g, temperature of water is T2 = 40°C, temperature of ice water mixture is T = 0°C, amount of unmolten ice in the ice-water mixture is m = 20g, specific heat of ice is J s1 = 2.1 , g ∞C specific heat of water is J s2 = 4.2 , g ∞C heat of fusion of water at 0°C is J L = 334 . g To find : The mass of ice added to water, m1. Heat lost by water : DQ = m2 s2 ( T2 - T ) = 50 ¥ 4.2 ¥ ( 40 - 0 ) = 8400 J Heat gained by ice : DQ¢ = m1s1 ( T - T1 ) + ( m1 - 20 )L = m1 ¥ 2.1 ¥ ( 0 - ( -20 )) + ( m1 - 20 ) ¥ 334 = 376m1 - 6680 As heat lost by water will be same as heat gained by ice : 376m1 - 6680 = 8400 ; m1 = 40 g Ans. 19. Option (a) is correct. Given : Reading of thermometer at 100°C : x0, x Reading of thermometer at 0°C : o . 3

PHYSICS

To find : The temperature T, at which the x thermometer reads o . 2 As the thermometer is graduated according to a linear scale : xo xo T-0 3 ; = 2 xo 100 - 0 xo 3 Ê 1ˆ Á ˜ 100 T = 100 ¥ Á 6 ˜ = = 25∞C 2 4 Á ˜ Ë 3¯

Ans. 20. Option (a) is correct. Given : At To = 30°C, length of rod A = length of rod B = l, rod A is heated to T¢ = 180°C, ratio of coefficients of linear expansion of rods A and B is a¢ : a = 4 : 3. To find : The temperature T¢ of rod B at which its linear expansion Dl is same as the linear expansion of rod A at T ¢. Linear expansion of rod A at T ¢ : Dl′ = lα ′( T¢ - To ) = lα ′(180 - 30 ) ...(i) Linear expansion of rod B at T : Dl = lα ( T - To ) = lα ( T - 30 )

...(ii)

Divide equation (ii) by (i), Dl′ lα ( T - 30 ) Ê α ˆ Ê T - 30 ˆ = = Dl lα ¢(180 - 30 ) ÁË α ¢ ˜¯ ÁË 150 ˜¯ 3 T - 30 ¥ = 1; T = 230∞C. 4 150 Ans. 21.  Option (d) is correct. Given : Mass of metal ball is m1 = 0.1 kg, initial temperature of ball is T1 = 500°C, the amount of water in the vessel into which the ball is dropped is m2 = 0.5 kg, the initial temperature of water in the vessel is T2 = 500°C, Heat capacity of vessel is J sV = 800 , K Specific heat capacity of water is J , sW = 4200 kg ◊ K Specific heat capacity of metal ball is J . sM = 400 kg ◊ K To find : The percentage increase in the temperature of water in the vessel. Let the final temperature of the vessel be T. Amount of heat lost by the ball : DQ = m1 sM ( T1 - T ) = . 0 1 ¥ 400 ¥ ( 500 - T ) = 40( 500 - T )

...( i )

145

THERMODYNAMICS

Amount of heat gained by water : DQ1 = m2 sW ( T - T2 ) = 0.5 ¥ 4200 ¥ ( T - 30 ) = 2100( T - 30 )

...(ii)

Amount of heat gained by the vessel : DQ2 = sV ( T - T2 ) = 800( T - 30 )

...(iii)

As amount of heat lost by the ball will be sum of heat gained by the water and the vessel : 40( 500 - T ) = 2100( T - 30 ) + 800( T - 30 ) 20000 - 40 T = 2100 T - 63000 + 800 T - 24000 2940 T = 107000

Ans. 22. Option (d) is correct. Given : Radius of a solid cylinder is R, thermal conductivity of the cylinder is K1, Inner radius of a cylindrical shell is R, outer radius of a cylindrical shell is 2R, thermal conductivity of the cylindrical shell is K2. To find : Thermal conductivity K, of the system of two cylinders for the heat flowing along the length of the cylinder. The equivalent thermal conductivity of the system of two cylinders will be : K 1 A1 + K 2 A 2 K = A1 + A 2

T = 36.39∞C

K 1 ( pR 2 ) + K 2 ( p( 2 R )2 - pR 2 )

=

Percentage increase in the temperature of water in the vessel : T - T2 36.39 - 30 = ª 20% T2 30

( pR 2 ) + ( p( 2 R )2 - pR 2 )

K 1 + 3K 2  4

=

Topic-2

Carnot Engine and P-V Diagrams Concept Revision (Video Based)         Part -1     Part -2

P-V Diagrams

Carnot Engine

                    Part -1     Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions (a) T = T1T2

(b) T = 0

2T1T2 T1 + T2 (d) T = T1 + T2 2 [JEE – Mains - 7th Jan. 2020 - Shift-2] Q.4. A thermodynamic cycle xyzx is shown on a V-T diagram. (c) T =

V

>

z

y

>

> x

T

The P-V diagram that best describes this cycle is : (Diagrams are schematic and not to scale) (a)

(b)

P x

>

z

P x

>

y >

>

>



>

Q.1. A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –QJ, during one cycle achieving and efficiency of 50.0%. The value of Q is : (a) 400 J (b) 980 J (c) 640 J (d) 40 J [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. A balloon filled with helium (32° C and 1.7 atm) bursts. Immediately afterwards the expansion of helium can be considered as : (a) reversible isothermal (b) irreversible isothermal (c) reversible adiabatic (d) irreversible adiabatic [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T1 and T2. The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1 and T2, if both the engines perform equal amount of work?

y

z

V

V

146 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (d)

P

a

P

x

x

>

>

> y >

z >

z

P

b

>

(c)

PHYSICS

c y

V

d V

V

th

[JEE (Main) - 8 Jan. 2020 - Shift-1] 1 is 10 being used as a refrigerator. If the work done on

Q.5. A Carnot engine having an efficiency of

the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is : (a) 99 J

(b) 100 J

(c) 90 J

(d) 1 J

(a) a d b c (c) a d c b

(b) d a c b (d) d a b c [JEE (Main) – 8th April 2019 - Shift-2] Q.8. Following figure shows two processes A and B for a gas. If DQA and DQB are the amount of heat absorbed by the system in two cases, and DUA and DUB are changes in internal energies, respectively, then :

[JEE (Main) – 8th January 2020 - Shift-2]

f

Q.6. Which of the following is an equivalent cyclic

A

P

process corresponding to the thermodynamic

B

cyclic given in the figure? Where, 1 Æ 2 is adiabatic. i

(Graphs are schematic and are not to scale)

V

(a)

1

> >

P

3

2

> V

(b)

>

2

>

V

1

>

3

(a) DQA < DQB, DUA < DUB (b) DQA > DQB, DUA > DUB (c) DQA > DQB, DUA = DUB (d) DQA = DQB; DUA = DUB [JEE (Main) – 9th April 2019 - Shift-1] Q.9. A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is – 180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is :

T (c)

2

>

>

V

a

1

>

3

T 3

>

V >

V

(b) 130 J (d) 140 J [JEE (Main) – 12th April 2019 - Shift-1] Q.10. A Carnot engine has an efficiency of 1 . When 6

1

T [JEE (Main) – 9th January 2020 - Shift-1] Q.7. The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by :

b

(a) 120 J (c) 100 J

2

>

(d)

c

P

the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are, respectively, (a) 62°C, 124°C (b) 99°C, 37°C (c) 124°C, 62°C (d) 37°C, 99°C [JEE (Main) – 12th April 2019 - Shift-2] Q.11. A gas can be taken from A to B via two different processes ACB and ADB.

147

THERMODYNAMICS

P

C

Q.14. For the given cyclic process CAB as shown for a gas, the work done is :

B

A



D

5 4 p(Pa) 3

When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If

2

path ADB is used work done by the system is 10 J.

1

B

The heat flow into the system in path ADB is : (a) 40 J

1

(d) 20 J [JEE (Main) – 9th Jan. 2019 - Shift-1]

(a) 30 J (c) 1 J

Q.12. Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1 (=600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3 (= 400 K). Calculate the temperature T2 if the work outputs of the two engines are equal : (a) 600 K

(b) 400 K

(c) 300 K

(d) 500 K [JEE (Main) – 9th Jan. 2019 - Shift-2]

Q.13. Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2 and T3, as shown, with T1 > T2 > T3 > T4. The three engines are equally efficient if :

3

4

5 3 V(m )

(b) 10 J (d) 5 J [JEE (Main) – 12th Jan. 2019 - Shift-1]

ANSWER – KEY 1. (b) 5. (c) 9. (b) 13. (b)

2. (d) 6. (b) 10. (b) 14. (b)

3. (c) 7. (d) 11. (a)

4. (b) 8. (c) 12. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. Given : A heat engine is involved with exchange of heat of Q1 = 1915 J, Q2 = –40 J, Q3 = 125 J, Q4 = –Q J during one cycle, the efficiency of engine is h = 50%. To find : Value of Q. Q + Q 2 + Q3 + Q 4 η = 1 Q1 + Q3 0.5 =

1915 − 40 + 125 − Q 1915 + 125

1020 = 2000 − Q T1 1 T2 2 T3 3 T4

(a) T2 = (T1T4 )1 / 2 ; T3 = (T12 T4 )1 / 3 (b) T2 = (T12 T4 )1 / 3 ; T3 = (T1 T4 )1 / 3 1/3 2 1/3 2 (c) T2 = (T1 T4 ) ; T3 = (T1 T4 ) 3 1/4 3 1/4 (d) T2 = (T1 T4 ) ; T3 = (T1 T4 )



2

(b) 80 J

(c) 100 J



A

C

6.0

[JEE (Main) – 10th Jan. 2019 - Shift-1]

Q = 980J Ans. 2. Option (d) is correct. Given : Temperature and pressure of helium gas inside a balloon is T = 32 °C and P = 1.7 atm, the balloon bursts. To find : The nature of expansion of helium. The expansion of helium gas after the balloon bursts is adiabatic and irreversible as the energy lost during the expansion cannot be restored. Ans. 3. Option (c) is correct. Given : Carnot engine 1 operates between two reservoirs of temperature T1 and T, T1 > T, Carnot engine 2 operates between two reservoirs of temperature T2 and T, T > T2, The two engines operate in cascade and perform equal amount of work. To find : T in terms of T1 and T2. Let heat taken by engine 1 from high temperature reservoir be Q1 and heat energy delivered by engine 1 to the low temperature reservoir be Q. So, heat taken by engine 2 from high temperature reservoir

148 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) will be Q. Let heat energy delivered by engine 2 to the low temperature reservoir be Q2.

1

> >

As work done by both the engines is same :

P

Q1 − Q = Q − Q2

3

Q1 Q2 + Q Q

2 =

T1 T2 + T T

T =

T1 + T2 2

V

Ans. 4. Option (b) is correct. Given : A VT diagram for thermodynamic cycle xyzx. V

>

z

2

>

2Q = Q1 + Q2 2 =

y

>

> x

T

To find : The corresponding VT diagram. For the given PV diagram : a. Process 1 Æ 2 : As the process is adiabatic, TVg –1 = constant. So, we have a non- linear dependence of V on T and also V is inversely proportional to T. b. Process 2 Æ 3 : Pressure is constant. So, V = kT, where k is a constant. That implies V changes linearly with T and is directly proportional to T. c. Process 3 Æ 1 : Volume is constant. Option (b) is the only graph which satisfies the points a, b, and c. Ans. 7. Option (d) is correct. Given : A P-V graph with four subplots (a, b, c, d) depicting four processes, isochoric, isobaric, isothermal and adiabatic process. a

To find : The corresponding PV diagram. For the given VT diagram :

P

b

a. Process x Æ y : Volume is linearly increasing with temperature that implies pressure is constant for this process.

c d

b. Process y Æ z : Volume is constant. c. Process z Æ x : Temperature is constant. So, pressure and volume both are varying. Option (b) is the only graph which satisfies the points a, b, and c. Ans. 5. Option (c) is correct. 1 , the 10 engine is used as a refrigerator, the work done on the refrigerator is W = 10 J. Given : Efficiency of Carnot engine is η =

To find : Q2, the amount of heat absorbed from the reservoir at lower temperature. Let heat taken by the engine from low temperature reservoir be Q2 and heat energy rejected by the engine to the high temperature reservoir be Q1. So, work done on the engine : W = Q1 − Q2 = 10J Q2 Q1 − Q2 1 = = Q1 Q1 10

V

To find : The correct assignment of processes. Isochoric : d, as for plot d volume is constant. Isobaric : a, as for plot a pressure is constant. Isothermal and adiabatic : Slope of plot b is less than slop of plot c. So, plot b is isothermal and plot c is adiabatic. Ans. 8. Option (c) is correct. Given : Case A : amount of heat absorbed by the system is DQA, change in internal energy of the system is DUA, Case B : amount of heat absorbed by the system is DQB, change in internal energy of the system is DUB. f A

P

...(i)

Efficiency of the engine :

η = 1−

PHYSICS

B i

...(ii)

From equations (i) and (ii) : = Q1 100 = , J Q2 90 J Ans. 6. Option (b) is correct. Given : A PV diagram for thermodynamic cycle 1231, process 1 Æ 2 is adiabatic.

V

To find : Compare DQA and DQB and DUA and DUB. For case A : DQA = DUA + WA ...(i) Here WA is the work done corresponding to process A, which is the area under the PV curve for process A. For case B : DQB = DUB + WB ...(ii)

149

THERMODYNAMICS

Here WB is the work done corresponding to process B, which is the area under the PV curve for process B. From the given graph it can be seen that area under the PV curve for process A is larger than area under the PV curve for process B. So, WA > WB ...(iii) Also, as for both the processes the initial and the final states are same, DUA = DUB ...(iv) From (i), (ii), (iii) and (vi), we can say : DQA > DQB Ans. 9. Option (b) is correct. Given : A cyclic process abca,

To find : The temperatures of the source and the sink. Use relation

Tsink  Tsource

Given efficiency of Carnot engine : T T 1 5 ...( i ) = 1 - sink ; sink = 6 Tsource Tsource 6 The efficiency gets doubled when temperature of sink is reduced by 62 K : T - 62 2 = 1 - sink 6 Tsource Tsink - 62 2 = ; Tsource 3

c

P

η = 1-

Tsink 62 2 = Tsource Tsource 3 a

b

62 5 2 1 = - = (using equation i) Tsource 6 3 6

V

Change in internal energy of gas along path ca is DUca = –180 J, amount of heat absorbed by the gas along path ab is DQab = 250 J, amount of heat absorbed by the gas along path bc is DQbc = 60 J. To find : Work done by the gas along the path abc. For process a ® b : DQab = DU ab + Wab

Tsource = 62 ¥ 6 = 372 K = 99∞C

...( ii )

Substitute from equation (ii) in equation (i), Tsink =

5 ¥ 372 = 310 K = 37∞C 6

Ans. 11. Option (a) is correct. Given : Case A : For path ACB, the heat flow in the system is DQACB = 60 J and work done by the 250 = DU ab + Wab ...(i) system is WACB = 30 J, For process b ® c : Case B : For path ADB, the work done by the system DQbc = DU bc + Wbc is WADB = 10 J. 60 = DU bc + 0 ...( ii ) To find : For path ADB, the heat flow in the system, DQADB. (DWbc = 0 as change in volume for process b ® c is Case A : zero) For process c ® a : DQca = DU ca + Wca DQca = -180 + Wca

DQACB = DU ACB + WACB

For the entire cycle abca : DU ab + DU bc + DU ca = 0

Substitute from equation (iv) in equation (i), Wab = 250 - 120 = 130J

DU ACB = DU ADB = 30 J ...( iv )

...( i )

...( ii )

Case B : DQADB = DU ADB + WADB

...( v )

Work done by the gas along the path abc, using equations (v) and (ii), Wab + Wbc = 130 + 0 = 130. J Ans. 10. Option (b) is correct. 1 , the 6 efficiency gets doubled when temperature of sink is reduced by DT = 62°C = 62 K. Given : Efficiency of a Carnot engine is η =

= 60 - 30 = 30J

As for both the cases the initial and final states of the system are same :

From equation (ii), DU ab + 60 - 180 = 0 ; DU ab = 120 J

DU ACB = DQACB - WACB

...( iii )

DQADB = 30 + 10 = 40 J Ans. 12. Option (d) is correct. Given : Carnot engine A : Operates between temperatures T1 = 600 K and T2, Carnot engine B : Operates between temperatures T2 and T3 = 400 K. Work outputs of the two engines are equal. To find : Temperature T2.

150 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Let 1. Q1 be the heat received by the engine A at T1. 2. Q2 be the heat rejected by the engine A at T2. So, Q2 will be the heat received by the engine B at T2. 3. Q3 be the heat rejected by the engine B at T3. Work done by engine A : Q1 – Q2 Work done by engine A : Q2 – Q3 Given : Q1 – Q2 = Q2 – Q3 Q1 + Q2 = 2Q2



Q1 Q3 + Q2 Q2



= 2

1

T3 = ( T1 T42 ) 3

6.0 5 4 p(Pa) 3

1

...( ii )

...( iii )

For equal efficiencies : T T T 1- 2 = 1- 3 =1- 4 T1 T2 T3

...( iv )

Also, from equation (iv), T34 = T22 T42 T33 = T1 T42

4

5 3 V(m )

To find : The work done. Work done = area enclosed by the cycle = area of the right triangle. 1 W = ¥ ( V2 - V1 ) ¥ ( p2 - p1 ) 2 =

1 ¥ ( 5 - 1) ¥ ( 6 - 1) = 10 J 2

Subjective Questions (Chapter Based) Q.1. A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with Vm volume (at 30°C) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If g(besker) = 6 × 10–6 °C–1 and g(mercury) = 1.5 × 10–4 °C–1, where g is the coefficient of volume expansion, then Vm (in cc) is close to _____________. [JEE (Main) – 3rd Sep. 2020 - Shift-1] Sol. Given : Volume of Bakelite beaker is V = 500 cc, temperature of the system is T = 30 °C, volume of mercury filled in the beaker is Vm, Volume of unfilled beaker Vu = V – Vm is constant with temperature, coefficient of volume expansion for the

To find : The value of Vm. Volume of beaker after DT increase in temperature : V′ = V + Vγ beaker ∆T ...(i) Volume of mercury after DT increase in temperature : Vm' = Vm + Vm γ mercury ∆T

T23 = T4 T12 T2 = ( T12 T4

3

−4 expansion for mercury is γ mercury = 1.5 × 10 / °C.

T24 = T32 T12 = T2 T4 T12

1 )3

2

−6 beaker is γ bea ker = 6 × 10 / °C. coefficient of volume

T2 T T = 3 = 4 T1 T2 T3 T22 = T3 T1 ; T32 = T2 T4

B 1

Ans. 13. Option (b) is correct. Given : Carnot engine 1, operates between T1 and T2 , Carnot engine 2, operates between T2 and T3, Carnot engine 3, operates between T3 and T4, T 1 > T 2 > T 3 > T 4. To find : The condition for the three engines to be equally efficient. Efficiency of engine 1 : T ...( i ) η1 = 1 - 2 T1

Efficiency of engine 1 : T η3 = 1 - 4 T3

A

C

2

1000 = 2; T2 = 500 K T2

Efficiency of engine 1 : T η2 = 1 - 3 T2

...(vi)

Equations (v) and (vi) are the required conditions. Ans. 14. Option (b) is correct. Given : A cyclic process CAB.

T1 T3 + = 2 T2 T2

Put given values in the equation above. 600 400 + = 2 T2 T2

PHYSICS

...(v)

...(ii)

As the volume of the unfilled beaker is constant, from equations (i) and (ii) : V′ − Vm′ = V − Vm + ∆T( Vγ beaker − Vmγ mercury ) Vu = Vu + ∆T( Vγ beaker − Vmγ mercury ) Vγ beaker = Vmγ mercury

151

THERMODYNAMICS

Vm = Vm =

Vγ beaker γ mercury 500 × 6 × 10 −6 1.5 × 10 −4

= 20 cc

Q.2. If minimum possible work is done by a refrigerator in converting 100 grams of water at 0°C to ice, how much heat (in calories) is released to the surroundings at temperature 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer? [JEE (Main) – 3rd Sep. 2020 - Shift-2] Sol. Given : Mass of water converted to ice by a refrigerator by doing minimum possible work is M = 100 g, temperature of ice is T2 = 0 °C = 273 K, temperature of surrounding is T1 = 27 °C = 300 K, latent heat of ice is L = 80 cal/g.

Q.4. Starting at temperature 300 K, one mole of an ideal diatomic gas (g = 1.4) is first compressed V adiabatically from volume V1 to V2 = 1 . It is 16 then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then the final temperature of the gas (in °K) is (to the nearest integer) [JEE (Main) – 9th Jan. 2020 - Shift-2] Sol. Given : Initial temperature of one mole of diatomic gas is T1 = 300 K, initial volume of gas is V1, g = 1.4, Process 1 : The gas is adiabatically compressed to V V2 = 1 , 16 Process 2 : The gas then expands isobarically to volume

To find : Q1, the heat released to the surrounding by the refrigerator in the process above. Heat consumed by the refrigerator in converting water in to ice : Q2 = mL = 100 × 80 = 8000 cal

V3 = 2 V2 , Both the processes are quasi static. To find : T, the final temperature of gas. For adiabatic process (process 1) : T1V1γ −1 = T2 V2γ −1

Efficiency of the refrigerator : T1 − T2 Q − Q2 = 1 T2 Q2

(T2 is temperature corresponding to volume V2) γ −1

V  T2 = T1  1   V2 

Q − 8000 300 − 273 = 1 273 8000

For isobaric process (process 2) : V2 V = 3 T2 T3

Q1 = 8791cal Q.3. A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy in J) delivered by the engine to the low temperature reservoir, in a cycle, is _______ . [JEE (Main) – 7th Jan. 2020 - Shift-1] Sol. Given : A Carnot engine operates between two reservoirs of temperature T1 = 900 K, T2 = 300 K, work done by the engine per cycle is W = 1200 J. To find : Q2, the heat energy delivered by the engine to the low temperature reservoir. Efficiency of a Carnot engine : T 300 2 ...(i) η = 1− 2 =1− = T1 900 3 Let heat taken by the engine from high temperature reservoir be Q1 and heat energy delivered by the engine to the low temperature reservoir be Q2. Then by conservation of energy : Q1 − Q2 = W ...(ii) Also,

η = Q1 =

3 3 W = × 1200 = 1800 J 2 2

Substitute from equation (iii) in equation (ii), Q2 = Q1 − W = 1800 − 1200 = 600 J

(T3 is temperature corresponding to volume V3) V  T3 = T2  3  = 300(16 )0.4 × 2 ≈ 1819 K  V2  Q.5. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy. Sol. Given : Initial volume of an ideal monoatomic gas is V1 = V, number of moles in the gas is n = 2, initial temperature of gas is T1 = 27°C = 300K, final volume of gas is V2 = 2V. To find : The final temperature T2 and change in internal energy DU of the gas if the gas expands adiabatically. For an adiabatic process :

Q1 − Q2 W 2 = = Q1 Q1 3

T2 V2γ -1 = T1 V1γ -1 5 Put the given values and γ = for a monoatomic 3 gas. 2

...(iii)

= 300(16 )0.4



2



2

T2 ( 2 V ) 3 = 300 ¥ V 3 2 3 T2 = 300 T2 = 189 K 

152 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)



Change in internal energy of a gas for an adiabatic process : nR( DT ) DU = - DW = γ -1 3 = -2 ¥ 8.3 ¥ ( 300 - 189 ) ¥ = -2.7 kJ 2

Q.6. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cp = b for nitrogen gas The correct relation between a and b is : Sol. Given : For hydrogen gas Cp – Cv = a, for nitrogen gas Cp – Cv = b. To find : Relation between a and b. Mayor’s relation for 1 g mole of gas : Cp - Cv = R Mayor’s relation for n g moles of gas : R Cp - Cv = n

Mayor’s relation for H2 gas : R Cp - Cv = =a 2 Mayor’s relation for N2 gas : R Cp - Cv = =b 28

...( i )

...( ii )





For case A : pi Viγ = p f Vγf ;



and



γ

...( i )

...( ii )

For case B step 1 : W = pi ( Vf - Vi ) = 10 5 ¥ 7 ¥ 10 -3



= 700J



...( iii )

Change in internal energy : 1 DU = ( p V - pi Vi ) γ -1 f f



=



3 Ê 100 3 ˆ 3 ¥ - 100˜ = ¥ - ¥ 100 ¯ 2 2 ÁË 4 4

= -



900 J 8

...(iv)

The amount of heat supplied to the system in the two step process : 900 Q = DU + W = + 700 = 588  J 8

Q.8. C onsider a spherical shell of radius R (numerically equal to the ideal gas constant) at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u =

U µ T­4 and pressure V

1 Ê Uˆ P Á ˜ . It shell now undergoes an adiabatic 3 Ë V¯ expansion the relation between T and R is : Sol. Given : Radius of spherical shell is R, temperature of the shell is T, internal energy per unit volume of an ideal gas of photons inside the shell is U u = µ T4 , ...( i ) V

pressure of the ideal gas is U P = . 3V



Vf = 8 ¥ 10 -3 m 3 ,

Case A : The gas goes from initial state to final state by an adiabatic quasi static process such that p3 V5 = constant. Case B : The gas goes from initial state to final state in two steps: an isobaric expansion at pi followed by an isochoric process at Vf. To find : The amount of heat Q, supplied to the system in the two step process.

pi Ê Vf ˆ = p f ÁË Vi ˜¯

Put given values in equation (i), 5 32 = 8γ ; γ = 3

From equations (i) and (ii), a = 14 b

Q.7. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10–3 m3 Ê 1ˆ changes to a final state at Pf = Á ˜ × 105 Pa Ë 32 ¯ and Vf = 8 × 10–3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps : an isobaric expansion at Pi followed by an isochoric (iso-volumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately. Sol. Given : Initial state of a gas is pi = 105 Pa and Vi =10–3 m3, final state of gas is 1 pf = ¥ 10 5 Pa 32



PHYSICS



...( ii )

To find : The relation between T and R if the shell undergoes an adiabatic expansion. Equation II : U P = 3V PV =

U 3

nRT =

U 3

nRT 1 Ê Uˆ = Á ˜ V 3 Ë V¯ nRT 1 µ T4 V 3

(from equation (i))

153

THERMODYNAMICS



3nR µ VT 3 (3nR is a constant)



VT 3 = constant



or

4 3 3 pR T = constant 3

or RT = constant . That gives, 1 T µ . R

Q.9. Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)...

Sol. Given : Initial temperature of an ideal helium gas in a rubber balloon is T1 = 30°C, number of moles of helium gas in the balloon is n = 2.

To find : Amount of heat required to raise the temperature of the helium gas to T2 = 35°C.



The process mentioned in the question may be assumed isochoric. So, amount of heat required to raise the temperature of the helium gas from T1 = 30°C to T2 = 35°C, will be :



Q = nC p ( T2 - T1 ) = 2¥

5R ¥ 5 = 208 J 2







3PV 3P 3 RT    M M

Cv

Cp

5 3

 

(4  f) (3  f )

s

b rted

De g re

Energy associated with each degree of freedom per molecule =

1 KT 2 B

Pa th

d om Fr ee

fre e M ea n

e of

(f)

and Ki ne ti c Ene



)

2nd 2

1

ssure ( P rgy (E)

Re l ati on be twee n Pre

es Pr

xe ee ur

s y ga

3 2



2  d2 P

kBT

For polyatomic gas: (i)at room temperature, f = 6 (ii)at high temperature, f = 8

For monoatomic gas: f = 3 For diatomic gas : (i) at room temperature, f = 5 (ii) at high temperature, f = 7

A

1 2 P  V rms 3

Total pressure of a mixture of non - reacting gases, P = P1 + P2+...... +Pn

3 PV= 2 KBTN R Here, KB= N

E

e Dal ton’s Law of Parti al Pre ssur

1 V

(If T = constant) PV = constant or P1V1=P2V2

P

y erg En

Hence, f is the degree of freedom

for polyatomic gases,

for diatomic



7  Cv 5

Cp

Cv

Cp

Theor y of Gases

on titi

for monoatomic

Gases

Kinetic Theory of

Ass ump tion of Kinetic

 All the molecules of a gas are identical.  The molecules of different gases are different.  The molecules of gases are in a state of random motion.  The collisions of gases molecules are perfectly elastic.

e

ed

ity ac p a tC

Sp

eed

a He fic i c Spe

s Mo

are Sp

le bab o r tp

s qu

of ar uip q E

Specific Heat Capacity for an ideal gas, Cp– Cv = R

2 RT  2    0.816 vrms  3  vrms M  

  vrms  0.92 v rms 

Root me a n

w La

vmp 

 va v  8 RT   8  M  3

v  v2  v3  ...  vN va v  1 N

rms

v

w La e’s l y Bo

An ideal gas satisfies equation PV = nRT at all pressure and temperature n = no. of moles, R = NAkB universal gas constant

Behaviour

Under the same condition of temperature and pressure equal volumes of all gases contain equal no. of molecules. i.e. N1= N2

Gu yL us sac ’s La w

V  T (If P = constant) V = constant T V 1 V2 or = T1 T2

Charles’s Law

of Gases

P T (If V = constant) P = constant T P P or 1  2 T1 T2

154 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Kinetic Theory of Gases

Chapter 9 Syllabus

Equation of state of a perfect gas; work done on compressing a gas; Kinetic theory of gases : assumptions, concept of pressure. Kinetic energy and temperature : rms speed of gas molecules : Degrees of freedom, Law of equipartition of energy, applications to specific heat capacities of gases  : Mean free path, Avogadro’s number.

Topic-1

LIST OF TOPICS : Topic-1 : Kinetic Theory of Gases

Kinetic Theory of Gases

.... P. 155

Topic-2 :  Degrees of Freedom and Specific Heat Capacities of Gases .... P. 162

Concept Revision (Video Based)    

Part -1   Part -2 Rms Speed of Gas Molecules

  

Kinetic Theory of Gases

Work Done on Compressing a Gas    

 

Mean Free Path     

          

     

Part -1  

Part -2

                        Part -1     Part -2

Avogadro’s Number                 

  

               Part -1           

 

Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true? (A) the mean free path of the molecules decreases (B) the mean collision time between the molecules decreases. (C) the mean free path remains unchanged. (D) the mean collision time relations unchanged. (a) (A) and (D) (b) (A) and (B) (c) (B) and (C) (d) (C) and (D) [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as A Æ B, A Æ C and

A Æ D. The change in internal energies during these process are taken as EAB, EAC and EAD and the work done as WAB, WAC and WAD. The correct relation between these parameters are :

(a) EAB > EAC > EAD, WAB < WAC 0, WAC = 0, WAD > 0 (c) EAB < EAC < EAD, WAB > 0, WAC > WAD (d) EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0 [JEE (Main) – 5th Sep. 2020 - Shift-1]

156 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.3. In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is : (a) 128 (b) 326 1 (c) (d) 32 32 [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.4. In a dilute gas at pressure P and temperature ‘T’, the mean time between successive collision of a molecule varies with T as : 1 (a) T (b) T (c)

T

(d)

1 T

[JEE (Main) – 6th Sep. 2020 - Shift-2] Q.5. Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from t1 to Cp = γ for this gas then a good estimate for t2. It Cv τ2 is given by : τ1 (a) 2g + 1/2

(b)

1 2

γ

1 (c)   2

(d) 2

[JEE (Main) – 7th Jan. 2020 - Shift-1] Q.6. The plot that depicts the behavior of the mean free time t (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is : (Graphs are schematic and not drawn to scale) (a)

(b)

PHYSICS

pressure exerted by the gas molecules will be of the order of : (a) 0 N/m2 (b) 103 N/m2 8 2 (c) 10 N/m (d) 1016 N/m2 [JEE (Main) – 8th April 2019 - Shift-1] Q.9. The temperature at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to : [Boltzmann Constant kB =1.38 × 10–23 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth : 6.4 × 106 m Gravitational acceleration on Earth = 10 ms–2] (a) 800 K (b) 3 × 105 K 4 (c) 10 K (d) 650 K [JEE (Main) – 8th April 2019 - Shift-2] Q.10. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127 °C. At 2 atm pressure and at 227 °C, the rms speed of the molecules will be : (a) 100 m/s (b) 80 5 m/s (c) 100 5 m/s (d) 80 m/s [JEE (Main) – 9th April 2019 - Shift-I] Q.11. A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300 K). The molecular diameter of O2 and its root mean square speed, are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule? (a) ~1012 (b) ~1011 (c) ~1010 (d) ~1013 [JEE (Main) – 10th April 2019 - Shift-1] Q.12. One mole of an ideal gas passes through a process where pressure and volume obey the relation 2 È 1Ê V ˆ ˘ P = Po Í1 - Á 0 ˜ ˙ . Here P0 and V0 are constants. 2Ë V ¯ ˙ ÍÎ ˚ Calculate the change in the temperature of the gas if its volume changes from V0 to 2V0. 5 P0 V0 1 P0 V0 (a) (b) 2 R 4 R 1 P0 V0 3 P0 V0 (d) 4 R 4 R [JEE (Main) – 10th April 2019 - Shift-2] Q.13. The number density of molecules of a gas depends 4 on their distance r from the origin as, n(r) = n0e–a r  . Then the total number of molecules is proportional to : (c)

(c)

(d)

[JEE (Main) – 8th Jan. 2020 - Shift-1] Q.7. Two gases - argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to : (a) 4.67 (b) 2.3 (c) 1.83 (d) 3.67 [JEE (Main) – 9th Jan. 2020 - Shift-2] Q.8. If 1022 gas molecules each of mass 10– 26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104 m/s, the

(a) n0a–3/4

(b) n0 α 1 / 2

(b) n0a 1/4 (d) n0a –3 [JEE (Main) – 12th April 2019 - Shift-2] Q.14. A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass= 40 u) is kept at 300 K in a container. The ratio of their rms È V ( helium ) ˘ speeds Í rms ˙ , is close to : Î Vrms ( argon ) ˚ (a) 3.16 (b) 0.32 (c) 0.45 (d) 2.24 [JEE (Main) – 9th Jan. 2019 - Shift-1]

157

KINETIC THEORY OF GASES

3 RDT 2

(d)

2k DT 3

[JEE (Main) – 11th Jan. 2019 - Shift-2] Q.19. An ideal gas occupies a volume of 2 m3 at a pressure of 3 × 106 Pa. The energy of the gas is : (a) 9 × 106 J (b) 6 × 104 J 8 (c) 10 J (d) 3 × 102 J [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.20. An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 × 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to : (a) 2 × 10–7 s (b) 4 × 10–8 s –8 (c) 0.5 × 10 s (d) 3 × 10–6 s [JEE (Main) – 12th Jan. 2019 - Shift-2] Q.21. A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is l1, and that below the piston is l2, such that l1 > l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by : (R is universal gas constant and g is the acceleration due to gravity)

RT È l1 - 3l2 ˘ Í ˙ ng Î l1 l2 ˚

(b)

RT È 2l1 + l2 ˘ Í ˙ g Î l1 l2 ˚

(c)

nRT È 1 1˘ Í + ˙ g Î l2 l1 ˚

(d)

nRT È l1 - l2 ˘ Í ˙ g Î l1 l2 ˚

[JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (c) 5. (a) 9. (c) 13. (a) 17. (c) 21. (d)

2. (b) 6. (b) 10. (c) 14. (a) 18. (a)

3. (a) 7. (c) 11. (a) 15. (c) 19. (a)

4. (b) 8. (a) 12. (b) 16. (b) 20. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : An ideal gas in a closed container. To find : changes in the characteristics of the gas as the temperature of the container is increased. As the mean free path of gas molecules is independent of temperature, option (C) is correct. Also, as the temperature of gas increases the velocity of the gas molecules will increase, thereby decreasing the mean collision time. So, option (B) is also correct. Ans. 2. Option (b) is correct. Given : a PV diagram for an ideal monoatomic gas,

P

...

D C

>

(c)

(a)

T1 >T2

>

Q.15. A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole] (a) 0.9 kJ (b) 6 kJ (c) 10 kJ (d) 14 kJ [JEE (Main) – 9th Jan. 2019 - Shift–2] Q.16. Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by gas is close to : (Gas constant R = 8.31 J/mol-K) (a) 581 J (b) 291 J (c) 146 J (d) 73 J [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.17. Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ? (a) 103 J (b) 105 J 4 (c) 10 J (d) 106 J [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.18. In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = k, where k is a constant. In this process the temperature of the gas is increased by DT. The amount of heat absorbed by gas is (R is gas constant) : 1 1 (a) RDT (b) kRDT 2 2

B A >

T1 T2

V

EAB is the change in internal energy during the process A Æ B, EAC is the change in internal energy during the process A Æ C, EAD is the change in internal energy during the process A Æ D, work done during the three processes is WAB, WAC, WAD and T1 > T2. To find : The relation between EAC, EAB, EAD and WAB, WAC, WAD As for all the three processes A Æ B, A Æ C, A Æ D, the change in temperature of the gas is (DT = T1 – T2) same. So, EAC = EAB = EAD As DV = 0, for process A Æ C, WAC = 0. As DV = +ve, for process A Æ B, WAB > 0. Ans. 3. Option (a) is correct. 7  Given : The density of a diatomic gas  γ =   5  changes with an adiabatic process as rf = 32ri, the pressure of the gas changes as Pf = nPi. To find : The value of n. For an adiabatic process : PVγ = constant

158 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) From equations (i) and (iii) :

γ

m P  ρ

τ 1 ∝ V × Vγ −(1 / 2 ) ,

= constant

P ∝ ργ

...(i)

From equation (i) : Pf ∝ ρ γf

...(ii)

Pi ∝ ρiγ

...(iii)

From equations (ii) and (iii) :  ρf Pf =   ρi

γ

  Pi 

Pf = ( 32 )7 / 5 Pi = Pf

= 27 Pi 128 Pi

n = 128 Ans. 4. Option (b) is correct. Given : Pressure of the gas is P, temperature of the gas is T. To find : The dependence of mean collision time t, on T. Let l be the mean free path and vrms be the root mean square velocity for the molecules of the given gas. Mean collision time : λ τ = vrms As l is independent of temperature and vrms ∝ T : 1  τ ∝ T Ans. 5. Option (a) is correct. Given : Initial volume of an ideal gas is V, volume of gas after an adiabatic process is 2V initial mean collision time for an ideal gas is t1, mean collision time for gas after an adiabatic process is C τ 2 , P = γ . CV To find :

PHYSICS

τ2 . τ1

Let T1 be initial temperature of the gas and T2 be the temperature after the adiabatic process. For an adiabatic process : 1 1 T1 ∝ (γ −1) ,T2 ∝ ...(i) V ( 2 V )(γ −1) Also, by definition of mean free path (l = vt) : λ ∝ Volume, vrms ∝ Temperature ...(ii) From equation (ii) : V 2V , τ2 ∝ τ1 ∝ T1 T2

...(iii)

τ 2 ∝ 2 V × ( 2 V )γ −(1 / 2 ) τ 1 ∝ Vγ +(1 / 2 ) , τ 2 ∝ ( 2 V )γ +(1 / 2 )

...(iii)

From equation (iii) : τ2 = 2γ +(1 / 2 ) τ1 Ans. 6. Option (b) is correct. Given : t is mean free time for an ideal gas, T is temperature of the gas. To find : The correct graph which depicts variation of t with temperature T. 1 1 As τ ∝ will be a , the plot of t versus T T straight line. So, option (b) is correct. Ans. 7. Option (c) is correct. Given : Atomic radius of Argon atom is r1 = 0.07 nm, atomic weight of Argon is m1 = 40, atomic radius of Xenon atom is r2 = 0.1 nm, atomic weight of Xenon is m2 = 140, both the gases are at same temperature and have same number density. τ To find : 1 , the ratio of their respective mean free τ2 times. Mean free path : V λ = ...(i) 2π d 2 N (d is atomic diameter) Rms velocity : 3k B T vrms = ...(ii) m Mean free time :

τ =

λ vrms

...(iii)

N As the temperature and the number density   V for both the gases is same, from equations (i), (ii) and (iii) : m1 m1 d22 r22 τ1 = × = × τ2 m2 m2 d12 r12

τ1 ( 0.1)2 40 = × ≈ 1.1 2 τ2 140 ( 0.07 ) Ans. 8. Option (a) is correct. Given : Number of gas molecules is n = 1022, mass of each molecule is m = 10–26 kg, area of the surface over which the gas molecules are colliding elastically per second is A = 1 m2, speed of gas molecules is v = 10 4 m/s To find : Order of pressure exerted by the gas molecules on the surface.

159

KINETIC THEORY OF GASES

Change in momentum of a gas molecule colliding elastically with the surface is : Dp = mv - ( - mv ) = 2 mv Change in momentum of n gas molecules colliding elastically with the surface is : Dp = mv - ( - mv ) = 2nmv ...( i ) Force exerted by n gas molecules on the surface : F =

Dp 2nmv = = 2nmv 1 Dt

...( ii )

(Dt = 1 s) Pressure exerted by n gas molecules on the surface: F 2mnv P = = A A =

2 ¥ 10 -26 ¥ 10 22 ¥ 10 4 = 2  1

Ans. 9. Option (c) is correct. Given : Boltzmann constant is kB = 1.38 ¥ 10 -23 J K , Avogadro number is N A = 6.02 ¥ 10 26 kg , 6

radius of earth is R = 6.4 × 10 m, gravitational acceleration on earth is

g = 10 m s2 . To find : The temperature at which the root mean square velocity of hydrogen molecules equals their escape velocity. Root mean square velocity of hydrogen molecules : 3k B T m

vrms 1 = 

...( i )

...( ii )

Equate equations (i) and (ii), 3kB T = m

2 gR

T =

2mgR 3k B

=

2 Ê ˆ ÁË m = 26 ˜ 6.02 ¥ 10 ¯

2 ¥ 2 ¥ 10 ¥ 6.4 ¥ 10 6 3 ¥ 1.38 ¥ 10 -23 ¥ 6.02 ¥ 10 26

= 10 4  Ans. 10. Option (c) is correct. Given :

P1 = 1atm ,vrms 1 = 200 m s , T1 = 127 ∞C = 400 K ,P2 = 2 atm ,

T2 = 227∞C = 500 K. To find : vrms 2 .

...( i )

3RT2 m

...( ii )

Similarly : vrms 2 = 

From equations (i) and (ii), vrms 2 T2 = ; vrms 1 T1 vrms 2

= vrms 1

vrms 2 = 200 ¥

T2 T1

...( iii )

500 = 100 5 m s 400

Ans. 11. Option (a) is correct. Given : volume of cylinder is V = 25 × 10–3 m3, number of moles of O2 gas in the cylinder is n = 1, temperature of gas is T = 300 K, root mean square velocity of O2 molecules is vrms = 200 m s , molecular diameter of O2 molecule is d = 0.3 nm = 0.3 ×10–9 m. To find : Average collision rate for an O2 molecule. Collision frequency : v ν = av ...( i ) λ (l is mean free path and nav is the average velocity) vav =  =

Escape velocity from earth : v E =  2 gR

3RT1 m

Put given values in equation (iii),

So, order of pressure exerted by gas molecules on the surface is 0.

vrms = 

The root mean square velocity of molecules :

λ =

8 vrms 3p 8 ¥ 200 = 184.3 m s 3p

...( ii )

V 2p( d )2 N

In the above equation, N is total number of molecules of oxygen occupying volume V. As n = 1, N = 6.023 ×1026 oxygen molecules.

λ =

25 ¥ 10 -3 2 ¥ 3.14 ¥ ( 0.3 ¥ 10 -9 )2 ¥ 6.023 ¥ 10 26

= 1.1 ¥ 10 -10 m

...( iii )

From equations (i), (ii) and (iii), 184.3 ν = = 1.7 ¥ 1012 s-1  1.1 ¥ 10 -10 Ans. 12. Option (b) is correct. Given : Pressure of an ideal gas as function of volume is È 1 Ê V ˆ2˘ P = Po Í1 - Á o ˜ ˙ , ÍÎ 2 Ë V ¯ ˙˚ P0, and Vo are constants, initial volume of gas is V = Vo, final volume of gas is V = 2Vo.

160 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) To find : The change in temperature DT, of the gas when its volume changes from V1 to V2. Ideal gas equation : PV = RT ...( i )

That gives, N = 4 pno Ú =

...( ii )

È 1 Ê V ˆ2˘ 7 P2 = Po Í1 - Á o ˜ ˙ = Po Í 2 Ë 2 Vo ¯ ˙ 8 Î ˚

• - t -1 / 4

From equations (i), (ii) and (iii), At V = Vo, temperature of the gas is PV PV T1 = 1 o = o o R 2R At V = 2Vo, temperature of the gas is 2 P2 Vo 7 Po Vo = T2 = R 4 R

...( iii )

...( iv )

...( v )

From equations (iv) and (v), 5 Po Vo DT = T2 - T1 = 4 R

4

To find : Total number of molecules in the gas. Consider a thin spherical shell of radius r and thickness dr centred at origin. Number of molecules in this thin spherical shell will be : dN = n( r ). dV ...(ii) In equation (ii), ...(iii)

•

Ú dN = Ú0 n(r ) . dV •

Ú0 no e

-α r 4

...(iv)

. 4p r 2 dr

•

= 4π no Ú e -α r r 2 dr  4

0

...(v)

In the equation above substitute :

α r 4 = t ;4α r 3dr = dt

...(vi)

dt = 4α r

...(vii)

r 2 dr =

dt 3 1 4α 4 t 4

t

dt

...(viii)

dt = k .

Ans. 14. Option (a) is correct. Given : Number of moles of Helium gas in a mixture is n1 = 2, atomic mass of Helium is m1 = 4 u, number of moles of Argon gas in a mixture is n2 = 1, atomic mass of Argon is m2 = 40 u, temperature of the mixture of gases is T = 300 K. vrms, He To find : Ratio of rms speeds . vrms, Ar 3RT m1

...(i)

3RT m2

...(ii)

For Argon : vrms, Ar = 

From equations (i) and (ii), vrms, He m 40 =  2 = = 3.16 vrms, Ar m1 4 Ans. 15. Option (c) is correct. Given : Mass of nitrogen gas is m = 15 g, temperature of nitrogen gas is T = 27 °C = 300 K. To find : Amount of heat transferred to the gas so that rms velocity of the molecules is doubled. Root mean square velocity of molecules : vrms = 

Substitute from equations (i) and (iii) in equation (iv). N =

e t

So, equation (viii) becomes πn N =  3 /o4 k ; N µ noα -3 / 4 α

is the volume of the spherical shell. Total number of molecules : N =

• - t -1 / 4

α 3/4 0

vrms, He = 

...(i)

dV = 4p r 2 dr ,

Ú

3 1

4α 4 t 4

For Helium :

Ans. 13. Option (a) is correct. Given : Number density of molecules of gas as function of distance r from the origin is n( r ) = no e -α r .

pno

e -t dt

As value of a definite integral is a constant. Let

Ú0 e

At V = 2Vo, pressure of the gas is

•

0

At V = Vo, pressure of the gas is È 1 Ê V ˆ2˘ P P1 = Po Í1 - Á o ˜ ˙ = o Í 2 Ë Vo ¯ ˙ 2 Î ˚

PHYSICS

3RT m

...(i)

From equation (i), for rms velocity to double the temperature of the gas should become four times its value. So, amount of heat transferred to the gas will be : m 5R Q = nC V DT = ¥ ¥ ( 4 T - T) M 2 m 15 = ,M is atomic mass of nitrogen and n is M 28 number of moles of the gas) m 5R Q = ¥ ¥ 3T M 2 (n =

=

15 5 ¥ ¥ 8.3 ¥ 3 ¥ 300 = 10 kJ 28 2

161

KINETIC THEORY OF GASES

Ans. 16. Option (b) is correct. Given : Number of moles of an ideal monoatomic 1 gas is n = , pressure of gas is P = 1 atm, initial 2 temperature of gas is T1 = 20 °C = 293 K, final temperature of gas is T2 = 90 °C = 363 K. To find : Work done by the gas. Work done : W = PDV = nRDT (PV = nRT ) ...( i ) Put given values in equation (i): 1 W = ¥ 8.34 ¥ ( 363 - 293) = 291J 2

Ans. 19. Option (a) is correct. Given : Volume of an ideal gas is V = 2 m3, pressure of the ideal gas is P = 3 × 106 Pa. To find : Energy of the gas. Internal energy of the gas : 3 3 U = nRT = PV 2 2 =

Ans. 17. Option (c) is correct. Given : Mass of a monoatomic gas is m = 2kg, pressure of gas is P = 4 ¥ 10 4 N m 2 , density of gas is

ρ = 8 kg m 3 . To find : Order of energy of gas due to its thermal motion. Volume of gas : m 2 1 3 V = = = m ...( i ) ρ 8 4 Internal energy of n moles of monoatomic gas : 3 3 U = nRT = PV (PV = nRT) 2 2 U =

So, amount of heat absorbed by gas : 1 DQ = DW + DU = RDT 2

3 1 ¥ 4 ¥ 10 4 ¥ = 1.5 ¥ 10 4 J 2 4

Ans. 20. Option (b) is correct. Given : Case a : Pressure of an ideal gas enclosed in a cylinder is P1 = 2 atm, temperature of the gas is T1 = 300 K, mean time between two successive collisions of gas molecules is t1 = 6 × 10–8 s, Case b : Pressure of an ideal gas enclosed in a cylinder is P2 = 2P1, temperature of the gas is T2 = 500 K. To find : t2, the mean time between two successive collisions in case b. Mean time between two successive collisions is given as : λ τ = ; v l is mean free path and v is average velocity. For case a : nRT1

τ1 =

Ans. 18. Option (a) is correct. Given : Temperature and volume of an ideal monoatomic gas are related as VT = k, ...( i ) k is a constant, change in temperature of gas is DT, number of moles of gas is n = 1. To find : DQ, amount of heat absorbed by gas. From equation (i) : 1 T µ V and from ideal gas equation, PV = nRT ; PV µ T From equations (i) and (ii), 1 PV µ ; PV 2 = constant V

...( ii )

...( iii )

Equation III represents a polytropic process (PVx = constant) with x = 2. Work done in a polytropic process : PDV nRDT DW = = 1- x 1- 2 = - nRDT = - RDT Change in internal energy of the gas : 3 DU = RDT 2

...( iii )

...(iv)

3 ¥ 3 ¥ 10 6 ¥ 2 = 9 ¥ 10 6 J 2

λ1 = v1

= For case b :

2 p d 2 P1

k B T1 m

8 3p T1 P1

¥ constant

...(i)

nRT1

λ τ2 = 2 = v2

=

T2 P2

2 p d 2 P1 8 3p

k B T2 m

¥ constant

...(ii)

From equations (i) and (ii), T τ2 P =  2 ¥ 1 τ1 P2 T1

τ2 = τ1 ¥

T2 P1 ¥ T1 P2

= 6 ¥ 10 -8 ¥

500 1 ¥ ª 4 ¥ 10 -8 s 300 2

Ans. 21. Option (d) is correct. Given : A vertical closed cylinder is divided into two parts by a frictionless piston of mass m, the length of cylinder above the piston is l1 and the

162 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

length of cylinder below the piston is l2, l1 > l2, number of moles of gas in each part of cylinder is n, temperature of ideal gas in both the parts of the cylinder is T. To find : The mass m of the piston if it is stationary. The force acting downwards on the piston is F1 = P1 A + mg ...( i )

the pressure of gas in the lower part of the cylinder and A is the area of cross section of the cylinder. In equilibrium : P1 A + mg = P2 A

P1A is the force exerted on the piston by the n moles of the gas from the upper part of the cylinder, P1 is the pressure of gas in the upper part of the cylinder and A is the area of cross section of the cylinder. The force acting upwards on the piston is F2 = P2 A ...( ii )

( pV = nRT )

Ê nRT nRT ˆ mg = ( P2 - P1 )A = Á A V1 ˜¯ Ë V2 Ê 1 Êl -l ˆ 1 ˆ = nRT Á 1 2 ˜ mg = nART Á ˜ A l A l Ë 2 Ë l1l2 ¯ 1¯ m =

P2A is the force exerted on the piston by the n moles of the gas from the lower part of the cylinder, p2 is

nRT Ê l1 - l2 ˆ g ÁË l1 l2 ˜¯

Topic-2

Degrees of Freedom and Specific Heat Capacities of Gases Concept Revision (Video Based) Law of Equipartition of Energy

Degrees of Freedom   

      Molar Heat Capacities of Gases

                            Part -1              Part -2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is : (a) 15 (b) 13 (c) 20 (d) 11 [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. Consider a gas of triatomic molecules. The molecules are assumed to be triangular and made of massless rigid rods whose vertices are occupied by atoms. The internal energy of a mole of the gas at temperature T is :

(a)

3 RT 2

(b) 3RT

(c)

5 RT 2

(d)

9 RT 2

[JEE (Main) – 3rd Sep. 2020 - Shift-1]

Q.3 To raise the temperature of a certain mass of gas by 50°C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)? (a) 7 (b) 6 (c) 5 (d) 3 [JEE (Main) – 3rd Sep. 2020 - Shift-2] Cp Q.4. Match the ratio for ideal gases with different Cv type of molecules : Cp Molecule Type Cv (A) Monoatomic

(I)

7 5

(B) Diatomic rigid molecules

(II)

9 7

(C) Diatomic non-rigid molecules (III)

4 3

(D) Triatomic rigid molecules

5 3

(IV)

163

KINETIC THEORY OF GASES

(a) (A) – (III), (B) – (IV), (C) – (II), (D) – (I) (b) (A) – (IV), (B)-(I), (C)– (II), (D) – (III) (c) (A) – (IV), (B) – (II), (C) – (I), (D) – (III) (d) (A) – (II), (B) – (III), (C) - (I), (D) – (IV) [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.5. Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm2, density of mercury = 13.6 g/cm3). (a) 5.8 × 1016 (b) 4.0 × 1016 18 (c) 4.0 × 10 (d) 5.8 × 1018 [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.6 Molecules of an ideal gas are known to have three translational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas and the value of  Cp  γ =  are given, respectively, by :  Cv  7 5 6 (a) U= and γ = (b) U = 5RT and γ = 5 2 5 (c) U = 5RT and γ =

7 5 6 (d) U = RT and γ = 5 2 5

[JEE (Main) – 6th Sep. 2020 - Shift-1] Cp 5 = are mixed Q.7. Two moles of an ideal gas with Cv 3 Cp 4 = . The with moles of another ideal gas with Cv 3 C value of p for the mixture is : (a) 1.50 C v (b) 1.45 (c) 1.47 (d) 1.42 [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.8. Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be Cp rigid) as an ideal gas. Its value will be : Cv 40 23 (a) (b) 27 15 (c)

19 13

(d)

67 45

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.9. Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid and have a mass m. Molecules of the gas B have an m additional vibrational mode and have a mass . 4 A B The ratio of the specific heats (CV and CV ) of gas A and B, respectively is : (a) 5 : 9 (b) 7 : 9 (c) 3 : 5 (d) 5 : 7 [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.10. An HCI molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v–, m is its mass

and kB is Boltzmann constant, then its temperature will be : (a)

mv 2 6 kB

(b)

mv 2 3kB

(c)

mv 2 7 kB

(d)

mv 2 5kB

[JEE (Main) – 9th April 2019 - Shift-1] Q.11. The specific heats, CP and CV of a gas of diatomic molecules, A, are given (in units of J mol–1 K–1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then : (a) A is rigid but B has a vibrational mode. (b) A has a vibrational mode but B has none. (c) A has one vibrational mode and B has two. (d) Both A and B have a vibrational mode each. [JEE (Main) – 9th April 2019 - Shift-2] Q.12. Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K) (a) 19.7 J/mol K (b) 15.7 J/mol K (c) 17.4 J/mol K (d) 21.6 J/mol K [JEE (Main) – 12th April 2019 - Shift-1] Q.13. A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is : 5 2 (a) (b) 3 5 2 3

3 5 [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.14. A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is : (a) 15 RT (b) 20 RT (c) 4 RT (d) 12 RT [JEE (Main) – 11th Jan. 2019 - Shift-1] (c)

(d)

ANSWER – KEY

1. (a) 5. (c) 9. (d) 13. (a)

2. (b) 6. (d) 10. (c) 14. (a)

3. (b) 7. (d) 11. (b)

4. (b) 8. (c) 12. (c)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Given : Number of moles of oxygen in a gas mixture is n1 = 3, number of moles of argon in the same gas mixture is n2 = 5, the temperature of gas mixture is T, the oxygen bond is rigid, the total internal energy of the mixture of gas is U = xRT. To find : The value of x. Number of degrees of freedom for diatomic oxygen considering the oxygen bond is rigid :

164 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) f1 = 5 Number of degrees of freedom for monoatomic argon : f2 = 3 Total internal energy of the mixture of gas : 1 1 U = n1 f1RT + n2 f2 RT 2 2

molecules of gas C are diatomic non-rigid molecules, molecules of gas D are triatomic rigid molecules. CP ,for gas samples A, B, C and To find : The ratio CV D. For gas A : f = 3 CP 2 2 5 = 1+ =1+ = CV f 3 3

1 1  U =  × 3 × 5 + × 5 × 3  RT = 15RT 2 2 

For gas B : f = 5

x = 15 Ans. 2. Option (b) is correct. Given : A triatomic gas molecule has a triangular shape, the temperature of the gas is T, number of moles in the gas sample are n = 1. To find : U, the internal energy of the gas. Number of degrees of freedom for a triatomic gas molecule in the shape of a triangle : f = 3 × number of atoms in a molecule − number of constraints f = 3×3 − 3 = 6

1 × 1 × 6 × RT = 3RT 2

Ans. 3. Option (b) is correct. Given : At constant pressure amount of heat required to raise the temperature of gas by DT1 = 50°C is Q1 = 160 cal, at constant volume amount of heat released when the gas is cooled by DT2 = 100°C is Q2 = 240 cal. To find : f, the number of degrees of freedom of a gas molecule. Let n be number of moles in the sample of gas. At constant pressure : Q1 = nC P DT1 ...(i) At constant volume : Q2 = nC V ∆T2

CP 2 2 7 = 1+ =1+ = f 5 5 CV For gas C : f = 7 CP 2 2 9 = 1+ =1+ = CV f 7 7 For gas D : f = 6 CP 2 2 4 = 1+ =1+ = CV f 6 3

Internal energy : 1 U = ( nfRT ) 2 U =

PHYSICS

...(ii)

From equations (i) and (ii) : C P 50 160 = × C V 100 240 CP 4 = CV 3

...(iii)

CP 2 = 1+ CV f

...(iv)

Also,

From equations (iii) and (iv) : f = 6 Ans. 4. Option (b) is correct. Given : Molecules of gas A are monoatomic, molecules of gas B are diatomic rigid molecules,

Ans. 5. Option (c) is correct. Given : Volume of a monoatomic gas is V = 4 cm3, temperature of the gas is T, pressure of the gas is P = 2 cm of Hg, kinetic energy of a molecule of gas is U = 4 × 10–14 erg, g = 980 cm/s2, density of mercury is r = 13.6 g/cm3. To find : N, number of molecules in the sample of gas. Number of molecules in a perfect gas : PV ...(i) N = kT Kinetic energy of one molecule of monoatomic gas : 3 U = kT ...(ii) 2 From equations (i) and (ii) : 3PV 3 × 13.6 × 980 × 2 × 4 = N = = 4 × 1018 2U 2 × 4 × 10 −14 Ans. 6. Option (d) is correct. Given : Number of translational degrees of freedom for the molecules of a sample of an ideal gas is f ¢ = 3, temperature of the gas is T, number of moles in the sample of gas is n = 1. To find : U, the total internal energy and the value of g for the gas. Let f be total number of degrees of freedom for the molecules of gas. For f ¢ = 3, the closest value of f is f = 5. f 5 = U = RT RT 2 2

γ =

CP 2 2 7 =1+ =1+ = CV f 5 5

165

KINETIC THEORY OF GASES

Ans. 7. Option (d) is correct. Given : Sample 1 contains n1 = 2 number of moles of an ideal gas, C P1 5 = γ1 = 3 C V1 Sample 2 contains n2 = 3 number of moles of an ideal gas, CP2 4 = γ2 = CV2 3 To find :

CP for the mixture of sample 1 and CV

sample 2. For sample 1 :

C P1 =

Rγ 1 5 = R, γ1 − 1 2

C V1 =

R 3 = R γ1 − 1 2

 ...(i)

For sample 2 : C P2 = C V2

Rγ 2 = 4R, γ2 −1

of sample 1 and 2 will be :

Ans. 9. Option (d) is correct. Given : A and B are two ideal diatomic gases at temperature T, molecules of A are rigid and have mass m, molecules of B have one additional m vibrational mode and have mass . 4 A To find : C V : C BV , the ratio of specific heats of gas A and B. As A is a diatomic gas, number of degrees of freedom associated with molecules of A will be :

f1 = 5 B is also diatomic with an additional vibrational mode (1 associated with each atom of molecule of B). So, number of degrees of freedom associated with molecules of B will be : f2 = 5 + 2 × 1 = 7

R = = 3R γ2 −1

From equations (i) and (ii), the

5 7 5 n × R + 2n × R +7 19 2 2 2 = = = 3 3 5 n × R + 2n × R + 5 13 2 2 2

...(ii) CP for the mixture CV

n C + n2 C P 2 CP = 1 P1 n1C V1 + n2C V2 CV 5 2 × R + 3 × 4R 2 = = 1.42 3 2 × R + 3 × 3R 2

Molar heat capacity at constant volume for A : f1R 5R A C = = ...(i) V 2 2 Molar heat capacity at constant volume for B : f2 R 7 R = C BV = ...(ii) 2 2 From equations (i) and (ii) : A CV : C BV = 5 : 7

Ans. 8. Option (c) is correct. Given : Sample 1 contains n1 = n number of moles of monoatomic He gas, Sample 2 contains n2 = 2n number of moles of diatomic Hydrogen gas is. CP To find : for the mixture of sample 1 and CV sample 2. For monoatomic He gas : 3 5 = C V1 = R , C P1 R ...(i) 2 2

Ans. 10. Option (c) is correct. Given : rms velocity of HCl molecules in its gaseous phase is v–, mass of HCl molecule is m, HCl molecule has rotational, translational and vibrational motions. To find : Temperature of the gas. As HCl is a diatomic molecule the total number of rotational, translational and vibrational modes will be n = 7 Average kinetic energy of an HCl molecule, by law of equipartition of energy will be : mv 2 1 7 mv 2 = kB T ; T = 2 7 kB 2 Ans. 11. Option (b) is correct.

...(ii)

Given : C pA = 29 J mol -1K -1 ,CVA = 22 J mol -1K -1 , are specific heats for gas A at constant pressure and constant volume respectively,

C From equations (i) and (ii), the P for the mixture CV of sample 1 and 2 will be : CP n C + n2C P2 = 1 P1 CV n1C V1 + n2C V2

C pB = 30 J mol -1K -1 ,CVB = 21J mol -1K -1 , are specific heats for gas B at constant pressure and constant volume respectively, Both the gases A and B are ideal diatomic gases. To find : Number of vibrational modes for gases A and B.

For diatomic Hydrogen gas : 5 7 = C V2 = R ,C P2 R 2 2

166 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) For gas A :

γA =

C pA C vA

=1+

2 29 = ; fA ª 7 fA 22

...(i)

(fA is number of degrees of freedom for gas A) For gas B : C pB 2 30 =1+ = ; fB ª 5 ...(ii) γB = C vB fB 21 (fB is number of degrees of freedom for gas B) Equations (i) and (ii) show that A has one vibrational mode but B has none. Ans. 12. Option (c) is correct. Given : In a mixture of two gases, number of moles of helium gas is n1 = 2 and number of moles of hydrogen gas is n2 = 3. To find : CV, molar specific heat of mixture of gas at constant volume. Molar specific heat for monoatomic helium gas at constant volume : 3R C V1 = ...(i) 2 Molar specific heat for diatomic hydrogen gas at constant volume : 5R C V2 = ...(ii) 2 Molar specific heat of mixture of gas at constant volume : ( n1 + n2 )C V DT = n1C V1 DT + n2C V 2 DT 3R 5R +3¥ ; 2 2 3R 3R 21R = + = = 17.4 J mol -1 K -1 5 2 10

( 2 + 3)C V = 2 ¥ CV

Ans. 13. Option (a) is correct. Given : A rigid diatomic ideal gas undergoes an adiabatic process at T = 300 K, for the process TV x = constant.

...( i )

To find : Value of x. For a diatomic gas, degrees of freedom at low temperature is f = 5. 2 7 ...(ii) γ = 1+ = f 5 For an adiabatic process, TVγ -1 = constant

...( iii)

Put value of g from equation (ii) in equation (iii). TV 2 / 5 = constant

...(iv)

To find : Total internal energy of the system. Internal energy of oxygen gas in the mixture, by law of equipartition theorem : n f RT U1 = 1 1 ...(i) 2 In equation (i), f1 is number of degrees of freedom for oxygen molecule. As oxygen is a diatomic molecule, f1 = 5( 3 translational + 2 rotational ) modes. That gives, 5 15 U1 = 3 ¥ RT = RT 2 2

...( ii )

Internal energy of argon gas in the mixture, by law of equipartition theorem : n f RT U2 = 2 2 ...(iii) 2 In equation (iii), f2 is number of degrees of freedom for argon molecule. As argon is a monoatomic molecule, f2 = 3. That gives, 3 15 U 2 = 5 ¥ RT = RT ...( ii ) 2 2 Total internal energy of the system : U = U1 + U 2 = 15RT

Subjective Questions (Chapter Based) Q.1. An engine takes in 5 moles of air at 20°C and 1 1 atm, and compresses it adiabatically to of the 10 th original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is : [JEE (Main) – 2nd Sep. 2020 - Shift-1] Sol. Given : Number of moles in a sample of diatomic ideal gas made up of rigid molecules is n = 5, temperature of gas is Ti = 20°C pressure of gas is P = 1 atm, the gas goes through an adiabatic V compression, Vf = i , the change in internal 10 energy of the gas during the adiabatic process is DU = X kJ To find : The value of X. For an adiabatic process : Ti Viγ −1 = Tf Vγf −1

On comparing equations (i) and (iv), 2 x = 5 Ans. 14. Option (a) is correct. Given : In a mixture of gas, number of moles of oxygen is n1 = 3 and number of moles of argon is n2 = 5, temperature of the mixture is T, the gas molecules of the mixture only have translational and rotational modes.

PHYSICS

V Tf = Ti  i  Vf 

γ =

γ −1

   

7 for diatomic gas with rigid molecules. 5 Tf = ( 20 + 273) × (10 )7 / 5 −1 = 736 K

167

KINETIC THEORY OF GASES

∆U =

1 1 fnR( Tf − Ti ) = × 5 × 5 × 8.31 × 443 2 2

∆U = 46016J ∆U = 46kJ

V T ∆P = ∆P nR P

∆T =

300 ∆P = 150 ∆P 2

C = 150

X = 46 Q.2. A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ____ [JEE (Main) – 4th Sep. 2020 - Shift-1] Sol. Given : Sample 1 contains n1 = 0.1 moles of monoatomic ideal gas at temperature T1 = 200 K, sample 2 contains n2 = 0.05 moles of the same monoatomic ideal gas at temperature T2 = 400 K. To find : T, the temperature of the mixture of two samples. n1T1 + n2 T2 = ( n1 + n2 )T 0.1 × 200 + 0.05 × 400 = ( 0.1 + 0.05) × T T ª 267K Q.3. The change in the magnitude of the volume of an ideal gas when a small additional pressure DP is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity DT at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively. If [DT] = C|DP| then value of C in (K/atm) is ........... [JEE (Main) – 4th Sep. 2020 - Shift-2] Sol. Given : Case I : At constant temperature T = 300 K, the pressure of an ideal gas is changed from P = 2 atm to P + DP, the corresponding change in volume is DV. Case II : At constant pressure P = 2 atm, the temperature of an ideal gas is changed from T = 300 K to T – DT, the corresponding change in volume is DV. ∆T = C ∆P ...(i) To find : The value of C. For case I : PV = nRT

...(ii)

P∆V + V∆P = 0 ∆V = −

V ∆P P

...(iii)

For Case II : PV = nRT P∆V = nR∆T ∆V =

∆T =

nR ∆T P

From equations (ii), (iii) and (iv) : V nR − ∆P = ∆T P P

...(iv)

Q.4. Nitrogen gas is at 300°C temperature. The temperature (in K) at which the rms speed of a H2 molecule would be equal to the rms speed of a nitrogen molecule, is ..........(Molar mass of N2 gas 28 g). [JEE (Main) – 5th Sep. 2020 - Shift-2] Sol. Given : Sample 1 contains nitrogen gas, the molar mass of nitrogen gas is MN = 28 g, temperature of nitrogen gas is TN = 300 °C = 573 K, sample 2 contains hydrogen gas, the molar mass of hydrogen gas is MH = 2 g. To find : TH, the temperature of hydrogen gas at which the rms speed of hydrogen molecules will be same at the rms speed of the nitrogen molecules of sample 1. Rms velocity for the molecules of sample 1 : N = vrms

3RTN MN

...(i)

Rms velocity for the molecules of sample 2 : H = vrms

3RTH MH

...(ii)

From equations (i) and (ii) : TN T = H MN MH TH =

MH 2 TN = × 573 MN 28

TH = 41K Q.5. Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a P volume 2V1 is given by P2. The ratio 2 is : P1 th [JEE (Main) – 6 Sep. 2020 - Shift-1] Sol. Given : Initial volume of a diatomic gas is V1, the initial pressure is P1, the initial temperature is T1 = 250 K and the number of molecules in the sample is n1, 25% of gas molecules dissociate, the final number of molecules in the sample is n2, the resulting temperature is T2 = 2000 K final volume is V2 =2V1, final pressure is P2, P To find : the ratio 2 . P1 n2 = 0.75n1 + 2 × 0.25n1 = 1.25n1 Ideal gas equation for initial state of gas : P1V1 = n1RT1

...(i)

168 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ideal gas equation for final state of gas : P2 V2 = n2 RT2

...(ii)

From equations (i) and (ii) : P2 n V T 1.25 2000 = 2× 1× 2 = × P1 n1 V2 T1 2 250





with which the molecules rebound after elastic collision is v = 103 m/s. To find : Pressure exerted by the molecules on the wall. Force Pressure = Area

P2 = 5 P1 Q.6. An engine operates by taking a monoatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to ____

PHYSICS

P =



=



n¥

changeinmomentum  collision Area

n ¥ 2mv cos 45∞ A 10 23 ¥ 2 ¥ 3.32 ¥ 10 -27 ¥ 10 3 ¥

=



[JEE (Main) – 6th Sep. 2020 - Shift-2] Sol. Given : The PV graph shown below depicts operation of an engine using monoatomic gas.

To find : h, the efficiency of the engine. Work done by the engine : W = areaofthesquareABCD = 2 Po Vo

...(i)

Heat intake by the engine : Q = WAB + WBC Q = nC V ∆TAB + nC P ∆TBC For monoatomic gas : 3 5 CV = and C P = 2 2 Q =

3 5 ( 3Po Vo − Po Vo ) + ( 6 Po Vo − 3Po Vo ) 2 2

21 Q = Po Vo 2

η =

W 4 = = 19% Q 21

Q.7. The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then what is the pressure on the wall? Sol. Given : Mass of hydrogen molecule is m = 3.32 × 10–27 kg, number of collisions per second is n = 1023, area of wall is A = 2 cm2, angle at which the molecules strike the wall is q = 45°, the speed

2 ¥ 10 -4

1 2

= 2.35 ¥ 10 3 N m 2  Q.8. The temperature of an open room of volume 30 m3 increases from 17 °C to 27 °C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then what is nf – ni? Sol. Given : Volume of an open room is V = 30 m3, initial temperature of room is Ti = 17 °C = 290 K, final temperature of room is Tf = 27 °C = 300 K, atmospheric pressure in the room during the heating from sunshine is p = 1 × 105 Pa, initial number of air molecules in the room is ni, final number of air molecules in the room is nf. To find : nf – ni Ideal gas equation, before the heating of the room : n pV = i RTi ...( i) NA Ideal gas equation, after the heating of the room : nf pV = RTf ...(ii) NA From equations (i) and (ii), pVN A pVN A n f - ni = RTf RTi pVN A Ê 1 1ˆ = - ˜ Á R Ë Tf Ti ¯ n f - ni =

1 ¥ 10 5 ¥ 30 ¥ 6.023 ¥ 10 23 8.3 1 ˆ Ê 1 ÁË 300 - 290 ˜¯

10 ˆ Ê = 2.2 ¥ 10 29 Á Ë 87000 ˜¯ Q.9. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, find (A) The average energy per mole of the gas mixture. (B) The ratio of the rms speed of helium atoms to that of hydrogen molecules.

169

KINETIC THEORY OF GASES

Sol. Given : In a mixture of gas, number of moles of hydrogen molecules is n1 = 1 and number of moles of helium atoms is n2 = 1, temperature of the mixture is T.



To find : E, the average energy per mole of the vrms ,He , the ratio of rms speed of gas mixture and vrms ,H 2 helium atoms to that of hydrogen molecules. Internal energy of hydrogen molecules : f 5 U1 = 1 n1 RT = RT ...(i) 2 2 Internal energy of helium atoms : f 3 U 2 = 2 n2 RT = RT 2 2

From equations (i) and (ii), total internal energy of the mixture : U = U1 + U 2 = 4RT ...(ii) Average energy per mole of the gas mixture : U 4 RT ...(iii) E = = = 2 RT 2 n1 + n2 rms speed of helium atoms : vrms ,He



...(ii)

3RT =  mHe

vrms ,H 2

=

mH 2

=



C - CV =

CP - CV 1-n

1-n =

CP - CV C - CV

n = 1-

CP - CV C - CP = C - CV C - CV

Q.11. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. What is the ratio of their densities? Sol. Given : In a gas mixture, ratio of atomic masses of two monoatomic ideal gases is

...(v)



1 2

...(v) vrms ,H 2 mHe Q.10. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then what is n? Sol. Given : During a quasi-static reversible process molar heat capacity, C = constant and PVn = constant.

To find : Value of n In a process where, PVn = constant, the molar heat capacity is given as : R R R C = + = CV + γ -1 1-n 1-n



...(iv)

From equations (iv) and (v), vrms ,He





rms speed of hydrogen molecules : 3RT =  mH 2





m1 2 =  m2 3 and ratio of their partial pressures is P1 4 = , P2 3 the temperature of the mixture is T.

ρ1 , ratio of densities of the two gases in ρ2 the gas mixture. ρ1 m V = 1 ¥ 2 (density is mass/volume) ρ2 V1 m2 To find :

ρ1 mP RT = 1 1¥ ρ2 RT P2 m2



(use ideal gas equation : rV = RT)

Êm ˆÊ ρ ˆ 2 4 8 ρ1 = Á 1˜Á 1˜ = ¥ = . ρ2 Ë m2 ¯ Ë ρ 2 ¯ 3 3 9



Frequency

F = spring constant x

o ue t ns d o i t a cill Os

ing a spr

x = xme-bt/2m sin('t+φ), here, '=

Damping force Fd= –bv, b = damping constant – Displacement of damped oscillator

x = xme-(b/2m)t

b2 m – 4m2 k

ns io at

en

SH

E

Velocity in SHM

dv =− 2A cos (t +φ) dt a max= 2 A a=

dx(t) = – A sin(t+φ ) d (t ) vmax = A v=

2

2

Fo 2 2 d

2 2 d

{m ( −  ) + b } −vo tanφ = d xo

Amplitude, A =

Displacement, x(t)=A cos (d t+φ ) 1 2

When frequency of external force is equal to natural frequency of oscillator.

= k m

F= −k ( x) , k = m2

Force law equation for SHM

Every oscillatory motion is periodic, Kinetic Energy but every periodic motion need not be K.E = m2A2 sin2 (t + ) oscillatory. To and fro motion repeatedly about a fixed point in a definite interval of time. Potential Energy 1 P.E = m 2 A2 cos2 (t +φ) 2 Total energy, T.E.=K.E.+ P.E. O sc illa 1 1 tio = kA2 cos 2 ( t+ φ) + sin2 ( t +φ)  = KA2 ns 2 2

Driving force, F (t) = Fo cosd t





Oscillations & Wave Part-1

d

Amplitude decreases with time as

m lace Disp

n ti

A= Amplitude i.e.,maximum displacement of particles.

Simple Harmonic Motion is The smallest interval of time T ) im e the simplest form of p eri o d ( T after which the motion oscillatory motion is repeated

Phase

ce

Oscillation of a body whose amplitude goes on decreasing with time.

Here, k =

m k

I l =2 π mgl g

Time period T =2 π

Time period T = 2 π

The number of oscillations per second i.e., 1  f= = T 2π

Time varying quantity  t   

SHM



x(t)= A cos(t+)

Fo r

D a mp ed

HM nS yi g r ne

O sc ill

M

Phase constant or Phase angle () It depends upon velocity  v and displacement of particle at t = 0

170 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

2l

m



4l

at

io n

(iii)  In open organ pipe, higher harmonics, both odd & even.  In closed organ pipe, higher harmonics, odd only.

v Fundamental freq. or I harmonic   2l (b) Closed at one end  v

(ii) In organ pipe: (a) Open at both ends,

(i) In string: Fundamental Frequency,   1 T

Waves associated with particles like-electrons, protons, neutrons, atoms, molecules etc.

tter Ma

po When two or more waves are Principle of super propagating with different s displacements then the net o f w a ve displacement of collective wave is given as Y = y1 + y2 + y3+.....+yn ves Wa

on siti

ar y Waves

1 Frequency

Wave number, v =

Types

Do not require any material medium for their propagation e.g., light waves

1 

)

V 

na udi t i g n Lo s wave

l

√

Individual particles of the medium oscillate along the direction of propagation of wave.

√ √

Speed of a longitudinal wave speed of sound, V = B  Y P =  =  for air, γ = 7/5

Require material medium for their propagation

rs

Individual particles of the medium oscillate perpendicular to the direction of propagation of wave

T = tension & m = mass/length

ve ns Tra

√ mT ,

Speed of transverse wave in a stretched string

Difference in frequencies of two superposing waves, beat= 1  2

(

 t) = A sin (kx +wt+) y (x, y (x,t) = Asin 2  t – x T    amplitude Velocity v0=A2 Acceleration Amplitude a0 = A2

Essential properties for propagation  Elasticity  Inertia  Minimum friction

a el t r ve n e si em res c la og ts sp pr Di Bea

Mech anic al W ave s

Oscillation & Waves Part-2

n gt tio len a l e Re wav

wavelength

1

=

= 1

Angular Frequency () = 2 π

v = λ Time period (T) =

b/ w h wa ( ve ) & v elo fre c it q u e y ( v), ncy ()

Stationary waves

Electromagnetic Waves

e la n

tio w n in av ap e

Properties

St

e

le

es

po

av

Do ct ffe E r

w

General Formula, Apparent Frequency  v  v  vm 0 n’ n   ∓ v   v s m  v Here, v = speed of sound, v0 = observer speed vs = source speed vm = medium speed

OSCILLATIONS AND WAVES

171

Oscillations and Chapter 10 Waves Syllabus Periodic motion - period, frequency, displacement as a function of time; Periodic functions; Simple harmonic motion (S.H.M.) and its equation; phase; oscillations of a spring - restoring force and force constant; energy in S.H.M. - kinetic and potential energies; Simple pendulum - derivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion - Longitudinal and transverse waves, speed of a wave; Displacement relation for a progressive wave; Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound. Linear and angular simple harmonic motion, progressive waves, resonance, speed of sound in gases.

Topic-1

LIST OF TOPICS : Topic-1 : Simple Harmonic Motion

Simple Harmonic Motion

Topic-2 :  Wave Motion

.... P. 172 .... P. 184

Concept Revision (Video Based)    

Part -1    Part -2

 

 

Simple Harmonic Motion

Simple Pendulum                  

Damped Oscillations and Resonance

   

Part -1   Part -2

Periodic Motion

displacement

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Two identical strings X and Z made of same material have tension Tx and Tz in then If their fundamental frequencies are 450 Hz and 300 Hz, T respectively, then the ratio x is : Tz (a) 1.25 (c) 1.4

(b) 0.44 (d) 2.25 [JEE (Main) – 2nd Sep. 2020 - Shift-1]

Q.2. The displacement time graph of a particle executing SHM is given in figure :

(sketch is schematic and not to scale)



2T 4 O. . . . . . T T 5T 4

3T 4

time (s)

4

Which of the following statements is/are true for this motion? 3T (A) The force is zero at t = 4 (B) The acceleration is maximum at t = T T (C) The speed is maximum at t = 4

173

OSCILLATIONS AND WAVES

(D) The P.E. is equal to K.E. of the oscillation at T t= 2 (a) (B), (C) and (D) (b) (A) and (D) (c) (A), (B) and (D) (d) (A), (B) and (C) [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope? (a) 12 (b) 3 (c) 9 (d) 6 [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.4. A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is : 1 1 (a) (b) 2 2 (c)

(d) 1 [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.5. An object of mass m is suspended at the end of a massless wire of length L and area of cross-section, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is : 2

1 (a) f = 2p

mL YA

1 (b) f = 2p

YL mA

1 2p

mA YL

(d) f =

1 2p

YA mL

(c) f =

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.6. When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion, is described by y(t) = y0sin2wt, where ‘y’ is measured from the lower end of upstretched spring. Then w is : (a)

1 g 2 y0

(b)

g y0

(c)

2g y0

(d)

g 2 y0

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.7. A wire of length L and mass per unit length 6.0 × 10–3 kg m–1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are 420 Hz and 490 Hz. Then L in meters is : (a) 8.1 m (b) 2.1 m (c) 1.1 m (d) 5.1 m [JEE (Main) – 9th Jan. 2020 - Shift-2]

A

Q.8.

B

L L A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is : (a) 3 : 5 (b) 4 : 9 (c) 1 : 2 (d) 1 : 4 [JEE (Main) – 8th April 2019 - Shift-1] Q.9. A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it 1 will take to drop to of the original amplitude 1000 is close to : (a) 50 s (b) 100 s (c) 20 s (d) 10 s [JEE (Main) – 8th April 2019 - Shift-2] Q.10. A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid 1 is th of the material of the bob. If the bob is 16 inside liquid all the time, its period of oscillation



in this liquid is : (a) 2 T

1 10

(b) 2 T

1 14

(c) 4 T

1 15

(d) 4 T

1 14

[JEE (Main) – 9th April 2019 - Shift-1] Q.11. A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin (0.157x) cos (200pt). The length of the string is : (All quantities are in SI units.) (a) 20 m (b) 80 m (c) 40 m (d) 60 m [JEE (Main) – 9th April 2019 - Shift-1] Q.12. A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K) (a) 10–4 K (b) 10–5 K –1 (c) 10 K (d) 10–3 K [JEE (Main) – 9th April 2019 - Shift-2] Q.13. A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is : (a) 180 m/s, 80 Hz (b) 320 m/s, 80 Hz (c) 320 m/s, 120 Hz (d) 180 m/s, 120 Hz [JEE (Main) – 9th April 2019 - Shift-2]

174 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.14. The displacement of a damped harmonic oscillator is given by x(t) = e–0.1t cos (10pt + j). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to : (a) 4 s (b) 7 s (c) 13 s (d) 27 s [JEE (Main) – 10th April 2019 - Shift-1] Q.15. A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constants, k1 and k2 will be : 1 (a) n (b) 2 n 1 (c) (d) n2 n [JEE (Main) – 12th April 2019 - Shift-2] Q.16. A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, v is equal to : (a) 332 m s–1 (b) 384 m s–1 –1 (c) 338 m s (d) 379 m s–1 [JEE (Main) – 12th April 2019 - Shift-2] Q.17. A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m (m t0 t th

t0 > t

[JEE (Main) – 6 Sep. 2020 - Shift-1] Q.7. A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer

185

OSCILLATIONS AND WAVES

hears 2 beats/sec. The oscillation frequency of each tuning fork is n0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to : 1 (a) 1 m/s (b) m/s 2 (c)

1 m/s 8

(d)

(a)1. 0

t(s) 2

y

(b) 2. 0

t(s) 2

t(s) 1

2

y

y

1

(c) 3. 0

1 m/s 4

[JEE (Main) – 7th Jan. 2020 - Shift-2] Q.8. Three harmonic waves having equal frequency n p and same intensity I0 have phase angles 0, and 4 p − respectively. When they are superimposed 4 the intensity of the resultant wave is close to : (a) 0.2 I0 (b) I0 (c) 3 I0 (d) 5.8 I0 [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.9. The pressure wave, P = 0.01 sin [1000t – 3x] N m–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 m s–1. Approximate value of T is : (a) 4°C (b) 11°C (c) 12°C (d) 15°C [JEE (Main) – 9th April 2019 - Shift-1] Q.10. Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 m s–1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B? (speed of sound in air = 340 m s–1) (a) 2250 Hz (b) 2060 Hz (c) 2300 Hz (d) 2150 Hz [JEE (Main) – 9th April 2019 - Shift-2] Q.11. A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in m s–1, (Given speed of sound = 300 m/s) (a) 12,16 (b) 12, 18 (c) 16, 14 (d) 8, 18 [JEE (Main) – 10th April 2019 - Shift-1] Q.12. The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz, is :

1

y

(d)

4. 0

t(s) 2

1

[JEE (Main) – 10th April 2019 - Shift-2] Q.13. A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s) (a) 750 Hz (b) 857 Hz (c) 1143 Hz (d) 807 Hz [JEE (Main) – 10th April 2019 - Shift-2] Q.14. A progressive wave travelling along the positive x-direction is represented by y(x, t) = Asin (kx – wt + f). Its snapshot at t = 0 is given in the figure. y A x



For this wave, the phase f is : p (a) - (b) p 2 (c) 0

(d)

p 2

[JEE (Main) – 12th April 2019 - Shift-1] Q.15. A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency n. The value of n is close to : (Speed of sound in water =1500 m s–1) (a) 504 Hz (b) 507 Hz (c) 499 Hz (d) 502 Hz [JEE (Main) – 12th April 2019 - Shift-1] Q.16. Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from sources S1, towards S2 with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u equals : (a) 5.5 m/s (b) 15.0 m/s (c) 2.5 m/s (d) 10.0 m/s [JEE (Main) – 12th April 2019 - Shift-2]

186 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.17. A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? (Given reference intensity of sound as 10–12 W/m2) (a) 40 cm (b) 20 cm (c) 10 cm (d) 30 cm [JEE (Main) – 12th April 2019 - Shift-2] Q.18. A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to : (a) 666 Hz (b) 753 Hz (c) 500 Hz (d) 333 Hz [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.19. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the speed of the train is reduced to 17 m/s, the frequency registered is f2. If speed of sound is 340 m/s, then the ratio f1/f2 is : 18 19 (a) (b) 17 18 (c)

20 19

(d)

21 20

[JEE (Main) – 10th Jan. 2019 - Shift-1] Q.20. A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz) (a) 6 (b) 4 (c) 7 (d) 5 [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.21. Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin (450t – 9x) where distance and time are measured in SI units. The tension in the string is : (a) 5 N (b) 7.5 N (c) 10 N (d) 12.5 N [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.22. A travelling harmonic wave is represented by the equation y(x, t) = 10–3 sin (50t + 2x), where x and y are in metre and t is in seconds. Which of the following is a correct statement about the wave? (a) The wave is propagating along the negative x-axis with speed 25 m s–1. (b) The wave is propagating along the positive x-axis with speed 100 m s–1. (c) The wave is propagating along the positive x-axis with speed 25 m s–1. (d) The wave is propagating along the negative x-axis with speed 100 m s–1 [JEE (Main) – 12th Jan. 2019 - Shift-1]

PHYSICS

ANSWER – KEY

1. (c) 5. (d) 9. (a) 13. (a) 17. (a) 21. (d)

2. (d) 6. (d) 10. (a) 14. (b) 18. (a) 22. (a)

3. (b) 7. (d) 11. (b) 15. (d) 19. (b)

4. (a) 8. (d) 12. (c) 16. (c) 20. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : Distance between two peaks of a wave is 5 m, distance between one crest and one trough is 1.5 m. To find : The possible values of the wavelength l. Distance between two peaks : 5 = n1λ ...(i) Distance between one crest and one trough : λ 1.5 = ( 2n2 + 1) ...(ii) 2 From equations (i) and (ii), where n1 and n2 are integers : n1 5 = 2 n2 + 1 3 3n1 − 10n2 − 5 = 0 n2 = 1, n1 = 5 ; λ = 1 n2 = 4 ,n1 = 15 ; λ =

1 3

n2 = 7 ,n1 = 25 ; λ =

1 5

Ans. 2. Option (d) is correct. Given : Velocity of sound is v = 330 m/s, frequency of bus horn is f = 420 Hz, frequency of the horn measured by the driver after reflection from the wall is f ¢ = 490 Hz. To find : v¢, velocity of the bus. Frequency of the horn as measured by the driver after reflection from the wall : v + v′ f′ = f v − v′ 490 =

330 + v′ × 420 330 − v′

7 330 + v′ = 6 330 − v′ 2310 − 7 v′ = 1980 + 6 v′ v′ = 25.38 m s ≈ 91 km hr Ans. 3. Option (b) is correct. Given : Displacement of air is proportional to the pressure difference created by sound wave, s ∝ DP, DP = 10 Pa, also s depends on speed of sound (v = 300 m/s) density of air (r = 1 kg/m3) and frequency of sound (f = 1000 Hz) To find : The order of s.

187

OSCILLATIONS AND WAVES

By dimensional analysis and taking multiplicative factor to be 1 : P s = ρ vf s =

10 1 = × 10 −4 m 1 × 300 × 1000 3

s ª

3 mm 100

Ans. 4. Option (a) is correct. Given : A tube filled with water resonates with a tuning fork when the height of the water column is h1 = 17.0 cm, next resonance occurs at h2 = 24.5 cm, velocity of sound in air is v = 330 m/s. To find : f, the frequency of the tuning fork. λ = h2 − h1 2 f = f

=

v v = λ 2( h2 − h1 )

330 × 100 = 2200 Hz 2( 24.5 − 17.0 )

Ans. 5. Option (d) is correct. Given : Velocity of sound is v = 345 m/s, frequency of car horn is f = 440 Hz, frequency of the horn measured by the driver after reflection from the wall is f ¢ = 480 Hz. To find : v¢, velocity of the car. Frequency of the horn as measured by the driver after reflection from the wall : v + v′ f′ = f v − v′ 480 =

345 + v′ × 440 345 − v′

12 345 + v′ = 11 345 − v′ 4140 − 12 v′ = 3795 + 11v′ v′ = 15 m s ª 54 km h Ans. 6. Option (d) is correct. Given : Velocity of sound source moving along a straight track is v, frequency of sound emitted by the source is n0, an observer is located at a finite distance at point O from the track. To find : Variation of the frequency of sound as measured by the observer n as function of time t. Let the distance between the observer and the source be minimum at t = to. For t < to, the source is approaching the observer and for t > to the source is receding from the observer. a. For t < to : value of n will be largest for t = 0 and decreases as the source approaches the observer, reaching no at t = to.

b. For t > to : value of n increases with time as the source recedes from the observer. Both points a and b are satisfied by graph 4. Ans. 7. Option (d) is correct. Given : Velocity with which source-1 approaches the observer is equal to velocity with which source-2 recedes from the observer = v¢, the observer hears fbeat = 2 beats/s, the oscillation frequency of both the sources is fo = 1400 Hz, velocity of sound in air is v = 350 m/s. To find : Value of v¢. The source-1 is approaching the observer, so the apparent frequency of sound coming from source-1 as measured by the observer will be :  v + v′  f1 = fo    v 

...(i)

The source-2 is receding away from the observer, so the apparent frequency of sound coming from source-2 as measured by the observer will be :  v − v′  f2 = fo    v 

...(ii)

Beat frequency as measured by the observer : fbeat = f1 − f2 From equations (i) and (ii) :  v + v′   v − v′  fbeat = fo   − fo  v   v     350 + v′   350 − v′  2 = 1400   − 1400  350     350  8 v′ = 2 v′ =

1 m s 4

Ans. 8. Option (d) is correct. Given : Three harmonic waves have same frequency n and same intensity Io, the phase angles for the three waves are

φ1 = 0 , φ2 =

p p rad, φ 3 = − rad. 4 4

To find : I, the intensity of the resultant of the three harmonic waves. A

/4

A

/4

A

Let the amplitude of the three harmonic waves be : A =

Io

...(i)

188 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

The resultant of the three amplitudes shown in the figure above will be : p  p A′ = A + A cos + A cos  −  4  4 2A = A+ = ( 2 + 1)A ...(ii) 2

To find : fo, the natural frequency of the sound source in car B. The apparent frequency as measured by an observer in car A: Ê v + vo ˆ fo f = Á Ë v - vs ˜¯

From equations (i) and (ii) :

Here vs is the velocity of the sound source (car B) and vo is the velocity of the observer (car A). Ê 340 - 20 ˆ 2000 = Á ¥ f ; f = 2250 Hz Ë 340 + 20 ˜¯ o o

I = A′2 = [( 2 + 1)A]2 = ( 2 + 1)2 A 2 = 5.8 I o Ans. 9. Option (a) is correct. Given : Pressure wave corresponding to the sound produced by a vibrating blade when atmospheric temperature is T = 0°C = 273 K is p = 0.01 sin (1000t - 3x ), ...(i) for some other atmospheric temperature T¢ the speed of the sound produced by the blade is v = 336 m s , for both the temperatures the frequency of the blade is same. To find : Approximate value of T¢. General equation of pressure wave of sound : p = p0 sin (ωt - kx )

...(ii)

Compare equations (i) and (ii) : ω = 1000 ,k = 3

...(iii)

Velocity of the wave at T = 0∞C = 273 K : v0 =

=

ω k

1000 m s 3

γ RT m

...(iv)

...(v)

γ R ¥ 273 1000 = m s m 3

...(vi)

Put T = T¢ in equation (v) : v0 = 

γ RT ' = 336 m s m

Ê 300 - v1 ˆ ; 480 = 500 Á Ë 300 ˜¯ 480 ¥ 3 = 12 m s. 5

As f2 > f0, observer-2 is moving towards the sound source. So, apparent frequency measure by observer-2 : Ê v + v2 ˆ f2 = f0 Á Ë v ˜¯ Ê 300 + v2 ˆ ; 530 = 500 Á Ë 300 ˜¯

Put T = 0°C = 273 K in equation (v) : v0 = 

To find : v1 and v2, speed of observer-1 and observer-2. As f1 < f0, observer-1 is moving away from the sound source. So, apparent frequency measure by observer-1 : Ê v - v1 ˆ f1 = f0 Á Ë v ˜¯

v1 = 300 -

Also, velocity of a sound wave at a temperature T  : v = 

Ans. 11. Option (b) is correct. Given : Frequency of sound waves emitted by a stationary source is f0 = 500 Hz, frequency measured by observer-1 is f1 = 480 Hz, frequency measured by observer-2 is f2 = 530 Hz, speed of sound in air is v = 300 m / s.

...(vii)

Divide equation (vii) by equation (vi) :

γ RT ' 336 m = 1000 γ R ¥ 273 3 m T' 1008 = ; T = 277 K = 4∞C 273 1000

Ans. 10. Option (a) is correct. Given : speed of car A = speed of car B = 20 m/s, both the cars are moving in opposite directions, frequency of sound source located in car B as measured in car A is f = 2000 Hz, speed of sound in air is v = 340 m/s.

v2 =

530 ¥ 3 - 300 = 18 m s. 5

Ans. 12. Option (c) is correct. Given : Frequency of wave 1 is f1 = 9 Hz and frequency of wave 2 is f2 = 11 Hz. To find : Choose the correct wave pattern produced by the superposition of the given two waves. Beat frequency : fbeat = f2 - f1 = 11 - 9 = 2Hz Time interval between beats : 1 1 T = = = 0.5 . s fbeat 2 So, the resultant wave from superposition of two given waves will have time period of 0.5 s. Ans. 13. Option (a) is correct. Given : Velocity of sound source moving towards an observer is vs = 50 m s ,

189

OSCILLATIONS AND WAVES

Frequency measured by the stationary observer is f = 1000 Hz, velocity of sound in air is v = 350 m s. To find : f ¢ apparent frequency of the sound source when it is moving away from the observer after crossing him. Apparent frequency measured by an observer when source is moving towards him : Ê v ˆ f = fo Á ; Ë v - vs ˜¯ fo = =

f ( v - vs ) 1000( 350 - 50 ) = v 350 6000 Hz 7

Ê v ˆ f ¢ = fo Á Ë v + vs ˜¯ 6000 Ê 350 ˆ ¥Á = 750Hz Ë 350 + 50 ˜¯ 7

Ans. 14. Option (b) is correct. Given : A progressive wave travelling along the x direction, y( x , t ) = A sin ( kx - ωt + φ ), ...(i) a snapshot of the wave at t = 0. To find : The phase f. For the given snapshot, at t = 0; x = 0 and y = 0. From equation (i) the above condition is only possible for f = p. Ans. 15. Option (d) is correct. Given : Speed of submarine A is 18000 v A = 18 km h = m s = 5 m s , 3600

ª 502Hz

f = 10 beats s ,  velocity of sound in air is v = 330 m s. To find : Value of u. The observer is moving away from S1, so apparent frequency of the sound coming from S2 for the observer : Ê v - uˆ f1 = fo Á Ë v ˜¯ Ê 330 - u ˆ = 660 Á Ë 330 ˜¯

...(i)

The observer is moving towards S2, so apparent frequency of the sound coming from S1 for the observer : Ê v + uˆ f2 = fo Á Ë v ˜¯ Ê 330 + u ˆ = 660 Á Ë 330 ˜¯

...(ii)

fbeat = f2 - f1

27000 = 27 km h = m s = 7.5 m s , 3600

frequency of sonar signal sent by B towards A is fo = 500 Hz, speed of sound in water is v = 1500 m s. To find : Frequency f ¢ of the sonar signal detected by submarine B after reflection from submarine A. Frequency of sonar signal received by submarine A: Ê v - vA ˆ f1 = fo Á Ë v - vB ˜¯

Ê 330 + u ˆ Ê 330 - u ˆ 10 = 660 Á - 660 Á ; Ë 330 ˜¯ Ë 330 ˜¯ u =

5 = 2.5 m / s 2

Ans. 17. Option (a) is correct. Given : Audio output of a speaker is P = 2 W, reference intensity of sound is I o = 10 -12 W / m 2 . To find : r, the distance from the speaker at which the loudness of sound is b = 120 dB. Loudness of sound :

Ê 1500 - 5 ˆ = 500 Á Ë 1500 - 7.5 ˜¯ = 500.8 Hz

Ê 1500 + 7.5 ˆ = 500.8 ¥ Á Ë 1500 + 5 ˜¯

Beat frequency as measured by the observer :

Speed of submarine B is

vB

Ê v + vB ˆ f ¢ = f1 Á Ë v + v A ˜¯

Ans. 16. Option (c) is correct. Given : Frequency of sound waves from source S1 = frequency of sound waves from source S2 = fo = 660 Hz, speed of observer walking from S1 towards S2 = u m/s, frequency received by the observer is

fo is the true frequency of the sound source. Apparent frequency measured by an observer when source is moving away from him :

=

Frequency of the sonar signal detected by submarine B after reflection from submarine A :

...(i)

The frequency reflected from submarine A is f1 = 500.8 Hz.

Ê Iˆ β = 10 log Á ˜  Ë Io ¯ Ê I ˆ 12 = log Á -12 ˜ Ë 10 ¯

190 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) 1012 = I (10 )12 ; fi

To find : Ratio

2

I = 1W/m .

Intensity of sound reaching at r is I = 1 W / m2. Also

I =

fi

r = =

P 4pr 2

f1 . f2

Apparent frequency for case a : Ê v ˆ f1 = fo Á Ë v - vs1 ˜¯ Apparent frequency for case b : Ê v ˆ f2 = fo Á Ë v - vs 2 ˜¯

;

P 4p I 2 = 0.398 m 4p ¥ 1

ª 40cm. Ans. 18. Option (a) is correct. Given : Length of flute is L = 50 cm, sound produced from the flute is of 2nd harmonics, speed of a person running towards the musician is 25 vo = 10 km / h = m / s, 9 Speed of sound is v = 330 m / s. To find : The frequency of sound as heard by the person. Wavelength of second harmonics as produced by the flute : λ = L Frequency of second harmonics as produced by the flute : v v 330 fo = = = = 660 Hz λ L 0.5 Frequency of second harmonics as heard by the person : f ( v + vo ) f = o v = 660 ¥

PHYSICS

330 + 330

25 9

= 2 ¥ ( 332.8 ) ª 666 Hz Ans. 19. Option (b) is correct. Given : Case a : Speed of train towards a stationary observer is vs1 = 34 m / s, Frequency of train whistle measured by the stationary observer is f1, Case b : Speed of train towards a stationary observer is vs2 = 17 m / s, Frequency of train whistle measured by the stationary observer is f2, Speed of sound is v = 340 m / s.

...(i)

...(ii)

In equations (i) and (ii) fo is the true frequency of the train whistle. Take ratio of equation (i) to equation (ii), f1 f2

Ê v ˆ ÁË v - v ˜¯ Ê v - v ˆ s1 S2 = = Ê v ˆ ÁË v - vS1 ˜¯ ÁË v - v ˜¯ s2

=

340 - 17 323 19 = =  340 - 34 306 18

Ans. 20. Option (a) is correct. Given : Fundamental frequency of an organ pipe is fo = 1.5 kHz, highest frequency a person can hear is f = 20,000 Hz. To find : Number of overtones that can be distinctly heard by a person. Let L be the length of an organ pipe, then : v fo = ...(i) 4L v is velocity of sound. Frequency of nth overtone : fn = ( 2n + 1)

v = ( 2n + 1) fo 4L

...(ii)

If number of overtones that can be distinctly heard by a person is n, then : f > fn f > ( 2n + 1) fo 20000 > ( 2n + 1)1.5 ¥ 10 3 ( 2n + 1) =

20 1 Ê 200 ˆ ;n= Á - 1˜ = 6 ¯ 1.5 2 Ë 15

Ans. 21. Option (d) is correct. Given : Equation of travelling wave on a stretched string is y = 0.03 sin ( 450t - 9 x ), ...( i ) linear density of string is m = 5 g m = 0.005 kg m . To find : T, the tension in the string. Equation (i) can be written as : 9x ˆ Ê y = 0.03 sin 450 Á t Ë 450 ˜¯ General equation of travelling wave : xˆ Ê y = A sin ω Á t - ˜ Ë v¯

...(ii)

...(iii)

191

OSCILLATIONS AND WAVES

Compare equations (i) and (ii) : 450 v = = 50 m s 9

f =

...(iv)

f = 35Hz

Tension in the string : v =

T ; m

T = v2 m

fi

= ( 50 )2 ¥ 0.005 = 12.5 N Ans. 22. Option (a) is correct. Given : A travelling harmonic wave is represented by y( x , t ) = 10 -3 sin ( 50t + 2 x ).

...(i)

To find : The magnitude and direction of propagation of wave. Equation (i) can be written as : È Ê -2 ˆ ˘ y = 10 -3 sin 50 Ít - Á ˜ x ˙ Î Ë 50 ¯ ˚ General equation of a travelling wave : xˆ Ê y = A sin ω Á t - ˜ Ë v¯ Compare equations (i) and (ii) : 50 v = = -25 m / s 2

...( ii )

Q.2. A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ____Hz. [JEE (Main) – 8th Sep. 2020 - Shift-1] Sol. Given : Length of the organ pipe is L = 1 m, density of gas inside the organ pipe is double the density of air at STP, speed of sound in air at STP is v = 300 m/s. To find : f1 – f0, the frequency difference between the fundamental and the second harmonic of the organ pipe. Let density of air at STP be r, that implies density of air inside the organ pipe is 2r. Let velocity of sound inside the organ pipe be v¢. So, 1 / 2ρ v′ = v 1/ ρ

...( iii)

v′ = ...(iv)

fn =

∆L = 4.9 × 10 −4 , Young’s modulus for L

the material of the wire is Y = 9 × 1010 N/m2. To find : f, the lowest frequency of transverse vibrations in the wire. YA∆L Let T = be the tension in the wire and m be L its weight per unit length. Fundamental frequency of transverse vibrations in the wire : f =

1 T 1 T 1 Y∆L = = 2L m 2L ρ A 2L ρ L

2 n+1 v ' 2L

Fundamental harmonic of the organ pipe : 0+1 300 fo = v′ = 2L 2 2

Subjective Questions (Chapter Based) Q.1. A wire of density 9 × 10–3 kg cm–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 × 10–4. The lowest frequency of the transverse vibrations in the wire (Young’s modulus of wire Y = 9 × 1010 N m–2), (to the nearest integer), _________ [JEE (Main) – 2nd Sep. 2020 - Shift-2] Sol. Given : Density of wire is r = 9 × 10–3 kg/cm3 = 9 × 103 kg/m3, length of wire is L = 1 m, strain in

v

Using :

From equation (iv) we can say that the given wave is travelling along the negative x axis with velocity v = 25 m / s.

the wire is

1 9 × 1010 × 4.9 × 10 −4 2×1 9 × 10 3

Second harmonic of the organ pipe : 1+1 300 f1 = v′ = 2L 2

...(i)

...(ii)

From equations (i) and (ii) : 300 f1 − f0 = = 106.1Hz 2 2 Q.3. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/ sec. What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1) Sol. Given : Frequency of oscillation of silver atom is

f = 1012 sec -1 , Molecular weight of silver is M = 108 Avogadro’s number is N A = 6.02 ¥ 10 23 g / mol. To find : The force constant of the bonds connecting the atoms. Angular frequency of oscillations : ω = 2p f

192 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Force constant :

Equate equations (iii) and (iv), k = mω 2 = =

M ¥ 10 NA

108 ¥ 10 -3 6.02 ¥ 10 23

-3

6 Ê 4A ˆ ω A = ω A ¢ 2 - Á Ë 5 ˜¯ 5

( 2 p f )2

A¢ 2 =

–2

Q.4. A magnetic needle of magnetic moment 7.3 × 10 A m2 and moment of inertia 12.5 × 10–6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.1 T. What will be time taken by the needle to complete 10 oscillations? Sol. Given : Magnetic moment of the needle is M = 7.3 × 10–2 Am2, moment of inertia of the needle is I = 12.5 × 10–6 kgm2, the needle is performing SHM in a magnetic field of B = 0.1 T. To find : t, time taken by the needle to complete 10 oscillations. Time period of SHM performed by the needle :

A¢ =





12.5 ¥ 10 -6 7.3 ¥ 10

-2

¥ 0.1

= 0.04 s

Hence, time for 10 oscillations will be : t = 10 ¥ T = 0.4 s Q.5. A particle performs simple harmonic motion with amplitude A. Its speed is doubled at the instant that it is at a distance 4A/5 from equilibrium position. What is the new amplitude of motion ? 4A , the Sol. Given : Amplitude of SHM is A, at x = 5 velocity of the particle is doubled. To find : A¢, the new amplitude of motion of the particle. Velocity of particle performing SHM : v = ω A2 - x2 x =

...(i)

4A 5 2

3 Ê 4A ˆ v = ω A2 - Á = ωA Ë 5 ˜¯ 5 Also, at

x =

...(ii)

4A 5

the velocity of the particle is doubled. So, new velocity will be : 6 3 ...(iii) v¢ = 2 ¥ ω A = ω A 5 5 But from equation (i), new velocity will be : Ê 4A ˆ v¢ = ω A¢2 - Á Ë 5 ˜¯

20 2 A 25 2 5

A

Q.6. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer ? (speed of light = 3 × 108 m s–1) Sol. Given : Speed of light is

I T = 2p MB = 2p

2

36 2 16 A = A¢2 - A 2 25 25

( 2 p ¥ 1012 )2

= 7.1 N / m

At

PHYSICS

c = 3 ¥ 10 8 m / s, speed of an observer towards a stationary microwave source is c v = , 2 true frequency of microwave emitted by the source is fo = 10 GHz. To find : f, the frequency of microwave as measured by the observer. Relativistic Doppler’s formula : 1

1

vˆ2 Ê Ê 3ˆ2 1+ Á ˜ Á ˜ c f = fo Á = 10 ¥ Á 2 ˜ v˜ 1 Á1- ˜ Á ˜ Ë Ë 2¯ c¯ = 10 ¥ 3 = 17.3 GHz Q.7. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 m s–1 and the man behind walks at a speed 2.0 m s–1. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 m s–1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is Sol. Given : Speed of man A is

2

...(iv)

vA = 1 m s , Speed of man B is vB = 2 m s , man C is stationary and is at height of 12 m, true frequency of the whistle blown by A and B is fo = 1430 Hz, speed of sound is v = 330 m s ,

193

OSCILLATIONS AND WAVES

Frequency of whistle blown by B as measured by C : Ê ˆ v fB = fo Á Ë v - v cos θ ˜¯

C

13 m

B

13 m 12 m

B





 5m

5m

A 1 m/s

2 m/s

To find : Frequency of beats as heard by C when C is equidistant from A and B and distance between A and B is d = 10 m. Frequency of whistle blown by A as measured by C: Ê ˆ v fA = fo Á Ë v + v cos θ ˜¯ A

Ê ˆ 330 Á ˜ = 1430 Á 5˜ Á 330 + 1 ¥ ˜ Ë 13 ¯

Ê ˆ 330 Á ˜ = 1430 Á 5˜ Á 330 - 2 ¥ ˜ Ë 13 ¯ Beat frequency : f = fB - fA Ê ˆ 330 Á ˜ = 1430 Á 5˜ Á 330 - 2 ¥ ˜ Ë 13 ¯ Ê ˆ 330 Á ˜ -1430 Á ª 5 Hz. 5˜ Á 330 + 1 ¥ ˜ Ë 13 ¯



194 Oswaal JEE (Main) Solved Papers (Chapter-wise & Topic-wise)

physics

Chapter 11

Electrostatics

Syllabus Electric charges : Conservation of charge, Coulomb’s law–forces between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field : Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell; Electric potential and its calculation for a point charge, electric dipole and system of charges; Equipotential surfaces, electrostatic potential energy of a system of two point charges in an electric field; Conductors and insulators, Dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, Energy stored in a capacitor.

Topic-1

list of Topics : Topic-1 : Electrostatic Force, Electric Field and Electrostatic Potential .... P. 195

Electrostatic Force, Electric Field and Electrostatic Potential

Topic-2 : Capacitors

.... P. 214

Concept Revision (Video Based) Electric Field

Coulomb’s Law

Part -1



Part - 2

    

Electric Field due to a Dipole

Part -1

         

Gauss’s Law

     

Part - 2 Problems on Gauss’s Law



Equipotential Surfaces



JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A charged particle (mass m and charge q) moves along X axis with velocity V0. When it passes through the origin it enters a region having  uniform electric field E = − E j which extends upto x = d. Equation of path of electron in the region x > d is :

Y E O

> V0

X d

196 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a) y =

qEd mV02

x

(b) y =

qEd  d  − x (c) y = 22 mV0  

(d) y =

qEd mV02 qEd 2 mV02

(x − d) x

[JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass? (Curves are drawn schematically and are not to scale)

.

E

surface charge densities of the concentric spheres are equal, the potential difference V(9R) – V(4R) is :

Q2 4p ε 0 R

(b)

(c)

3Q1 16p ε 0 R

3Q1 (d) 4p ε R 0

[JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.6. Two charged thin infinite plane sheets of uniform charge density s+ and s– , where |s+| > |s–|, intersect at the right angle. Which of the following best represents the electric field lines for the system :

>> > > >>

(b)

y

y

x

x

(c)

(d)

y

– >> >> > >

y

(a)

+

–

(b)

y

> > > > > > > >> > >

+

>

–

(c)

> >

.

> > >

> >

>

>

>

x

+

> >

x

[JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

3Q2 4p ε 0 R

(a)

(a)

x

PHYSICS

r R

+

> >

[JEE (Main) – 4th Sep. 2020 - Shift-1] Q.7. A two point charges 4q and –q are fixed on the d −d x-axis at x = and x = , respectively. If the 2 2 third point charge ‘q’ is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will : >

4q

>

[JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.4. Two isolated conducting spheres S1 and S2 of 1 2 radius R and R have 12 mC and –3mC charges, 3 3 respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S1 and S2 are respectively : (a) 6 mC and 3 mC (b) 4.5 mC on both (c) + 4.5 mC and –4.5 mC (d) 3 mC and 6mC [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.5. Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that

>

> >

1 (R + r ) Q (d) 4p ε 0 2(R 2 + r 2 )

>

> >

>

1 (R + r ) Q 4p ε 0 2(R 2 + r 2 )

>

1 (R + 2r )Q (c) 4p ε 0 2(R 2 + r 2 )

(b)

>

1 2(R + r ) Q 4p ε 0 (R 2 + r 2 )

–

>

(a)

(d)

–q

197

ELECTROSTATICS

(a) Increase by

2q2 3p ε 0 d

(b) Increase by

2q2 4p ε 0 d

(c) decrease by

q2 4p ε 0 d

(d) decrease by

4q2 3p ε 0 d

field E at the centre of the circle are respectively : [Take V = 0 at infinity] (a) V = 0; E = 0 10 q (b) V = ;E=0 4p ε 0 R

[JEE (Main) – 4th Sep. 2020 - Shift-1] Q.8. A particle of charge q and mass m is subjected to an electric field E = E0(1 – ax2) in the x-direction, where a and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is  : (a) a (c)

10 q 10 q ;E= 4p ε 0 R 4p ε 0 R 2 10 q (d) V = 0 , E = 4p ε 0 R 2 [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.11. Charges Q1 and Q2 are at point A and B of a right angle triangle OAB. The resultant electric field at point O is perpendicular to the hypotenuse, then Q1 is proportional to : Q2 (c) V =

3 a

(b)

1 a

(d)

2 a

[JEE (Main) – 4th Sep. 2020 - Shift-2] Q.9 A solid sphere of radius R carries a charge Q + q distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed v when it has fallen through a vertical height y (see figure), then (assume the remaining portion to be spherical)

Q

Q1 x1 O

(a) (c)

  qQ + g (c) v 2 = y  4 p ε R(R + ) y m 0     qQ 2 + g (d) v = 2 y   4p ε 0 R(R + y )m  th [JEE (Main) – 5 Sep. 2020 - Shift-1] Q.10 Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric



(b)

x2 x1

(d)

x12

1

(a)

q

  QqR + g (a) v 2 = 2 y  3  4p ε 0 (R + y ) m    qQ 2 + g (b) v = y  2  4p ε 0 R ym 

B

x13

x 23 x1 x2

2

the x-axis at distance ‘a’ from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is ‘m’, their speed when they are infinitely far apart is :

.

v

x 22

Q2

x2

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.12. Two identical electrical point dipoles have dipole   moments p = pi and p = − pi and held on

R

y

A

3 P a 2p ε 0 ma

(b)

1 P a 2p ε 0 ma

2 P 1 P (d) a p ε 0 ma a p ε 0 ma [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.13. Consider the force F on a charge ‘q’ due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if ‘q’ is placed at distance r from the centre of the shell? (c)

(a) F =

1 Qq for all r 4p ε 0 r 2

1 Qq for r < R 4p ε 0 R 2 1 Qq > F > 0 for r < R (c) 4p ε 0 R 2 (b) F =

(d) F =

1 Qq for r > R 4p ε 0 r 2 [JEE (Main) – 6th Sep. 2020 - Shift-2]

198 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.14. Two infinite planes each with uniform surface charge density +s are kept in such a way that the angle between them is 30°. The electric field in the region shown between them is given by : + 30°

x

(b)

σ 2ε 0

 x  (1 + 3 ) y +  2  

 x  (1 + 3 ) y −  2  

(d)

σ 2ε 0

 3   x   1 − y −  2  2  

σ ε0

(c)

σ 2ε 0

v x

x

(d) v

v

 3   x   1 + y +  2  2  

(a)

(b) v

(c)

y

+

(a)

[JEE (Main) – 7th Jan. 2020 - Shift-I] Q.15. In finding the electric field using Gauss law the  q formula |E|= enc is applicable. In the formula e0 |A| e0, is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation? (a) For any choice of Gaussian surface. (b) Only when the Gaussian surface is an equipotential surface. (c) Only when the Gaussian surface is an  equipotential surface and |E| is constant on the surface.  (d) Only when |E| = constant on the surface [JEE (Main) – 8th Jan. 2020 - Shift-I] Q.16. Three charged particles A, B and C with charges –4q, 2q and –2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is :

PHYSICS

x

x

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.18. Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such V1 (on S1 ) E R that 1 = 1 . Then the ratio of the V2 (on S 2 ) E2 R 2 electrostatic potentials on each sphere is : 3

 R2  (b)    R1 

R  (a)  1   R2 

2

R  (d)  1   R2  [JEE (Main) – 8th Jan. 2020 - Shift-2]  Q.19. An electric dipole of moment p = ( − i − 3 j + 2 k ) –29 × 10 C-m is at the origin (0, 0, 0). The electric  kk m field due to this dipole at r r==++i i++33j j++5    (note that r ⋅ p = 0) is parallel to : (a) ( +i − 3 j − 2 k ) (b) ( −i − 3 j + 2 k ) (c)

R1 R2

(c) ( +i + 3 j − 2 k ) (d) ( −i + 3 j − 2 k ) [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.20. Consider a sphere of radius R which carries a uniform charge density r. If a sphere of radius  R |E | is carved out of it, as shown, the ratio  A 2 |EB |   of magnitude of electric field E A and EB , respectively, at points A and B due to the remaining portion is :

y 2q B d

150° O

(a) (c)

3q 4p ε 0 d 2 3q p ε 0d 2





–4q A d 30° 30° d

(b) (d)

R 2 x

A R

C

. 3 3q

4p ε 0 d 2

(a)

21 34

(b)

18 54

(c)

17 54

(d)

18 34

2 3q p ε 0d 2

[JEE (Main) – 8th Jan. 2020 - Shift-1] Q.17 A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)

B

[JEE (Main) – 9th Jan. 2020 - Shift-1] Q.21. The bob of a simple pendulum has mass 2 g and a charge of 5.0 mC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is :

199

ELECTROSTATICS

(take g = 10 m/s2) (a) tan–1 (2.0) (b) tan–1 (0.2) –1 (c) tan (5.0) (d) tan–1 (0.5) [JEE (Main) – 8th April 2019 - Shift-1] Q.22. A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 4 Q, the new potential difference between the same two surfaces is : (a) – 2V (b) 2 V (c) 4 V (d) V [JEE (Main) – 8th April 2019 - Shift-1] Q.23. A positive point charge is released from rest at a distance ro from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to :



If D  d, the potential energy of the system is best given by :

1 4p εo

(b)

1  q 2 2 qQd  − +  D2  4p εo  d

(c)

1  q 2 qQd  + + 2  D  4p εo  d

1  q 2 qQd  − − 2  D  4p εo  d [JEE (Main) – 9th April 2019 - Shift-1] Q.27. Four point charges –q, + q, + q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as : 1 1 (a) E ∝ 3 (b) E ∝ D D (d)

(c) E ∝ ro

r (b) v ∝ ln    ro 

+r / r (a) v ∝ e o

r r (c) v ∝ ln   (d) v ∝   r  ro   o [JEE (Main) – 8th April 2019 - Shift-2] Q.24. An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency w is : (a) (c) 2

qE md qE md

(b)

2qE md

(d)

qE 2md

[JEE (Main) – 8th April 2019 - Shift-2] Q.25. The electric field in a region is given by  E = (Ax + B) i , where E is in N C–1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = –5 is V2, then V1 – V2 is : (a) 320 V (b) – 48 V (c) 180 V (d) – 520 V [JEE (Main) – 8th April 2019 - Shift-2] Q.26. A system of three charges are placed as shown in the figure :

+q •

• –q d

• Q

1 D4

(d) E ∝

1 D2

[JEE (Main) – 9th April 2019 - Shift-2] Q.28. In free space, a particle A of charge 1 mC is held fixed at a point P. Another particle B of the same charge and mass 4 mg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is :   1 = 9 × 10 9 Nm 2C −2   Take 4p ε o   (a) 1.0 m/s (b) 3.0 × 104 m/s 3 (c) 2.0 × 10 m/s (d) 1.5 × 102 m/s [JEE (Main) – 10th April 2019 - Shift-2] Q.29. A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by : – + – + – + – + – L + – + – + – + – + m – + – + – + q – + – + – + — — — — E (a) 2p

(c) 2p D

 q 2 qQd  − − 2   d 2D 

(a)

L qE   g + m   

(b) 2p

L  2 q 2 E2  g − 2  m  



L L (d) 2p 2 qE    qE  g − m  g2 +     m [JEE (Main) – 10th April 2019 - Shift-2]

200 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.30. Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries  charge Q. At its centre is a dipole p as shown. In this case :

p

•

(a) surface charge density on the inner surface is (Q/ 2 ) uniform and equal to 4p a 2 (b) electric field outside the shell is the same as that of a point charge at the centre of the shell (c) surface charge density on the outer surface  depends on | p | (d) surface charge density on the inner surface of the shell is zero everywhere [JEE (Main) – 12th April 2019 - Shift-1]  Q.31. A point dipole p = - p x is kept at the origin. The 0

potential and electric field due to this dipole on the y-axis at a distance d are, respectively : (Take V = 0 at infinity)    | p| p −p , (a) (b) 0 , 2 3 4p ε 0 d 4p ε 0 d 4p ε 0 d 3    p | p| −p (c) 0 , (d) , 3 2 4p ε 0 d 4p ε 0 d 4p ε 0 d 3 [JEE (Main) – 12th April 2019 - Shift-1] Q.32. Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then : 3R (a) a = 8–1/4 R (b) a = 1 / 4 2 (c) a = 2–1/4 R

(d) a =

R

3 [ JEE (Main) – 12 April 2019 - Shift-2] Q.33. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is : R R (a) (b) 2 5 th

(c) R

(d) R 2

[JEE (Main) – 9th Jan. 2019 - Shift-1] Q.34. Three charges +Q, q, +Q are placed respectively, d at distance, 0, and d from the origin, on the 2 x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is : Q Q (a) − (b) + 2 4

(c) +

Q 4

(d) −

PHYSICS

Q 2

[JEE (Main) – 9th Jan 2019 - Shift-1] Q.35. Two point charges q1 ( 10 mC ) and q2(–25 mC) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,   1 = 9 × 10 9 Nm 2C −2   take 4p ε 0   (a) ( 63i − 27 j ) × 10 2 (b) ( −63i + 27 j ) × 10 2 (c) ( 81i − 81j ) × 10 2 (d) ( −81i + 81j ) × 10 2 [JEE (Main) – 9th Jan 2019 - Shift-2] Q.36. Charge is distributed within a sphere of radius A R with a volume charge density ρ ( r ) = 2 e −2 r / a , r where A and a are constants. If Q is the total charge of this charge distribution, the radius R is : Q   (a) a log  1 − 2p aA  

    a 1 (b) log   Q 2   1 − 2p aA  

    1 a Q   (c) a log  (d) log  1 −  Q 2 2 p aA     1 − 2p aA   [JEE (Main) – 9th Jan 2019 - Shift-2] Q.37. A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be : Q ab + bc + ca (a) 12p ε 0 abc (c)

Q 4p ε 0 ( a + b + c )

(b)

Q( a 2 + b 2 + c 2 ) 4p ε 0 ( a 3 + b 3 + c 3 )

(d)

Q( a + b + c ) 4p ε 0 ( a 2 + b 2 + c 2 )

[JEE (Main) – 10th Jan 2019 - Shift-1] Q.38. Two electric dipoles, A, B with respective dipole   moments dA = −4 qai and dB = −2 qai are placed on the x-axis with a separation R, as shown in the figure • A



R

• B

X

The distance from A at which both of them produce the same potential is : (a) (c)

R 2 +1 R 2 −1



(b)



(d)

2R 2 +1 2R 2 −1

[JEE (Main) – 10th Jan 2019 - Shift-1] Q.39. Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be :

201

ELECTROSTATICS

(a) (c)

1  Q2  1 +  4p ε 0  3

Q2 2 2 p ε0



(b)

1  Q2  1 +  4p ε 0  5

(d)

Q2 4p ε 0

[ JEE (Main) – 10th Jan 2019 - Shift-2] Q.40. Charges –q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line y to P¢ such that OP′ =   , the force on Q will be 3 y  close to :  >> 2 a  3 

(c) Potential of a uniformly charged sphere (d) Electric field of a uniformly charged spherical shell [JEE (Main) – 11th Jan 2019 - Shift-1] Q.43. An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10–29 C-m. What is the potential energy of the electric dipole ? (a) – 20 × 10–18 J (b) – 7 × 10–27 J –29 (c) – 10 × 10 J (d) – 9 × 10–20 J [JEE (Main) – 11th Jan 2019 - Shift-2] Q.44. Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure : –2q y

P

l

l +q

+q l Q P' O

A

B

+q

–q

(a) 3 ql

j − i 2

(b) ( ql )



i + j 2

(d) − 3 ql j

(c) 2 ql j F (a) 3F (b) 3 (c) 9F (d) 27F [JEE (Main) – 10th Jan 2019 - Shift-2] Q.41. Three charges Q, +q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is :

x

[JEE (Main) – 12th Jan 2019 - Shift-1] Q.45. There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed v(R(t)) of the distribution as a function of its instantaneous radius R(t) is : (a)

(b)

u(R[t])

u(R[t])

Q

+q

(c) (a)

−q 1+ 2

(b) +q



(c) –2q

(d)

R0

(d)

u(R[t]) V0

R(t)

u(R[t])

− 2q 2 +1

Ru R(t)

[JEE (Main) – 11th Jan 2019 - Shift-1] Q.42. The given graph shows variation (with distance r from centre) of :

•

R(t)

R0

+q

ro

ro

r

(a) Electric field of a uniformly charged sphere (b) Potential of a uniformly charged spherical shell



R0

R(t)

th

[JEE (Main) – 12 Jan 2019 - Shift-1]

ANSWER – KEY 1. (c) 5. (c) 9. (d) 13. (d) 17. (c) 21. (d) 25. (c) 29. (d) 33. (b)

2. (c) 6. (b) 10. (a) 14. (d) 18. (d) 22. (d) 26. (d) 30. (b) 34. (a)

3. (d) 7. (d) 11. (d) 15. (c) 19. (c) 23. (b) 27. (c) 31. (b) 35. (a)

4. (a) 8. (c) 12. (b) 16. (c) 20. (d) 24. (b) 28. (c) 32. (a) 36. (b)

202 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) 37. (d) 41. (d) 45. (c)

38. (b) 42. (b)

39. (b) 43. (b)

40. (d) 44. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : Mass of a particle is m, charge on particle is q, initial velocity of particle before entering the  region of electric field is V i , electric field E = −E j 0

exists in the region x = 0 to x = d. To find : The equation of the particle for the region x > d. y x t=0

t Vy

x 1 2 2 at Vx (x1–y) Straight line

After travelling through the electric field the components of velocity of the particle will change as : qE Vx = Vo , Vy = at , a = − . m In the region x > d, the equation of the particle will be a straight line. 1 y − at 2 at 2 = x−d Vo 1 y − at 2 x−d 2 = at Vo y 1 d x d − = − Vo Vo at 2 Vo

qEd  d   − x   mVo2  2 

Ans. 2. Option (c) is correct. Given : Charge on a point mass released from the edge of a table is +q, initial velocity of the point mass is u = 0. To find : The trajectory of the point mass in presence of horizontal electrical and vertical gravitational field. E x

y

Resultant acceleration of the point mass : 2

 qE  a =   + g 2 = constant m As initial velocity of the point mass was u = 0 and the acceleration is constant its trajectory will be straight line. Ans. 3. Option (d) is correct. Given : Radii of two concentric spherical shells are r and R, r < R, surface charge density on both shells is same s, total charge on the shells is Q. To find : V, the electric potential at the centre of the shells. Charge on the inner shell : Q1 = σ ( 4p r 2 ) = 4pσ r 2 Charge on the outer shell : Q 2 = σ ( 4p R 2 ) = 4p σ R 2 Total charge : 

Q = Q1 + Q2 = 4pσ ( r 2 + R 2 )

Electric potential at the centre of the shell : 1 Q1 1 Q2 +  V= 4p ε o r 4p ε o R V=

1  4p σ r 2 4p σ R 2  +   4p ε o  r R 

V=

1 × 4p σ ( r + R ) 4p ε o

V=

1 Q( r + R ) × 4p ε o r 2 + R 2

2 R, charge on 3 R sphere S1 is q1 = 12 mC, radius of sphere S2 is r2 = , 3 charge on sphere S2 is q2 = –3 mC, the spheres are at very large distance from each other, the spheres are connected by a conducting wire. To find : The charge on the spheres after a long time. Total charge on the two spheres : q = q1 + q2 = 9µC Given : Radius of sphere S1 is r1 =

− ymVo x 1 d = − qEd Vo 2 Vo

.

Components of acceleration of the point mass, m : qE = ax = ,a y g m

Ans. 4. Option (a) is correct.

y x 1 d  = − at Vo 2 Vo

y =

PHYSICS

After connecting a wire, the spheres will attain same electric potential. Let the final charge on the spheres be q1′ and q2′ . 1 q1′ 1 q2′ = 4p ε o r1 4p ε o r2 q1′ q′ = 2 2 R R 3 3

203

ELECTROSTATICS

q1′ = 2 q2′

...(i)

As total charge will be conserved in the process : q1′ + q2′ = q = 9 µC

...(ii)

From equations (i) and (ii) : q1′ = 6 µC , q2′ = 3 µC Ans. 5. Option (c) is correct. Given : Radii of two concentric spherical spheres are R and 4R, charge on both the spheres is Q1 and Q2 respectively, surface charge density on both spheres is same s. To find : The potential difference V(R) – V(4R). 1  Q1 Q2  + ...(i) V( R ) = 4p ε o  R 4 R  1  Q1 Q2  V( 4 R ) = + 4p ε o  4 R 4 R 

...(ii)

From equations (i) and (ii) : 3Q1 1  3 Q1  =  V( R ) − V( 4 R ) = 4p ε o  4 R  16p ε o R Ans. 6. Option (b) is correct. Given : Charge density on infinite plane sheet-1 is s+, charge density on infinite plane sheet-2 is s– , |s+| > |s–|, the two sheets intersect at right angles. To find : The best representation of electric field lines in the region of space between the two sheets. The electric field lines coming from an infinite plane sheet of charge will be straight lines and the electric field will be uniform in space. Graph-2 depicts these two qualities correctly. Ans. 7. Option (d) is correct. d Given : Location of charge q1 = 4q is x = − , 2 d location of charge q2 = –q is x = , a third charge 2 q3 = q is fired from x = 0 to x = d following a semicircular path. To find : DU, the change in energy of the charge q. Potential energy of q at x = 0 :   1  4q2 q2  2 3q 2 Ui = − = ×  d  4p ε o 4p ε o  d d   2   2 Potential energy of q at x = d : Uf

  1  4q2 q2  2 q2 = − = ×  d  4p ε o 3d 4p ε o  3d   2   2

Change in potential energy of q : 8 2 q2 4q2 ∆U = U f − Ui = × × = 3 4p ε o d 3p ε od Ans. 8. Option (c) is correct. Given : Charge on a particle is q, mass of the particle is m, the particle is subjected to an electric field

E = Eo (1 – ax2), initial position of particle is x = 0 initial velocity of particle is u = 0. To find : x¢, the position of particle where its kinetic energy is zero. Kinetic energy of a particle at x = 0 is : Ki = 0 Kinetic energy of a particle at x = x¢ is : Kf = 0 Work done on the particle as it moves from x = 0 to point x = x¢ is the change in its kinetic energy : W = K f − Ki = 0 W = x′

∫0

x′

∫0

qEdx = 0

qEo (1 − ax 2 )dx = 0 x′

qEo ∫ (1 − ax 2 )dx = 0 0

x′

 ax 3  x −  = 0 3   0 1−

ax′2 = 0 3 x′ =

3 a

Ans. 9. Option (d) is correct. Given : Radius of solid sphere is R, charge on sphere is Q + q, mass of small piece which gets detached from the sphere is m, charge on small piece q, the vertical distance through which the piece falls is y, the velocity acquired by the piece is v. To find : Expression for v. Energy of the piece after it falls a distance y : 1 2 1 Qq mv + 2 4p ε o R + y Energy of the piece before it detaches itself from the sphere : 1 Qq mgy +  4p ε o R By law of conservation of energy : 1 2 1 Qq 1 Qq mv + = mgy +  2 4p ε o R + y 4p ε o R Qq  1 1 2 1  mv = mgy +  −  2 4p ε o  R R + y  v 2 = 2 gy +

2Qq 4p ε o m

  y    R (R + y) 

  Qq v2 = 2 y  g +   4p ε omR ( R + y )   Ans. 10. Option (a) is correct. Given : Radius of circle is R, ten charges are uniformly distributed on the circumference of the circle, charges 1, 3, 5, 7, 9 have charge +q and charges 2, 4, 6, 8, 10 have charge –q. To find : The electric potential V and the electric field E at the centre of the circle.

204 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) The net charge on the circumference of the circle : Q = 5q − 5q = 0

mv 2 =

As the opposite charges are equidistant from the centre of the circle their electric field components cancel out and electric field at the centre of the circle is E = 0. Ans. 11. Option (d) is correct. Given : Charges Q1 and Q2 are located on vertices of a right angled triangle OAB,

.

Q1

x1 O

F2

x2 

.



Q2

F1

The resultant electric field at O due to Q1 and Q2 is perpendicular to side AB. Q To find : 1 . Q2 From the construction : tan θ =

x1 F2 = x 2 F1

Ans. 12. Option (b) is correct.

  Given : Dipole moment of dipole 1 is p1 = pi,  dipole moment of dipole 2 is p2 = − pi, initial distance between the dipoles is x = a, initial speed of both the dipoles is u = 0 mass of both the dipoles is m. To find : v, the speed of the dipoles when they are infinitely far apart. Energy of the two dipole system when they are x = a distance apart and at rest : r

a3 2 kp 2 ma3

=

p 1 a 2p ε oma

Ans. 13. Option (d) is correct. Given : Radius of a spherical shell is R, charge on spherical shell is Q. To find : The force on charge q, located at a radial distance r from the centre of the shell. For r < R, the force on charge q due to the shell is F = 0. For r ≥ R, the force on charge q due to the shell is 1 Qq F = . 4p ε o r 2 Ans. 14. Option (d) is correct. Given : Surface charge density on two infinite planes is +s, angle between two planes is q = 30°.  To find : E, the electric field in the region between the planes. +

y 30°

x

...(ii)

Electric field vector due to plane-2 :  σ σ E2 = − sin 30 x − cos 30 y 2ε o 2ε o

x Q1 = 1 x2 Q2

3

2 kp 2

Electric field vector due to plane-1 :  σ  E1 = y 2ε o

Q x2 = 2 Q1 x1

cos 180 =

...(ii)

Magnitude of electric field due to a single infinite plane of charge density +s : σ E = ...(i) 2ε o

Q2

2 kp1 p2

v =

+

x1 x2 = 2 Q1 x2 x12

Ei = 0 −

1 2 1 2 mv + mv = mv 2 2 2

By law of conservation of energy :

Electric potential at the centre of the circle : 1 Q V = = 0 4p ε o R

A

Ef =

PHYSICS

2 kp 2 a

3

...(i)

Energy of the two dipole system when they are at infinite distance from each other :

 3σ  σ  E2 = − x− y 4ε o 4ε o

...(ii)

Net electric field in the region between two planes is :    E = E1 + E2  E = =

3σ  σ  σ  y− x− y 2ε o 4ε o 4ε o 3   x  σ   1 − y −  2ε o  2  2 

Ans. 15. Option (c) is correct. Given : Gauss’s law  q E = enc , εo A

...((iii)

...(i)

205

ELECTROSTATICS

here qenc is the charge enclosed by a Gaussian surface, e0 is the permittivity of free space and |A| is the magnitude of area enclosed by the Gaussian surface. To find : Condition on Gaussian surface, when equation (i) is valid. Equation (i) is valid only for a Gaussian surface which is an equipotential and the magnitude of electric field is constant over the surface. Ans. 16. Option (c) is correct. Given : Three charges A, B and C are arranged on the circumference of a circle as shown below. y 2q B d

d 30° 30° d

O

ExB = − ExC = −

4p ε o d 2 2q 4p ε o d

2

2q 4p ε o d

2

Ex( net ) = − Ex( net ) =

8p ε od

2

8 3q 8p ε od 2

Ans. 18. Option (d) is correct. Given : Radius of metallic sphere S1 is R1, electric field on the surface of S1 is E1, radius of metallic sphere S2 is R2, electric field on the surface of S2 is E2,

...(ii)

cos 30

...(iii)

cos 30

...(iv)

2 3q 8p ε od

=−

2

−

...(i)

V1 , the ratio of electrostatic potentials on V2

Electrostatic potential on the surface of sphere

cos 30

−

So, option c is correct.

the surface of two spheres S1 and S2.

C –2q

4pe od 2 4 3q

v 2 ∝ x ,v ∝ x

To find :

x

From equations (i), (ii), (iii) and (iv) : 4q 2q Ex( net ) = − cos 30 − cos 30 − 2 4pe od 4pe od 2 2q

Ex( net ) = −

...(i)

E1 R = 1 E2 R2

To find : |Ex(net)| the magnitude of x component of net electric field due to the three charges at point O. The x component of net electric field at O, will be sum of x components of electric fields due to A, B and C. Ex( net ) = ExA + ExB + E xC ...(i) ExA = −

1 2 mv 2

From equation (i) :

–4q A

150°

4q

qEx =

S1 : V1 = E1R1

...(ii)

Electrostatic potential on the surface of sphere S 2 : V2 = E2 R 2

...(iii)

From equations (i), (ii) and (iii) : R  V1 =  1 V2  R2 

2

Ans. 19. Option (c) is correct. Given : The dipole moment of an electric dipole located at origin is  p = ( −i − 3 j + 2 k ) × 10 −29 C -m.  To find : E,the electric field due to this dipole at  r = +i + 3 j + 5k .

cos 30

2 3q 8p ε od 2

3q p ε od 2

3q p ε od 2

Ans. 17. Option (c) is correct. Given : Mass of particle is m, charge on particle is q, the particle is released from rest in a uniform electric field. To find : Dependence of speed v of particle on distance x travelled by it. By work energy theorem, the work done on the particle by the electric field will be equal to the change in its kinetic energy.

Since,

 p.r = ( −i − 3 j + 2 k ).( +i + 3 j + 5k ) × 10 −29 = ( −1 − 9 + 10 ) × 10 −29 = 0.

 The electric field vector at r will be antiparallel to  p.   ( here k is a constant.) E = −kp  Among the given options, only option c( c ) fits the criteria.  c = −

 p 10 −29

= −( −i − 3 j + 2 k ) = ( +i + 3 j − 2 k )

Ans. 20. Option (d) is correct. Given : Radius of sphere is R, uniform volume charge density of the sphere is r, a sphere of radius R is carved out of the original sphere as shown. 2

206 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

From equations (vi) and (vii), net electric field at B : 17 ρ R 54ε o

EB = R 2

From equations (v) and (viii) :  EA 9 18 = = 17 34 EB

A R

.

B

 EA To find :  , the ratio of electric fields at point EB A (centre of original sphere) and B due to the remaining portion of the sphere. Electric field inside a sphere of volume charge density r :

ρr E = 3ε o

Ans. 21. Option (d) is correct. Given : Mass of pendulum bob is m = 2 g, charge on bob is q = 5.0 mC, intensity of applied electric field is E = 2000 V / m. To find : q, the angle that the pendulum makes with the vertical at equilibrium.



...(i)

E =

...(ii)

3r 2ε o

In equations (i) and (ii), r is the distance from the centre and R is the radius of the sphere. Using equations (i), electric field at A due to sphere of radius R : E A1 = 0

...(iii)

Using equations (ii), electric field at A due to the R cut out sphere of radius with –r volume charge 2 density : 3

E A2

R ρ   2  = −ρR = 2 6ε o R 3  εo 2

...(iv)

From equations (iii) and (iv), net electric field at A : EA =

−ρR 6ε o

...(v)

Similarly, electric field at B due to sphere of radius R: EB1 =

ρR 3ε o

...(vi)

Electric field at B due to the cut out sphere of radius R with –r volume charge density: 2 3 R ρ   2  = −ρR EB2 = ...(vii) 2 54ε o  3R  3  εo  2 

T

Tcos  qE

Tsin 

Electric field outside a sphere of volume charge density r :

ρ R3

...(viii)

mg

Balance the forces acting on the pendulum bob at equilibrium : T cos θ = mg ; T sin θ = qE ...(i) (T is the tension in the pendulum string) That gives, tan θ =

qE mg

...(ii)

Put given values in equation (ii).  5.0 × 10 −6 × 2000  −1 θ = tan −1   = tan ( 0.5) −3  2 × 10 × 10 

Ans. 22. Option (d) is correct. Given : Charge on solid conducting sphere is Q, charge on the surrounding conducting spherical shell is 0, the potential difference between the two spherical surfaces is V. To find : What will be the new potential difference between the two spherical surfaces if the shell is given a charge –4Q. Let a be the radius of solid conducting sphere and b be the outer radius of the surrounding conducting spherical shell. –4Q +Q

+Q

a

a

b

(i)

b

(ii)

207

ELECTROSTATICS

For diagram (i) : Potential at the surface of solid conducting sphere is

v2 =

  1 is a constant  k = 4p ε o  

v ∝

Vs =

kQ a

Potential at the surface of surrounding conducting spherical shell : kQ Vsh = b So, potential difference between two spherical surfaces is : 1 1 V = Vs − Vsh = kQ  −  a b For diagram (ii) : Potential at the surface of solid conducting sphere is kQ k ( 4Q) − V 's = a b Potential at the surface of surrounding conducting spherical shell : kQ k ( 4Q) V 'sh = − b b So, potential difference between two spherical surfaces is :  kQ k ( 4 Q )   k Q k ( 4 Q )  V 's − V 'sh =  − − − =V b   b b   a Ans. 23. Option (b) is correct. Given : Charge density on uniform positive line charge is l, initial distance between the positive point charge q and the line charge l is ro. To find : The speed of the point charge as function of instantaneous distance r from the line charge. Electric field of positive line charge measured at distance x from the line charge : 2k λ E = ...( i ) x

  1 is a constant k = 4p ε o   Force acting on a point charge q located a distance x from the line charge : 2 qk λ F = qE = ...(ii) x Work done in moving a point charge q from distance ro to r towards the line charge : r 2 qk λ r dx W = ∫ dx = 2 qk λ ∫ ro x r0 x r = 2kqλ ln    ro 

...(iii)

By work energy theorem, the work done on a particle is equal to change in its kinetic energy. ∆K = K final − K initial = W

r 1 2 mv − 0 = 2 kqλ ln   2  ro 

4 kqλ  r  ln   m  r0  r ln    ro 

Ans. 24. Option (b) is correct. Given : Two charges forming an electric dipole are q and – q, separation between the two charges is d, mass of the two charges is m, intensity of applied electric field is E. To find : w, angular frequency of the dipole if it is slightly rotated from its equilibrium orientation. Let at equilibrium the dipole make an angle q with the direction of applied electric field E. Now, the torque on the dipole will be : τ = Iα = − pE sin θ ...(i) In equation (i), I is moment of inertia of the system : 2

2

md 2 d d I = m  + m  = , 2 2 2 a is the angular acceleration and p = qd , is the dipole moment. From equation (i) : pE α = − θ I

...(ii)

...(iii)

...(iv)

(small angle approximation, sin q ª q) Substitute from equation (ii) and (iii) in equation (iv). 2 qEθ 2 α = − qdEθ × =− ...(v) md md 2 Comparing equation (v) with general equation of SHM :

α = −ω 2θ

...(vi)

We get :

ω =

2qE md

Ans. 25. Option (c) is correct. Given : Electric field in a region is E = (Ax + B) i , A = 20, B = 10, potential at x = 1 is V1, potential at x = – 5 is V2, all the values are in SI units. To find : V1 – V2. Electric potential is given as : dV = − E.dx ...(i) From equation (i) : V1 − V2 = − ∫

x =1

x =−5

( Ax + B) dx

Put values of A and B. 20 V1 − V2 = − [ x 2 ]xx ==−1 5 − 10[ x]xx ==−1 5 2 = −10( −24 ) − 10( 6 ) = 240 − 60 = 180V

208 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 26. Option (d) is correct. Given : A system of three charges, D  d

Dependence of electric field of a dipole is : 1 E ∝ 3 r

D +q •

• –q

Dependence of electric field of a quadrupole is : 1 E ∝ 4 r

• Q

d

To find : Potential energy of the system. Potential energy of the dipole formed by two charges +q and –q separated by a distance d : U1 = −

PHYSICS

kq 2 d

...(i)

and so on. So, option c is correct. 1 E ∝ D4

Ans. 28. Option (c) is correct. Given : Charge on particle A is qA = 1 mC, particle   1 A is fixed at point P, charge on particle B is qB = is a constant  k = 4p ε o   1 mC, mass of particle B is mB = 4 mg, distance between two particles is d = 1 mm. Potential energy of the charge Q and the dipole : To find : v, velocity of B at distance d¢ = 9 mm from kQqd A. U 2 = − 2 ...(ii) D Potential energy of the system of two given particles when they are d = 1 mm, apart : Total potential energy of the system: U = U1 + U 2 = −

= −

kq 2  kQqd  + −  d  D2 

U1 =

Potential energy of the system of two given particles when they are d¢ = 9 mm, apart :

1  q 2 qQd  + 2   4p ε o  d D 

Ans. 27. Option (c) is correct. Given : Four point charges are located as charge –q at y = –2d, charge +q at y = –d, charge +q at y = +d and charge –q at y = +2d. To find : Magnitude of electric field at point x = D, D  d.

U2 =

kq A qB 1 (1 × 10 −6 )2 = = 1J d' 4p ε o 9 × 10 −3

Potential energy lost by the system of particles will be equal to the kinetic energy gained by particle B. 1 U1 − U 2 = mB v 2 2 9 −1 =

–q Y d +q

1 ( 4 × 10 −6 )v 2 ; 2

v = 2 × 10 3 m / s

d P D d

kq A qB 1 (1 × 10 −6 )2 = = 9 J d 4p ε o 1 × 10 −3

E1 x E2

+q d –q

The 4 charges are arranged in the xy plane as shown above. The combination of charges can be viewed as pair of dipoles or a quadrupole. Due to symmetry of problem the y component of E1 and E2 will cancel out and net electric field due to quadrupole will be along the x-axis, which is perpendicular to the quadrupole. We know, Dependence of electric field for a point charge is : 1 E ∝ 2 r

Ans. 29. Option (d) is correct. Given : Length of pendulum placed between plates of a parallel plate capacitor is L, electric field produced by a parallel plate capacitor is E, mass of pendulum bob is m, charge on pendulum bob is q. + + + + + + + + + + + + + + +

L m q

E

– – – – – – – – – – – – – – –

qE

Fnet=manet mg — — — —

To find : T, time period of the pendulum. Forces acting on the pendulum bob when it is oscillating between the capacitor plates :

• Gravitational force, F1 = mg, acting downwards. • Electrostatic force, F2 = qE, acting towards the right.

209

ELECTROSTATICS

Net acceleration of the pendulum due to the two forces : anet =

 qE  a12 + a22 = g 2 +   m

2

Time period of pendulum : L = 2p anet

T = 2p

L  qE  g2 +   m

2

Ans. 30. Option (b) is correct. Given : A conducting spherical shell of inner radius a and outer radius b, charge on the shell is Q, a dipole p is at the centre of the shell. To find : The correct statement among the given set. +

+ +

–

+

–

– p – •+q – + •–q + + + + +

+

Electric field at point P due to the dipole : kp E = − 3 3 cos2 θ + 1 d q is the angle between the y-axis (the line at which point P is located) and the axial line of the dipole. From the diagram above, q = 90°. That gives, 1 p E = −  4p ε o d 3 Ans. 32. Option (a) is correct. Given : Total charge on a sphere is 2Q, radius of sphere is R, charge density on the sphere is ρ ( r ) = kr , ...(i) r is the radial distance from the centre, two charges A and B of magnitude –Q are at diametrically opposite points on the sphere, separation between A and B is 2a. To find : Value of a, if A and B do not experience any force. R

–

+ +

+

+ –Q

+

outer surface of the shell.

• Due to presence of dipole p, there will be a non uniform distribution of charges on the inner surface of the shell. That gives, options a and d are incorrect.

• As the net charge on the dipole is zero, the net electric field outside the shell is same as that of a point charge at the centre of the shell. So, option b is correct and option c is incorrect. Ans. 31. Option (b) is correct. Given : A point dipole p = − po x is located at origin. To find : The potential and the electric field on the y-axis at a distance d from the origin, due to the dipole. P

•

r dr

B

Consider a spherical shell of thickness dr located at a radial distance r within the give sphere of radius R. Charge dQ present in the shell : dQ = 4p r 2 dr × ρ ( r ) = 4π kr 3 dr ...(ii) Total charge on the sphere : 2Q =

a

• –q

Potential at point P due to the two charges shown in the figure above : k( + q ) k( −q ) V = V+ q + V− q = + =0 2 2 a +d a2 + d 2

R

R

∫0 dQ = ∫0 4p kr = p kR 4 ; k

=

3

dr

2Q p R4

...(iii)

Using Gauss’ law, electric field on the surface of the sphere of radius a : a 1 a E ∫ dA = 4p kr 3dr 0 ε o ∫0 (dA is the surface area of the sphere of radius a) 1  2Q  a 3 E( 4p a 2 ) = 4p   ∫ r dr εo  p R4  0

E =

d a

2a

+

+

• Charge Q will be uniformly distributed on the

• +q

A

–Q

=

1  2Q  a 4   ε 0 a2  p R4  4 1  2Qa 2    4p ε o  R 4 

...(iv)

Now, force of attraction on charge –Q at point A, due to positive charge on sphere of radius a : Fa = QE =

1  2Q 2 a 2    4p ε o  R 4 

...(v)

210 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Force of repulsion on charge –Q at point A, due to negative charge at point B. 1 ( −Q)( −Q) Fr = 4p ε o ( 2 a )2 1 Q2 =  4p ε o 4 a 2

PHYSICS

Ans. 35. Option (a) is correct. Given : Two charges q1 = 10 mC and q2 = –25 mC are placed on the x-axis at x1 = 1 m and x2 = 4 m respectively. To find : The electric field at point y = 3 cm, on the y-axis.

...(vi)

y-axis

As A does not experience any force, equate equations (v) and (vi). 1  2Q 2 a 2  1 Q2 =   4p ε o  R 4  4p ε o 4 a 2 8a

4

R4

y=3

= 1 1

a = 8 −1 / 4 R

x=1

Ans. 33. Option (b) is correct. Given : Radius of a ring is R, the ring is uniformly charged, the electric field on its axis has the largest magnitude at a distance h from its centre. To find : Value of h. Electric field on the axis of a uniformly charged ring at a distance h from its centre : kQh E = ...(i) 2 ( h + R 2 )3 / 2

E1 =

=

( h 2 + R 2 )3 / 2

To maximize the electric field at distance h, we should have dE = 0 dh That gives, 3h 2 = h 2 + R 2 ; h =

R 2

.

Ans. 34. Option (a) is correct. Given : Three charges are located along the x-axis d as, +Q at x = 0, q at x = and +Q at x = d, the net 2 force experienced by the charge at x = 0 is 0. To find : The value of q. The net force on the charge at x = 0 : kqQ d 2  

2

+

kQ 2 d2

= 0

x-axis

q1  1  9 × 10 9 × 10 × 10 −6 =  4p ε o  x12 + y 2  12 + 32

Direction of E1 :  E1 = 9 10 × 10 2 (cos θ1 ( −i) + sin θ1 j )  −1  3 = 9 10 × 10 2  i+  2 2 2 1 + 32  1 +3

 E1 = 900( −i + 3 j ) N C −1

 j    ...(i)

Magnitude electric field at y = 3 cm due to charge q2 : E2 = =

q2  1    4p ε o  x 22 + y 2  9 × 10 9 × 25 × 10 −6 4 2 + 32

= 9 × 10 3 N C −1

Direction of E2 :  E2 = 9 × 10 3 (cos θ 2 i + sin θ 2 ( − j ))  4 3 i − = 9 × 10 3   2 2 2 4 + 32  4 +3

 j   

 3  4 E2 = 9000  i − j  = 7200i − 5400 j NC −1 ...(ii) 5  5

Total electric field at y = 3 cm due to charges q1 and q2 :    E = E1 + E2 = 900( −i + 3 j ) + 7200i − 5400 j = ( 63i − 27 j ) × 10 2 N C −1

4q + Q = 0 q = −

x=4

= 9 10 × 10 2 N C −1

3   h( 2 h )   1 dE 2 = kQ  2 − 2  2 3/2 dh ( h + R 2 )5 / 2   (h + R )    3h 2  1 − 2 2  h + R 

q2

2

Magnitude electric field at y = 3 cm due to charge q1 :

( Let Q be the total charge on the ring.) Differentiate equation (i) with respect to h.

kQ

q1

Q 4

Ans. 36. Option (b) is correct. Given : Radius of solid sphere is R, volume charge density within the sphere is

211

ELECTROSTATICS

ρ (r ) =

A r

2

e −2 r / a ,

...(i)

total charge on the sphere is Q. To find : Radius R. Consider a spherical shell of thickness dr located at a radial distance r within the given sphere of radius R. Charge in the spherical shell will be : dQ = ρ ( r )( 4p r 2 dr )

...(ii)

Total charge on the sphere : R

∫ dQ = ∫0 Q=

R

∫0

2

ρ ( r )( 4p r dr ) A r2

R

0

4p A  −2 R / a  e −1   2 −  a  

Q = −2p aA  e −2 R / a − 1  

−

Q 2p aA

   a  1  R = ln Q  2    1 − 2p aA    

Ans. 37. Option (d) is correct. Given : Three concentric spherical shells are of radii a, b, c (a < b < c), surface charge densities on all the three shells is same and is equal to s, total charge on the three shells is Q. To find : Total potential at a point, at a radial distance r from the common centre of the shells (r < a). Let the three shells have charge : Qa, Qb, Qc on them. That gives, Q a + Qb + Q c = Q ...(i) and

4p a 2

=

Qb

4p b 2

=

Qc

4p c 2

=σ

Qa = Qb =

a2 a2 + b2 + c 2 b 2

2

a + b2 + c2

Q,  Q,

...(v)

 1  k =  4p εo   Potential at r due to shell with radius b : kQb kbQ Vb = = 2 a a + b2 + c2

...(vi)

...(vii)

From equations (v), (vi) and (vii), total potential at r: Q  a+b+c  V = Va + Vb + Vc = 4p ε o  a 2 + b 2 + c 2 

Potential at P due to B : VB =

2 qa

...(ii)

( R − x )2

Equate equations (i) and (ii). 4 qa 2 qa = 2 x ( R − x )2 2 x

2

=

1 ( R − x )2

2( R 2 + x 2 − 2 Rx ) = x 2 x 2 − 4 Rx + 2 R 2 = 0 x =

...(iii)

Equations (i) and (iii) together give :

...(iv)

Potential at r due to shell with radius a : kQa kaQ Va = = 2 a a + b2 + c2

...(ii)

Equation (ii) can be rewritten as : Q a : Qb : Q c = a 2 : b 2 : c 2

Q

 Given : Dipole A has dipole moment d A = −4 qai, ,  dipole B has dipole moment dB = −2 qai, separation between two dipoles is R. To find : Distance from A at which both the dipoles produce same potential. Both the dipole moments point towards negative x axis so, the point where both the dipoles produce same potential will lie between A and B. Let the point be P which is at distance x from A. Then potential at P due to A : 4 qa VA = 2 ...(i) x

Q  2R  = ln  1 − a 2p aA  

Qa

a2 + b2 + c 2

Ans. 38. Option (b) is correct.

Q = 1 − e −2 R / a    2p aA e −2 R / a = 1 −

c2

Potential at r due to shell with radius a : kQc kcQ Vc = = 2 a a + b2 + c2

e −2 r / a ( 4p r 2 )dr

= 4p A ∫ e −2 r / adr =

Qc =

4 R − ( 4 R )2 − 4( 2 R 2 ) 2

= 2 R − 2 R = 2 R( 2 − 1) =

2 R( 2 − 1) 2 +1

=

2R 2 +1

Ans. 39. Option (b) is correct. Given : Four point charges of magnitude Q are placed in an xy plane at points (0, 2), (4, 2), (4, –2), (0, –2).

212 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) To find : The work required to put a fifth charge Q at the origin of the coordinate system. Electrostatic potential at (0, 0) due to charge Q located point (0, 2) : kQ kQ = ...(i) V1 = 2 [( 0 − 0 )2 + ( 2 − 0 )2 ]1 / 2

 1  k =  4p ε o   Electrostatic potential at (0, 0) due to charge Q located point (4, 2) : kQ kQ = V2 = ...(ii) 2 2 1/2 20 [( 4 − 0 ) + ( 2 − 0 ) ] Electrostatic potential at (0, 0) due to charge Q located point (4, –2) : kQ kQ = V3 = ...(iii) 20 [( 4 − 0 )2 + ( −2 − 0 )2 ]1 / 2 Electrostatic potential at (0, 0) due to charge Q located point (0, –2) : kQ kQ = V4 = ...(iv) 2 [( 0 − 0 )2 + ( −2 − 0 )2 ]1 / 2 Total potential at (0, 0) due to presence of four charges of equal magnitude Q : 1   V = V1 + V2 + V3 + V4 = kQ  1 +  5 

...(v)

Now, work done to bring another charge Q at (0, 0) will be : 1  1  Q2  W = QV = kQ  1 + = 1 +  5 5  4p ε o   2

Ans. 40. Option (d) is correct. Given : An arrangement of a dipole located along the x-axis, the distance AB = 2a, P

O –q

kp

...(i)

y3

Force on charge Q placed at point P : kQp = F QE = P y3

...(ii)

Electric field at P’ due to the dipole : kp EP' = 3/2 2  y   2   + a   3   (p = 2qa, is the dipole moment) As

y  2a, 3 EP' =

27 kp y3

...(iii)

Force on chare Q placed at point P’ :  kQp  F ' = QEP' = 27  3  = 27 F  y    Ans. 41. Option (d) is correct. Given : Three charges are arranged as shown in the figure. Q a +q a To find : Value of Q, if total electrostatic energy of the configuration is zero (U = 0). From the diagram, electrostatic energy between charges +q and +q separated by distance a : +q

U1 =

kq 2 a

...(i)

Electrostatic energy between charges +q and Q separated by distance a : kQq U2 = ...(ii) a

Q P' A

EP =

PHYSICS

+q

B

O is the midpoint of the dipole and distance OP = y  2a, charge Q when located at P experiences a force F. To find : F¢, force on charge Q when it is located at y   P¢  OP' =  2 a  . 3   Electric field at P due to the dipole : kp EP = 2 ( y + a 2 )3 / 2 (p = 2qa, is the dipole moment) As y  2a,

Electrostatic energy between charges +q and Q separated by distance 2 a : kQq U3 = ...(iii) 2a Total Electrostatic energy of the system of three charges : U = U1 + U 2 + U 3 =

kq  Q  q + Q + =0 a  2

1   Q1 +  = −q 2  Q = −

2q 2 +1

213

ELECTROSTATICS

Ans. 42. Option (b) is correct. Given : A graph showing variation of a quantity with distance r from the centre. To find : What quantity is plotted on the y-axis. Potential inside a uniformly charged spherical shell is : kQ = Vinside V = = constant ...(i) surface ro

Charge –2q can be thought of as two charges of magnitude –q each. p= 1 pnet =

angle between the direction of electric field and the equatorial line of a dipole is q = 45°, value of dipole moment is p = 10–29 Cm. To find : U, potential energy of the dipole. Potential energy of the dipole : U = − p.E = pE cos θ

negative y-axis. So,  pnet = − 3ql j Ans. 45. Option (c) is correct. Given : Initial radius of a uniform spherically symmetric surface charge density is Ro, the initial velocity of the charge density is 0. To find : The best representation of velocity v of the charge density as function of its radius R(t), as it expands due to repulsion. = t 0= , radius R 0 ,v = 0

At

= −7 × 10 −27 J

By law of conservation of energy : 0+

dv kQ2 1 m( 2 v ) − 2 = 0 dR 2 R 2 dv kQ2 = dR 2mvR 2

–2q l pnet l

kQ 2 1 kQ 2 = mv 2 + 2Ro 2 2R

Differentiating the above equation :

Ans. 44. Option (d) is correct. Given : A system of three charges placed on the vertices of an equilateral triangle.

l

...(i)

and let at t = t, radius = R.

U = −10 −29 × 1000 × cos 45

+q

...(iii)

Direction of net dipole moment is along the

From equation (i) and (ii), we can see option (b) is correct. Ans. 43. Option (b) is correct. Given : Applied electric field is E = 1000 V / m ,

p1

...(ii)

in equation (ii). pnet =  3ql

kQ r

(r is the radial distance from the centre of shell) 1 ...(ii) Voutside ∝ r

y

p12 + p22 + 2 p1 p2 cos 60

...(i)

Substitute the values of p1 and p2 from equation (i)

Outside the shell : Voutside =

p= ql 2

...(ii)

From equations (i) and (ii), the graph should start p2 +q x

To find : Electric dipole moment of the system of charges.

dv dR decreases with increasing v and R and should tend from origin and increase with R, the slope to 0 as R ® ∞.

214 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Topic-2

Capacitors Concept Revision (Video Based) Capacitor Circuits

Capacitors

Part -1

Part - 2 Dielectrics

                 

Part -1

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1 A 10 mF capacitor is fully to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is : (a) 10 mF (b) 20 mF (c) 30 mF (d) 15 mF [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. In the circuit shown in the figure, the total charge is 750 mC and the voltage across capacitor C2 is 20 V. Then the charge on capacitor C2 is : C2

C1 =15 F

C3 =8 F

..

+

V

–

(a) 650 mC (c) 590 mC

(b) 450 mC (d) 160 mC [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3 A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged C capacitor of capacitance . The energy loss in the 2 process after the charge is distributed between the two capacitors is : 1 1 (a) CV02 (b) CV02 6 2 1 2 (c) CV0 4

1 (d) CV02 3 [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.4. Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one

is connected to the negative terminal of the other. The final energy of this configuration is : 9 25 2 CV 2 (a) (b) CV 2 6 (c) zero

(d)

3 CV 2 2

[JEE (Main) – 5th Sep. 2020 - Shift-2] Q.5. A parallel plate capacitor has plate of length ‘l’ width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. at what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored? l l (a) (b) 4 2 (c)

l 3

(d)

2l 3

[JEE (Main) – 5th Sep. 2020 - Shift-2] Q.6. In the circuit shown, charge on the 5mF capacitor is : 2 F

4 F

5 F

6V

O

6V

(a)

120 mC 11

(b)

150 mC 11

(c)

180 mC 11

(d)

90 mC 11

[JEE (Main) – 5th Sep. 2020 - Shift-2]

215

ELECTROSTATICS

Q.7.



(a) 3 V (c) 5 V

A parallel plate capacitor has plates of area A separated by distance ‘d’ between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + ax) where ‘x’ is the distance measured from one of the plates. If (ad) rC (a) rC > rA > rT (c) rA > rM > rC (d) rM > rA > rC [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. Model a torch battery of length l to be made up of a thin cylindrical bar of radius ‘a’ and a concentric thin cylindrical shell of radius ‘b’ filled in between with an electrolyte of resistivity r (see figure). If

the battery is connected to a resistance of value R, the maximum joule heating in R will take place for  :

231

CURRENT ELECTRICITY

(a) R =

ρ b 2pl  a 

(b) R =

(c) R =

ρ b ln pl  a 

(d)

2ρ  b  ln pl  a 

ρ b ln 2pl  a 

[JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. Two resistors 400 W and 800 W are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 kW across 400 W resistor is close to : (a) 2 V (b) 1.95 V (c) 2.05 V (d) 1.8 V [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.4. A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is : (a) 0.125 W (b) 0.50 W (c) 0.10 W (d) 0.072 W [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.5. The value of current i1 flowing from A to C in the circuit diagram is :

(a) 0.4 (c) 0.2

(b) 0.25 (d) 0.5 [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.9. In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be : (a) 25 A (b) 15 A (c) 10 A (d) 20 A [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.10. In the given circuit diagram, a wire is joining points B and D. The current in this wire is :

(a) 0.4A (c) 2A

(b) zero (d) 4A [JEE (Main) – 9th Jan. 2020 - Shift-1]

(a) 5 A (c) 2.25 A

(b) 0.75 A (d) 1 A [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.6. An electrical power line, having a total resistance of 2W, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately : (a) 85% (b) 91% (c) 96% (d) 72% [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.7. A circuit to verify Ohm’s law uses ammeter and voltmeter in series or parallel connected corrected correctly to the resistor. In the circuit : (a) ammeter is always used in parallel and voltmeter is series (b) ammeter is always connected in series and voltmeter in parallel (c) Both ammeter and voltmeter must be connected in parallel (d) Both ammeter and voltmeter must be connected in series [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.8. The current I1 (in A) flowing through 1W resistor in the following circuit is :

Q.11. A 200 W resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be : (a) 100 W (b) 400 W (c) 300 W (d) 500 W [JEE (Main) – 8th April 2019 - Shift-1] Q.12. For the circuit shown, with R1 = 1.0 W, R2 = 2.0 W, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points ‘a’ and ‘b’ is approximately (in V) : R1 R1 a

.

R2

E1

. R1

(a) 2.7 (c) 3.7

E2

E3 R1

b

(b) 2.3 (d) 3.3 [JEE (Main) – 8th April 2019 - Shift-1] Q.13. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when : (a) R = 0.001 r (b) R = 1000 r (c) R = 2r (d) R = r [JEE (Main) – 8th April 2019 - Shift-2]

232 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.14. In the figure shown, what is the current (in Ampere) drawn from the battery? You are given : R1 = 15 W, R2 = 10 W, R3 = 20 W, R4 = 5 W, R5 = 25 W, R6 = 30 W, E = 15 V R3

.

R1 E

+ –

R2

R4

PHYSICS

and mean free time is 25 fs (femto second), it’s approximate resistivity is : (me = 9.1 × 10–31 kg) (a) 10–6 Wm (b) 10–7 Wm –8 (c) 10 Wm (d) 10–5 Wm [JEE (Main) - 9th April 2019 - Shift-2] Q.19. In the given circuit, an ideal voltmeter connected across the 10 W resistance reads 2 V. The internal resistance r, of each cell is : 15  2

R6

R5

(a) 13 24 9 (c) 32

(b)

7 18

(d)

20 3

10 

1.5 V, 1.5 V r r

[JEE (Main) – 8th April 2019 - Shift-2] Q.15. Determine the charge on the capacitor in the following circuit :

. .

I2

I2 I1 2  6 10  4

10 F

(b) 0.5 W (d) 0 W [JEE (Main) – 10th April 2019 - Shift-1] Q.20. In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line.

>

72 V

. .

> > >

I

(a) 1 W (c) 1.5 W

(b) 2 mC (d) 200 mC [JEE (Main) – 9th April 2019 - Shift-1] Q.16. A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is : (E is mid-point of arm CD)

lnR(T)

(a) 60 mC (c) 10 mC

D

(a) R (c)

3 R 4

E

C

(b) (d)

7 R 64 1 R 16

[JEE (Main) – 9th April 2019 - Shift-1] Q.17. A metal wire of resistance 3 W is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be : 5 12 W (a) (b) W 2 5 5 (c) W 3

One may conclude that : 2 2 R (a) R(T) = 20 (b) R(T) = R 0 e - T0 / T T (c) R(T) = R 0 e - T

B

A

1/T2

[JEE (Main) – 9 April 2019 - Shift-2] Q.18. In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3

/ T02



T (d) R(T) = R 0 e

2

/ T02

[JEE (Main) – 10th April 2019 - Shift-1] Q.21. Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity r. The resistance between the two spheres will be : (a)

ρ Ê 1 1ˆ 4p ÁË a b ˜¯

ρ Ê 1 1ˆ (b) 2p ÁË a - b ˜¯

(c)

ρ Ê 1 1ˆ + 2p ÁË a b ˜¯

(d)

ρ Ê 1 1ˆ + 4p ÁË a b ˜¯

[JEE (Main) – 10th April 2019 - Shift-2] Q.22. The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is : 4R 6R R R 4R

7 (d) W 2 th

2

(a) 6 W (c) 1 W

12 R =16 V (b) 8 W (d) 16 W [JEE (Main) – 12th April 2019 - Shift-1]

233

CURRENT ELECTRICITY

Q.23. A resistance is shown in the figure. Its value and tolerance are given respectively by : Red

R3

Orange

R4

R1 R2

Violet

Silver

(a) 270 W, 10 %

(b) 27 kW, 10 %

(c) 27 kW, 20 %

(d) 270 W, 5 % [JEE (Main) - 9th Jan. 2019 - Shift-1]

Q.24. Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is v. If the electron density in copper is 9 × 1028/m3 the value of v in mm/s is close to

(Take charge of electron to be = 1.6 × 10–19 C) (a) 0.02

(b) 3

(c) 2

(d) 0.2 [JEE (Main) – 9th Jan. 2019 - Shift-1]



Q.25. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is : (a) 2.0 %

(b) 2.5 %

(c) 1.0 %

(d) 0.5 % [JEE (Main) – 9th Jan. 2019 - Shift-1]

Q.26. When the switch S, in the circuit shown, is closed, then the value of current i will be :

.

20 V A

i1

i2

.

C 2

.10 V

18 V

(a) 300 W (c) 550 W

(b) 450 W (d) 230 W [JEE (Main) - 9th Jan. 2019 - Shift-2] Q.29. A uniform metallic wire has a resistance of 18 W and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is : (a) 4 W (b) 8 W (c) 12 W (d) 2 W [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.30. A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is : (a) 20 mA (b) 100 mA (c) 0.4 mA (d) 63 mA [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.31. The actual value of resistance R, shown in the figure is 30 W. This is measured in an experiment as shown using the standard formula R = V/I, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is :

4

i

V

2

A

.. S

R

V=0

(a) 3 A

(b) 5 A

(c) 4 A

(d) 2 A [JEE (Main) - 9th Jan. 2019 - Shift-1]

Q.27. A carbon resistance has a following colour code. What is the value of the resistance ?

GOY



Golden

(a) 530 kW ± 5%

(b) 5.3 MW ± 5%

(c) 6.4 MW ± 5%

(d) 64 kW ± 10%

[JEE (Main) - 9th Jan. 2019 - Shift-2]

Q.28. In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 W, R3 =100 W and R4 = 500 W and the reading of an ideal voltmeter across R4 is 5 V, then the value of R2 will be :

(a) 600 W (c) 35 W

(b) 570 W (d) 350 W [JEE (Main) - 10th Jan. 2019 - Shift-2] Q.32. A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is : (a) 11 × 10–5 W (b) 11 × 10–3 W –4 (c) 11 × 10 W (d) 11 × 105 W [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.33. Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be : (a) 30 W (b) 60 W (c) 120 W (d) 240 W [JEE (Main) - 11th Jan. 2019 - Shift-1]

234 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.34. In the circuit shown, the potential difference between A and B is : 1

1V

1

2V

PHYSICS

Consider a cylindrical shell at distance x from the axis of the battery and the thickness of the shell be dx.

M

.

5

A

D

10  C

1

. B

3V

N

(a) 1 V (c) 3 V

(b) 2 V (d) 6 V [JEE (Main) - 11th Jan. 2019 - Shift-2] Q.35. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then : (a) P1 =16 W, P2 = 4 W (b) P1 = 16 W, P2 = 9 W (c) P1 = 9 W, P2 = 16 W (d) P1 = 4 W, P2 = 16 W [JEE (Main) – 12th Jan. 2019 - Shift-1]

Resistance of the shell : ρ dx dr = 2p xl Resistance of the battery : b ρ dx ∫ dr = ∫a 2p xl r =

ρ b dx 2p l ∫a x

r =

ρ b ln 2p l  a 

ANSWER – KEY 1. (d) 5. (d) 9. (d) 13. (d) 17. (c) 21. (a) 25. (c) 29. (a) 33. (d)

2. (d) 6. (c) 10. (c) 14. (c) 18. (c) 22. (b) 26. (b) 30. (a) 34. (b)

3. (b) 7. (b) 11. (d) 15. (d) 19. (b) 23. (b) 27. (a) 31. (b) 35. (a)

4. (c) 8. (c) 12. (d) 16. (b) 20. (b) 24. (a) 28. (a) 32. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. Given : Resistivity of copper is rC, resistivity of tungsten is rT, resistivity of mercury is rM and resistivity of aluminium is rA. To find : Arrange the given resistivities in correct order. We know,

For maximum joule heating R = r. R =

ρ b ln 2pl  a 

Ans. 3. Option (b) is correct. Given : Resistors R1 = 400 W and R2 = 800 W are connected in series, emf of the battery is E = 6 V internal resistance of voltmeter is r = 10 kW. To find : The reading of the voltmeter, when connected across R1.

ρC = 1.7 × 10 −8 Ω m ρ T = 5.6 × 10 −8 Ω m ρ M = 98 × 10 −8 Ω m ρ A = 2.6 × 10 −8 Ω m From the above data : ρ M > ρ T > ρ A > ρC Ans. 2. Option (d) is correct. Given : Length of torch battery is l, radius of inner thin cylindrical bar is a, radius of concentric outer thin cylindrical shell is b, resistivity of the electrolyte material filled between the two concentric cylinders is r, the battery is connected to an external resistance R. To find : The value of R corresponding to maximum joule heating.

Equivalent resistance of the circuit : rR 1 + R2 R = r + R1 R =

10000 × 400 + 800 = 1184.6 Ω 10000 + 400

Current through the circuit : E 6 = = = 5.07 mA I R 1184.6 Voltage drop across R2 : V2 = R 2 I = 800 × 5.07 × 10 −3 = 4.06 V Voltage drop across R1 : V1 = E − V2 = 6 − 4.06 = 1.94 V

235

CURRENT ELECTRICITY

Ans. 4. Option (c) is correct. Given : Power dissipated in a resistor R is PR = 0.5 W, emf of the battery connected across the resistor is E = 3 V, terminal voltage of the battery is VR = 2.5 V To find : Pr, power dissipated in the internal resistor r of the battery. Let I be the current in the circuit. Power dissipated in resistor R : PR = VR I = I

Ans. 8. Option (c) is correct. Given : Four resistors and a 1 V battery are connected as shown below,

PR 0.5 = = 0.2 A VR 2.5

Power dissipated in resistor r : Pr = Vr I Pr = ( E − VR )I = ( 3 − 2.5) × 0.2 = 0.5 × 0.2 = 0.1W Ans. 5. Option (d) is correct. Given : A resistive circuit.

To find : The current I1 through the 1W resistor. Effective resistance of branch AB : 1×1 1 5 R AB = +2= +2= Ω 1+1 2 2 Effective resistance of branch CD : RCD = 2Ω Equivalent resistance of the entire circuit : 5 2× 5 10 2 R eq = = = Ω 5 9 9 2+ 2 2 Total current through the circuit : 1V 9 = = A = 0.9 A I R eq 10

To find : The current i1 in branch AC. 8 = 1A i1 = 4+4 Ans. 6. Option (c) is correct. Given : Total resistance of an electrical power line is R = 2 W, power delivered by the power line is 1 kW, voltage applied across the power line is V = 220 V. To find : h, the efficiency of the power line. Current through the power line : P 1000 = = I V 220 Power loss in the resistor : = PR

= I 2 R 41.3 W

Efficiency of power line : P η = P+ PR

η =

1000 = 96% 1000 + 41.3

Ans. 7. Option (b) is correct. Given : A circuit is designed to verify Ohm’s law. To find : How to connect a voltmeter and an ammeter in the circuit. An ammeter is always connected in series and a voltmeter is always connected in parallel.

...(i)

The current I gets divided into branches AB and CD. The voltage drop across both the branches is 1 V. Current through branch CD : 1V 1V I2 = = = 0.5 A ...(ii) RCD 2 From equations (i) and (ii), current through branch AB : I AB = I − I 2 = 0.4 A IAB will get equally divided between the two 1W resistors. So, I AB = I1 = 0.2 A 2 Ans. 9. Option (d) is correct. Given : Voltage rating of the electric main supply of a building is V = 220 V, Following are the electric appliances installed in the building : 1. 15 bulbs of 45 W 2. 15 bulbs of 100 W 3. 15 fans of 10 W 4. 2 heaters of 2 kW To find : I, the minimum fuse capacity of the building. Total power consumed by the electric appliances installed in the building : P = 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000 = 4325W

236 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) The minimum fuse capacity of the building : P 4325 = = ª 20 A I V 220

Ans. 12. Option (d) is correct. Given : for the circuit below, R1 = 1.0 W ,R 2 = 2.0 W ,E1 = 2 V ,E2 = E3 = 4 V.

Ans. 10. Option (c) is correct. Given : A circuit diagram as shown below,

.

R1

R1

a

.

As 2 W and 3 W resistors are in parallel, their effective resistance will be : 6 Ω R″ = 5 The equivalent resistance of the circuit will be : 10 R = R′ + R ″ = = 2Ω 5 Current through the circuit : V 20 = i = = 10A R 2 The current i will be distributed among the various resistors in a ratio opposite to the value of resistors :

E3

R2

E1

To find : I, current through wire BD. As 1 W and 4 W resistors are in parallel, their effective resistance will be : 4 R′ = Ω 5

PHYSICS

R1

E2

b R1 To find : The potential difference between point a and b. The above circuit can be viewed as a parallel combination of three cells. 2V 2

.

4V

.

2

a

b 4V

Vab = R eq Ieq

2

Ê 2 4 4ˆ ÁË 2 + 2 + 2 ˜¯ 10 = = = 3.3 V 1 1 1 3 + + 2 2 2

Ans. 13. Option (d) is correct. Given : Internal resistance of a cell is r, value of external resistance connected to the cell is R. To find : Value of R, when the power delivered by the cell to the external resistance will be maximum. Net current through the circuit will be : E I = ...(i) R+r In equation (i), E is the voltage provided by the cell. Power delivered to R : 2

Ê E ˆ P = I2 R = Á R Ë R + r ˜¯ From the above illustration, we can see the current through BD will be : 4i 3i i 10 I = − = = = 2 A 5 5 5 5 Ans. 11. Option (d) is correct. Given : R = 200 W. To find : New value of resistance R¢, if in the colour code of R the red colour is replaced by green. For R = 200 = 20× 101 W, the colour code of the given resistor will be : 2 = Red, 0 = Black ,101 = Brown. New colour code is : Green, black and brown. Green = 5,Black = 0 ,Brown = 10

1

The new colour code will correspond to : R ¢ = 50 ¥ 101 = 500 W

...(ii)

Differentiate equation (ii) : 2

dP Ê E ˆ + RE2 ( -2 ) ( R + r )-3 = Á Ë R + r ˜¯ dR For value of R when P is maximum dP = 0. dR So, equation (iii) becomes : 2

Ê E ˆ -3 2 ÁË R + r ˜¯ + RE ( -2 )( R + r ) = 0 Ê E ˆ ÁË R + r ˜¯

2

=

2 RE2 ( R + r )3

2R = R + r R = r

...(iii)

237

CURRENT ELECTRICITY

Ans. 14. Option (c) is correct. Given : An arrangement of 6 resistors and a battery, R1 = 15 W, R2 = 10 W, R3 = 20 W, R4 = 5, R5 = 25 W, R6 = 30 W, E = 15 V. R3

.

R1 + E –

So, voltage drop across the 4 W resistor is V4 = 72 - 48 = 24 V That gives, I1 =

24 = 6 A ; I 2 = I - I1 = 8 - 6 = 2 A. 4

Hence, voltage drop across the 10 W resistor is R2

R4

V10 = 10 ¥ I 2 = 10 ¥ 2 = 20 V Voltage drop across the capacitor : VC = 20V.

R6 R5 To find : I, current drawn from the battery. Calculating the equivalent resistance of the circuit: • R3, R4, R5 are in series, R eq1 = R 3 + R 4 + R 5 = 20 + 5 + 25 = 50 W  • Resistor R2 is connected parallel to the series combination of R3, R4, R5, R 2 R eq 1 10 ¥ 50 50 R eq 2 = = = W R 2 + R eq 1 10 + 50 6 • Now, resistances R1, Req2, R6 are in series. So, 50 160 R eq 3 = R1 + R eq 2 + R 6 = 15 + + 30 = W 6 3 So, current drawn from the battery : E 15 15 ¥ 3 9 I = = = = A R eq 3 160 160 32 3 Ans. 15. Option (d) is correct. Given : A combination of 4 resistors, 1 capacitor and a battery.

I2

. .

I2 I1 2  6 10  4

10 F

>

72 V

. .

> > >

I

To find : Q, the charge on the capacitor. In steady state the capacitor will act as open circuit. Calculating the equivalent resistance of the circuit : • The 2W, 10 W resistors are in series. R eq1 = 2 + 10 = 12 W. • Req1 is connected parallel to the 4 W resistor. 4 ¥ 12 R eq 2 = = 3 W. 4 + 12 • Req2 and the 6 W resistor are in series. R eq3 = 3 + 6 = 9 W. So, current through the circuit will be : 72 I = = 8A 9 Voltage drop across the 6 W resistor is V6 = 8 ¥ 6 = 48 V.

The charge on the capacitor will be : Q = CVc = 10 ¥ 10 -6 ¥ 20 = 200 mC Ans. 16. Option (b) is correct. Given : Resistance of a wire is R, the wire is bent to form a square ABCD, E is the mid point of arm CD. B A

D

E

C

To find : Reff, effective resistance between points E and C. Let length of each segment of square be x. R So, resistance per unit length of wire will be : . 4x Resistance per unit length of segment EC : Ê R ˆ ¥ length of segment EC˜ R1 = Á Ë 4x ¯ =

R x R ¥ = 4x 2 8

...(i)

Resistance per unit length of segment EDABC : R 7x 7R R2 = ¥ = ...(ii) 4x 2 8 Effective resistance between points E and C will be the equivalent resistance of R1 and R2 that are connected in parallel. R eff

R ¥ R1R 2 = = 8 R1 + R 2 R + 8

7R 8 = 7 ¥ R = 7R 7 R 8 8 64 8

Ans. 17. Option (c) is correct. Given : Resistance of a metal wire is R = 3 W, length of the wire is l, the wire is elongated such that the new length of the wire is 2l, the wire is then bent to form a circle. To find : Req, the equivalent resistance between the two points on the circle that make an angle of q = 60° at the centre. Resistance of the wire in terms of resistivity : ρl R = A where A is area of cross section of wire.

238 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) In process of elongation of wire the volume of the

15 

> >

I

A Al = A ¢ ¥ 2l ; A ¢ = 2

>

of cross section of wire, then :

I2

terms of resistivity :

The wire after elongation is bent to form a circle as shown below :

. .

Total current through the circuit is I = I1 + I2 As the 10 W and the 15 W resistors are parallel, voltage drop across the 15 W resistor = 2 V. 2 2 6+4 1 I = + = = A 10 15 30 3 Now, voltage drop across the series combination of 2 W resistor and the two r W resistors with effective resistance R¢ = (2 + 2r) W will be : V¢ = 1.5 + 1.5 - 2 = 1V

P

By Ohm’s law : V¢ = R ¢I; 1 = ( 2 + 2r )

60°

Resistance of segment AB : R1 =

60 ¥ 12 = 2 W 360

Resistance of segment APB : R2 =

1 3

3 = 2 + 2r r = 0.5 W

B

A

10 

1.5 V, 1.5 V r r

New resistance of the wire (after elongation) in

ρ 2l 4 ρ l = = 4 R = 4 ¥ 3 = 12 W A A 2

2

I1 

wire will remain conserved. Let A¢ be the new area

R¢ =

PHYSICS

360 - 60 ¥ 12 = 10 W 360

Ans. 20. Option (b) is correct. Given : An experimental plot of resistance R of a material as function of temperature T. lnR(To) lnR(T)

The equivalent resistance between points A and B : R eq =

R1R 2 20 5 = = W R1 + R 2 12 3

1/T02 1/T02

Ans. 18. Option (c) is correct. Given : Number of conduction electrons per unit volume of a conductor is n = 8.5 × 1028 m–3, mean free time is t = 25 fs = 25 × 10–15 s, mass of electrons is m = 9.1 × 10–31 kg

To find : The approximate resistivity of the conductor.

ρ =

=

To find : An expression for R(T). From the plot we can say : 1 2 ln R( T ) T + = 1 1 ln R( To ) To2

Ê T2 ˆ ln R( T ) = ln[ R( To )] Á 1 - o2 ˜ T ¯ Ë

m ne 2τ 9.1 ¥ 10 -31 8.5 ¥ 10 28 ¥ (1.6 ¥ 10 -19 )2 ¥ 25 ¥ 10 -15

ª 10 -8 Wm Ans. 19. Option (b) is correct. Given : The voltage drop across the 10 W resistor is 2 V for the circuit below. To find : Value of internal resistance r of each cell.

Ê T2 ˆ R( T ) = R( To )exp Á - o2 ˜ Ë T ¯ Ê T2 ˆ = R o exp Á - o2 ˜ Ë T ¯ Ans. 21. Option (a) is correct. Given : Radii of two concentric conducting spheres are a and b, b > a, the space between the spheres is filled with a medium of resistivity r.

239

CURRENT ELECTRICITY

To find : R, the resistance between two spheres. Consider a spherical shell of radius r with thickness dr (a < r < b). Resistance of the shell : ρ dr dR = 4p r 2

From the table for colour code of carbon resistor, the given colour code corresponds to : Red (value) : 2 Violet (value) :7 Orange (multiplier) : 103 Silver : ±10% So, red and violet together become 27 and order of magnitude of resistance is 10­3.

Integrate the above equation to get the total resistance between the two spheres. b ρ dr ρ b dr R = ÚdR = Ú = a 4p r 2 4p Úa r 2 b

= =

R = 27 ¥ 10 3 W ± 10% R = 27 kW ± 10% Ans. 24. Option (a) is correct. Given : current through copper wire is I = 1.5 A, cross sectional area of copper wire is A = 5 mm2, electron density in copper is n = 9 × 1028 m–3 = 9 × 1019 mm–3, charge on electron is e = 1.6 × 10–19 C. To find : vd, drift speed of electrons in copper wire. Current through a conductor : I = neAvd

a

ρ Ê 1ˆ ρ Ê 1ˆ = 4p ÁË r ˜¯ a 4p ÁË r ˜¯ b

ρ 4p

Ê 1 1ˆ ÁË a - b ˜¯

Ans. 22. Option (b) is correct. Given : Power consumed by a resistive network shown below is P = 4 W. 4R 1

6R

vd =

2 R

R

= 4R

12 R

I neA 1.5 9 ¥ 1019 ¥ 1.6 ¥ 10 -19 ¥ 5

= 0.02 mm / s =16 V

Ans. 25. Option (c) is correct. Given : Original length of copper wire is l, length of copper wire after it is stretched is 0.5 l¢ = l + l = 1.005l, 100

To find : Value of R. Equivalent resistance of component-1 of the given circuit : 4¥4 R eq1 = = 2 R  4+4

volume of wire remains same after stretching. To find : Percentage change in electrical resistance of the copper wire. Let the resistance of copper wire of length l and area of cross section A be R. Resistivity of copper : RA ρ = l

Equivalent resistance of component-2 of the given circuit : 6 ¥ 12 R eq 2 = = 4 R  6 + 12 Equivalent resistance of the given circuit : R eq = 2 R + R + 4 R + R = 8 R Power consumed by the resistive network :

R =

E2 (16 )2 P = = =4 R eq 8R 8R =

256 4

R = 8W Ans. 23. Option (b) is correct. Given : A resistor with a colour code as shown below. R = 8W To find : Value of resistance and tolerance of the resistor. The colour code for a resistor is read from left to right and the last band gives the tolerance. The first two colours give the significant figures of the resistance and the third colour gives the order of magnitude of the resistance.



ρl A

...(i)

New length of copper wire is l¢. As the volume of wire remains same after stretching, new cross sectional area will be : Al Al A A¢ = = = ...(ii) l ¢ 1.005l 1.005 Resistance of the stretched wire : ρ l ¢ ρ 1.005l R¢ = = A A¢ 1.005 = 1.005 ¥ 1.005

ρl = 1.01 R A

Percentage change in electrical resistance of the copper wire : 1.01 R - R R¢ - R ¥ 100 = ¥ 100 = 1% R R

240 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 26. Option (b) is correct. Given : An arrangement of three resistors as shown below. i i2 10 V C 20 V 1 4

i

I

I2




I3

dR 1 ∝ . Two dl l

equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?

R'

I6

>

>

of its resistance R with length l is

P

>

Q.14. In a meter bridge, the wire of length 1 m has a nondR uniform cross-section such that, the variation dl

l



1. (d)

2. (a)

3. (d)

4. (d)



5. (c)

6. (c)

7. (a)

8. (c)



9. (a)

10. (c)

11. (d)

12. (a)

13. (c)

14. (c)

15. (a)

16. (b)

17. (b) (a) 0.2 m (c) 0.25 m

(b) 0.3 m (d) 0.35 m [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.15. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5W. The value of R to give a potential difference of 5 mV across 10 cm of potentiometer wire is: (a) 395 W (c) 490 W

(b) 495 W (d) 480 W [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.16. The galvanometer deflection, when key K1 is closed but K2 is open, equals q0 (see figure). On closing K2 also and adjusting R2 to 5 W, the θ deflection in galvanometer becomes 0 . The 5 resistance of the galvanometer is, then, given by (Neglect the internal resistance of battery) :

R1=220 

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. Given : Length of potentiometer wire PQ is l = 1m, emf of cell 2 is E2 = 1.02 V, null point is located at l¢ = 49 cm from point Q. To find :

E2 , the potential gradient in the l′

potentiometer wire. E2 1.02 V = = 0.02 l′ 49 cm Ans. 2. Option (a) is correct. Given : A circuit diagram

K2 ( )

G

To find : VB – VA From the circuit diagram shown above : Current through the 2 W resistor will be I = 1 A

( ) K1

(a) 5 W (c) 25 W

(b) 22 W (d) 12 W [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.17. In the given circuit diagram, the currents, I1 = –0.3 A, I4=0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are :

VD − VB = 2V VC − VA = 1V VD − VC − ( VB − VA ) = ( 2 − 1)V VB − VA = ( VD − VC ) − 1 VB − VA = 2 × 1 − 1 = 1V

246 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 3. Option (d) is correct. Given : A circuit diagram

0.5 

PHYSICS

V

0.5  A

J

5

50 cm i1+i2

1

10 

20V 2 

2

i1 10 V R

1 B

4

To find : i1, the current in the 10 V battery. For loop-1 : 20 = 5i2 + 10(i1 + i2 ) + 2i2 20 = 17i2 + 10i1

...(i)

For loop-2 : 10 = 10(i1 + i2 ) + 4ii 10 = 10i2 + 14i1 From equations (i) and (ii) : i1 = −0.217 A

...(ii) 

So current through the 10 V battery is from positive to negative terminal. Ans. 4. Option (d) is correct. Given : Length of potentiometer wire is l = 1200 cm, current through the potentiometer wire is I = 60 mA, emf of cell connected to the potentiometer is E = 5 V, internal resistance of the cell is r = 20 W, null point for the cell on the potentiometer wire is at l¢ = 1000 cm. To find : R, the resistance of the wire.

100 cm

To find : Reading of an ideal voltmeter with jockey J at l = 50 cm from A. Total resistance of the potentiometer wire is R¢ = r ¢ ¥ l ¢ = 0.01 ¥ 400 = 4 W Net internal resistances of the two cells : r = 0.5 + 0.5 = 1W Equivalent resistance of the circuit : R eq = R + r + R ¢ = 1 + 1 + 4 = 6W Current through the circuit : Net emf 3 1 I = = = A R eq 6 2 Voltage drop across the potentiometer wire : 1 V¢ = R¢ ¥ I = 4 ¥ = 2W 2 Voltage drop at l = 50 cm from A : 50 l V = ¥ V¢ = ¥ 2 = 0.25 V l¢ 400 Ans. 6. Option (c) is correct. Given : A circuit diagram for a meter bridge experiment and a set of 4 corresponding observations. To find : Which among the 4 readings is inconsistent. R Resistance box

X

.. . G

UNknown resistance

l

Let terminal voltage of the primary battery of the potentiometer be V. The potential gradient on the wire will be : E V = l′ l 5 V = 1000 1200 V = 6V Resistance of the potentiometer wire : V 6 R = = = 100 Ω I 60 × 10 −3 Ans. 5. Option (c) is correct. Given : Length of potentiometer wire is l¢ = 400 cm, the resistance per unit length of the wire is r¢ = 0.01 W/cm.

. E

K

In meter bridge experiment unknown resistance 100 - l x = R ...(i) l From reading 1 : 100 - 60 x = ¥ 1000 = 667W 60 From reading 2 : 100 - 13 x = ¥ 100 = 669W 13 From reading 3 : 100 - 1.5 x = ¥ 10 = 667 W 1.5

247

CURRENT ELECTRICITY

From reading 4 : 100 - 1 x = ¥ 1 = 99W 1 So, reading 4 is inconsistent. Ans. 7. Option (a) is correct. Given : The given circuit is used by a student to verify Ohm’s law, the given graph shows the measured voltage plotted as a function of the current, Vo is almost zero.

To find : The length AJ at which the galvanometer shown no deflection. Current through the potentiometer wire : ε ε i = = ...( i ) r + 12r 13r Let length of wire AJ be x. So, potential drop across the balance length AJ : Ê 12r ˆ VAJ = iR AJ = i Á ¥ x˜ ...(ii) Ë L ¯ As the null point occurs at J, ε VAJ = 2

V internal resistance

From equations (i), (ii) and (iii) : ε Ê 12r ε ˆ ¥ x˜ = ¯ 13r ÁË L 2

Ammeter

R

x =

To find : Identify the correct statement among the given set of 4 statements. Effective resistance of the circuit : R eq = r + R , where r is internal resistance of the cell. Current through the circuit : E I = R+r

13 L 24

Ans. 9. Option (a) is correct. Given : In a Wheatstone bridge the resistor R1 has colour code “orange, red, brown”, R2 = 80 W, R4 = 40 W. R2

R1

E = IR + Ir

G

...(i)

Voltage difference across R : E = V + Ir

...(iii)

...(ii)

R3

R4

From graph at : I = 0 ,V = 1.5 V From equation (ii) at : I = 0 ,E = V = 1.5 V

...(iii)

From graph at : I = 1000 mA = 1A ,V = Vo ª 0 V From equation (ii) at : I = 1A ,E = Ir = r

–

To find : Colour code for resistor R3. The colour code “orange, red, brown” corresponds to : ...(iv)

From equations (iii) and (iv) : r = E = V = 1.5 W Ans. 8. Option (c) is correct. Given : Length of potentiometer wire is L, resistance of wire is RAB = 12r emf of cell D is e, ε internal resistance of cell D is r, emf of cell C is , 2

internal resistance of cell C is 3r. C

+

R1 = 32 ¥ 101 = 320 W Condition for balancing of a Wheatstone bridge : R1 R = 3 R2 R4 R 320 = 3 80 40 R 3 = 160 W Colour code as read from left to right, for R3 =160 W = 16 × 101 W will be : 1 Æ brown ,6 Æ blue,101 Æ brown.

A

B J

6V

Rh

(brown, blue, brown) Ans. 10. Option (c) is correct. Given : An arrangement of 2 resistors and 2 batteries, the internal resistance of the batteries is zero.

248 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) I1

Case 1 :

I2

P Q = 400 X

I1+I2

+

–

R2

20 

R1

+

10V

20 

–

P 400 = Q X

Q P  = 405 X

To find : The current through R1 and R2. Applying Kirchhoff ’s loop rule to the left loop of the given circuit : 10 - ( I1 + I 2 ) R1 = 0 10 - ( I1 + I 2 ) ¥ 20 = 0

P X = Q 405 Equate equations (ii) and (iii) : 400 X = X 405

...(i)

Applying Kirchhoff ’s loop rule to the right loop of the given circuit : 10 - ( I1 + I 2 )R1 - I 2 R 2 = 0 (Substitute from equation (i)) 20 I 2 10 = 0 2 20 I2 = 0

...(ii)

Substitute from equation (ii) in equation (i).

X = 402.5 W Ans. 12. Option (a) is correct. Given : The resistance of meter bridge wire AB is R=4W Case a : For emf of cell is e = 0.5 V, rheostat resistance a is R h = 2W , the null point is obtained at some point J, Case b : For emf of cell is e2, rheostat resistance is

R bh = 6W , the null point is obtained at the same point J. C

I1 = 0.5 A Ans. 11. Option (d) is correct. Given : In a Wheatstone bridge resistance P ≈ Q, the bridge is balanced at R = 400 W, on interchanging P and Q the bridge is balanced at R = 405 W.

Q

G

K2

A

B J

6V

B

A

..

C

X

R

...(iii)

X 2 = 400 ¥ 405

1 2

P

...(ii)

Case 2 : on interchanging P and Q, equation (i) becomes, Q P = R X

10V

( I1 + I 2 ) =

PHYSICS

Rh

To find : Value of e2. Let length of wire AB be L and length of wire AJ be x. Case a : Current through wire AB : 6 6 ia = = = 1A a R + Rh 4 + 2 Potential drop across wire AB :

D

.. K1

To find : The value of X. Condition for a balanced Wheatstone bridge : P Q = ...(i) R X

a VAB = Ria = 4 ¥ 1 = 4V

Potential drop across wire AJ : x x a VAJ = Ria = 4 L L As null point is at J : x a VAJ = 4=ε L x 4 = 0.5 L

249

CURRENT ELECTRICITY

1 x = L 8

...(i)

Case b : current through wire AB : 6 6 ib = = = 0.6 A b 4 + 6 R + Rh

Put values of R1 and R¢ in the equation above : 30 R¢¢  = 20 30 + R¢¢

Potential drop across wire AJ : x x Ê xˆ b VAJ = Rib = ¥ 4 ¥ 0.6 = 2.4 Á ˜ Ë L¯ L L

R¢¢ = 60W Ans. 14. Option (c) is correct. Given : Variation of resistance of the meter bridge wire with its length is given as, dR 1 ...(i) µ dl l

As null point is at J : Ê xˆ b VAJ = 2.4 Á ˜ = ε 2 Ë L¯

R¢¢ ( R1 + R ¢ ) = R1  ( R¢¢ + R1 + R¢ )

[substitute from equation (i)] 1 2.4 ¥ = ε 2 8

R'

R'

ε 2 = 0.3 V Ans. 13. Option (c) is correct. Given : The null point in a meter bridge is obtained at a distance l = 40 cm from A. Case a : The null point shifts by 10 cm if a R¢ = 10 W is connected in series with R1, Case b : A resistance R≤ is connected in parallel with R1 + R¢ such that the null point shifts back to its initial position. R1 R2

G A

B

G

.

.

P

A

1–l

l

B l

To find : Length AP at which the galvanometer shows zero deflection. Condition for zero deflection in galvanometer in a meter bridge experiment : R¢ R¢ = R AP R PB R AP = R PB

• To find : Value of R≤. For a null point in a meter bridge : R1 l 40 2 = = = R2 100 - l 100 - 40 3

From equation (i) : dR k = dl l ...(i)

Case a : R1 + R ¢ 50 = R2 100 - 50 R1 + 10 = 1 R2 R1 10 + = 1 R2 R2

...(ii)

(k is a constant) Integrate equation (iii) : l k R AP = Ú dl 0 l R AP = 2k l R PB =

[substitute from equation (i)]

10 2 1 = 1- = R2 3 3 R2 = 30 W and from equation (i), R1 = 20 W Case b : Equivalent resistance when resistance R≤ is connected in parallel with R1 + R¢ : R¢¢ + R1 + R¢ 1 1 1 = + = R eq R1 + R ¢ R ¢¢  R¢¢ ( R1 + R ¢ ) As the null point shifts back to its original position Req = R1

...(iii)

1

Úl

k l

...(iv) dl = 2 k (1 - l )

...(v)

Substitute values from equations (iv) and (v) in equation (ii). 2 k l = 2 k (1 - l ) l = (1 - l )

2 l = 1 l =

1 = 0.25 m 4

Ans. 15. Option (a) is correct. Given : Length of potentiometer wire is L = 1 m, resistance of potentiometer wire is R¢ = 5W, the potentiometer is connected in series with an ideal

250 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) battery and a resistor R, emf of battery is e = 4V. To find : Value of R which gives a potential difference of DV = 5 mV across l = 10 cm of potentiometer wire. Potential difference per unit length of the potentiometer wire : DV 5 ¥ 10 -3 = = 0.05 V / m l 10 ¥ 10 -2



Potential difference across the total length of potentiometer wire : DV VL = ¥ L = 0.05 ¥ 1 = 0.05 V l

Current through the circuit : R2 θ V i gb = ¥ =C o R g R2 R2 + R g 5 R1 + R2 + R g

Ê R g R2 ˆ (R2 + R g ) = 5 R1 + 5 R g Á R1 + ˜¥ R2 + R g ¯ R2 Ë R2

R g = 22W

4 ¥R 5+R

Ans. 17. Option (b) is correct. Given : An arrangement of 6 resistors, I1 = –0.3 A, I4 = 0.8 A, I5 = 0.4 A.

0.05R = 19.75 R = 395W

P

>

>

>

> S

I2

>

>

I5

I1

R

I4

To find : Value of I2, I3, I6. Applying Kirchhoff ’s junction rule at S : I3 = I 4 - I 5 = 0.8 - 0.4 = 0.4 A

...(i)

I1 = –0.3 A, the negative sign means the direction of I1 is opposite to as shown in the circuit above. Applying Kirchhoff ’s junction rule at R :

( ) K1

To find : Rg, the resistance of the galvanometer. Current through galvanometer : ig µ θ

I 2 = I1 + I 4 = 0.3 + 0.8 = 1.1A

...(ii)

Applying Kirchhoff ’s junction rule at Q :

(C is constant of proportionality )

I3 + I6 + I1 = I 2 I6 = I 2 - I1 - I3 = 1.1 - 0.3 - 0.4

Case a, the equivalent resistance of the circuit : R a = R1 + R g

Case b, the equivalent resistance of the circuit : R g R2 R b = R1 + R2 + R g

Q

>

I 4 = I3 + I 5

G

Current through the circuit : V i ga = = Cθ o R1 + R g

I3

>

K2 ( )

I6

>

>

Ans. 16. Option (b) is correct. Given : R1 = 220 W Case a : When key K1 is closed but K2 is open the deflection in galvanometer is q0, Case b : When key K2 is also closed and R2 = 5W the θ deflection in galvanometer is o . 5 R2

i g = Cθ

R1R g

40 R g = 880

19.75 + 3.95R = 4 R

R1=220 

4 R1 + 4 R g =

Put R1 = 220 W and R2 = 5 W in the above equation. 220 R g 880 + 4 R g = 5

Applying Ohm’s law for R : VR = iR 3.95 =

...(ii)

Put value of Cqo from equation (i) in equation (ii). R2 V V 1 ¥ = R g R2 R2 + R g 5 R1 + R g R1 + R2 + R g

Potential drop across R : VR = ε - VL = 4 - 0.05 = 3.95 V Current through the circuit : ε 4 i = = R¢ + R 5 + R

PHYSICS

= 0.4 A ...(i)

...(iii)

Subjective Questions (Chapter Based) Q.1. An ideal cell of emf 10V is connected in circuit shown in figure. Each resistance is 2W. The potential difference (in V) across the capacitor when it is fully charged is _______

251

CURRENT ELECTRICITY

[JEE (Main) – 2nd Sep. 2020 - Shift-2] Sol. Given : R1 = R 2 = R 3 = R 4 = R 5 = 2 Ω , E = 10 V. To find : VC, voltage drop across the capacitor when it is fully charged.

Sol. Given : Resistors R1 = 40 Ω ,R 2 = 60 Ω ,R 3 = 90 Ω ,R 4 = 110 Ω R4 = 110 W are connected in a circuit, emf of the battery connected across the resistors is E = 40 V. To find : The potential difference VB – VD. Current through branch ABC : I1 =

E 40 = = 0.4 A R1 + R 2 60 + 40

Voltage drop across AB : VA − VB = R1 × I1 = 40 × 0.4 = 16 V Equivalent resistance of the circuit when the capacitor is fully charged (that is no current flows through R5) :  1 1  R′ =  +  R R R + 2 3  1

−1

1  1 + R4 =  +  2 + 2 2

−1

+2 10 = Ω 3

Current through the circuit : E 10 i = = = 3A R′ 10 3 Voltage drop across R4 : i × R 4 = 3 × 2 = 6V So, voltage drop across R3 : V3 = 10 − 6 = 4 V i1 =

V3 4 = = 2 A R3 2

i2 = i − i1 = 3 − 2 = 1A Potential at point B : VB = 0V Potential at point A : E − i2 R1 = 10 − 2 = 8V VC = VA − VB = 8V Q.2. Four resistance 40W, 60W, 90W, and 110W make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40V and internal resistance negligible. The potential difference across BD in V is .......... [JEE (Main) – 4th Sep. 2020 - Shift-2]

Current through branch ADC : I2 =

E 40 = = 0.2 A R 3 + R 4 ( 90 + 110 )

Voltage drop across AD : VA − VD = R 3 × I 2 = 90 × 0.2 = 18 V VB − VD = ( VA − VD ) − ( VA − VB ) = 2 V Q.3. The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 W is connected in parallel to the cell, the balancing length changes by 60 cm. If the N internal resistance of the cell is W, where N is 10 an integer then value of N is ____________. [JEE (Main) – 7th Jan. 2020 - Shift-2] Sol. Given : Balancing length for a cell in a potentiometer as connected in figure 1 is l = 560 cm, the balancing length changes to l¢ = 500 cm when an external resistance R = 10 W is connected parallel to the cell as shown in figure 2, the internal resistance of the cell is r =

N Ω 10

...(i)

252 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

To find : The value of N. If x is the potential gradient across the potentiometer wire, then for figure 1 : E = 560x ...(ii) For figure 2 : E × 10 = 500 x 10 + r

PHYSICS

Q.5. The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20 W and 5 W, is connected to the parallel combination of two resistors 30 W and R W. The voltage difference across the battery of internal resistance 20 W is zero, the value of R (in W) is _________ . [JEE (Main) – 8th Jan. 2020 - Shift-2] Sol. Given : A circuit as shown below, emf of battery 1 is equal to emf of battery 2, E1 = E2 = 10 V, the internal resistance of battery 1 is r1 = 20 W the internal resistance of battery 2 is r2 = 5 W, the terminal voltage across battery 1 is V1 = 0,

...(iii)

560 × 10 = 500 10 + r 56 = 50 + 5r r =

6 12 = Ω 5 10

...(iv)

On comparing equations (i) and (iv), we get : N = 12 Q.4. Four resistances of 15 W, 12 W, 4 W and 10 W respectively in cyclic order to form Wheatstone’s network. The resistance that is to be connected in parallel with the resistance of 10 W to balance the network is ___________ W. [JEE (Main) – 8th Jan. 2020 - Shift-1] Sol. Given : A Wheatstone network connected as shown below,

To find : The value of R. Terminal voltage across battery 1 : V1 = E1 − Ir1 = 10 − I × 20 = 0 = I

1 = 0.5 A 2

Terminal voltage across battery 2 : V2 = E2 − Ir2 = 10 − 0.5 × 5 = 7.5 V Voltage drop across branch AB = voltage drop across branch CD = V′ = V2 + V1 = 7.5 V Current through branch AB : V′ 7.5 = = 0.25 A I1 = 30 30 Current through branch CD : I 2 = I − I1 = 0.5 − 0.25 = 0.25 A Value of resistance R : V ' 7.5 V R = = = 30 Ω I 2 0.25 A

To find : The value of resistance R, such that the network is balanced. As the 10 W and R resistors are in parallel, their equivalent resistance will be : 10 R R′ = 10 + R

Q.6. In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l = 25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance l’ (in cm) will now be _________.

Condition for balancing of the Wheatstone network  : R′ × 12 = 15 × 4 10 R × 12 = 15 × 4 10 + R R = 10Ω

[JEE (Main) – 9th Jan. 2020 - Shift-2]

253

CURRENT ELECTRICITY

Sol. Given : Balancing length in a meter bridge experiment for wire 1 of resistance R is l = 25 cm, S is a standard resistance. Q

To find : The new balancing length l¢, for wire 2 of same material but of half the length and half the diameter as that of the wire 1.

Al 50 mm

Fe 2 mm P 7 mm

Condition for balancing for wire 1 : 100 − l 75 S = = =3 R l 25

...(i)

Resistance of wire 1 : R =

ρl 4 ρl = A pd2

...(ii)

(r is the resistivity of material of wire 1 and 2, 2

d A = p   is area of cross section of wire 1) 2

In equation (i) : A Al = (7 2 - 2 2 ) ¥ 10 -6 m 2

Resistance of the wire 2 with half the length and half the diameter as wire 1 : l 4ρ    2  = 8 ρl = 2 × 4 ρl = 2R R′ = 2 pd2 pd2 d p  2

Sol. Given : A composite bar of aluminium and iron with dimensions as shown in the figure above, electrical resistivity of Al is rAl = 2.7 ×10–8 Wm, electrical resistivity of Fe is rFe = 1.0 ×10–7 Wm. To find : Electrical resistance between two faces P and Q of the composite bar. The equivalent resistance of the composite bar as measured between two faces P and Q will be : A A 1 1 1 = + = Al + Fe ...(i) R eq R Al R Fe ρ Al l ρ Fe l

...(iii)

is area of cross section of aluminium bar. AFe = 22 × 10–6 m2 is area of cross section of iron bar, l = 50 × 10–3 m is the length of the composite bar as shown in the given figure. Put given values in equation (i) : 1 45 ¥10 -6 = R eq 2.7 ¥ 10 -8 ¥ 50 ¥ 10 -3 +

Condition for balancing for wire 2 : S 100 − l′ = R′ l′

[ Use equations (i) and (iii)]

S 100 − l′ = 2R l′

4 ¥ 10 -6 1.0 ¥ 10 -7 ¥ 50 ¥ 10 -3

= 4.1 ¥ 10 4 Q.8. For the arrangements of resistors shown, find the equivalent resistance between points A and B. A

3 100 − l′ = 2 l′

R R

l′ = 40cm

R

R

Q.7. In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10

–8

W m and 1.0 ×

10–7 W m, respectively. What will be the electrical resistance between the two faces P and Q of the composite bar,

B

R

C

Sol. Given : 6 resistors of equal resistance are arranged in form of an equilateral triangle. To find : Equivalent resistance between points A and B.

254 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Resistance of arm BC :

From equation (ii), error in value of X : DX Dl Dl 0.1 0.1 = + = + X l 100 - l 40 60

1 1 1 2 = + = R Bc R R R R BC =

DX = 0.25

R 2

R AC

Q.10. In a big house there are 7 bulbs of 100 W, 4 bulbs of 40 W, 5 fans of 80 W and 1 water heater of 1 kW. The voltage of the electric mains is 220 V. What should be the minimum capacity of the main fuse of the house? Sol. Given : In a house there are 7 bulbs of 100 W, 4 bulbs of 40 W, 5 fans of 80 W and 1 water heater of 1 kW, the voltage of the electric mains is 220 V. To find : The minimum capacity of the main fuse. Total power consumed when all the appliances are working :

1 1 1 3 = + + = R R R R

R BC =

R 3

Resistance of arm AB : R AB = R Equivalent resistance between points A and B : 1 1 1 1 6 11 = + = + = R R R eq R R 5R 5R + 2 3 R eq

P = 7 ¥ 100 + 4 ¥ 40 + 5 ¥ 80 + 1 ¥ 1 ¥ 10 3 = 700 + 160 + 400 + 1000 = 2260W

5 = R 11

Q.9. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 W, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. Find the value of unknown resistance. Also include the error in your answer.

R

X = ( 60 ± 0.25) W

So,

Resistance of arm AC : 1

PHYSICS

.

90 

40.0 cm

Sol. Given : In a meter bridge experiment the null point is measured at a distance l = 40.0 cm, value of standard resistance is R = 90 W, least count of the scale used in the meter bridge is Dl = 1 mm = 0.1 cm. To find : X, the value of unknown resistance. Condition for null deflection in galvanometer in a meter bridge experiment : X l = R 100 - l X 40 = 90 100 - 40

...(i)

So, minimum capacity of the main fuse should be 11 A. Q.11. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 W. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb ? Sol. Given : Supply voltage of the room is V = 120 V, resistance of the lead wires is R = 6 W, power rating of the bulb is Pb = 60 W, power rating of the heater which is connected in parallel to the bulb is Ph = 240 W. To find : The decrease in voltage across the bulb when the heater is switched on. Resistance of the bulb : Rb =

V 2 (120 )2 = = 240 W Pb 60

The lead wire and the bulb are in series. So, equivalent resistance of the circuit when only the bulb is on : R eq1 = R + R b = 6 + 240 = 246 W Current through the bulb : V 120 i1 = = = 0.5 A R eq1 246 Voltage drop across the bulb when only bulb is on : Rb 240 V1 = ¥ 120 = ¥ 120 = 117 V R eq 1 246

X = 60W

Resistance of the heater :

From equation (i) : Ê l ˆ X = RÁ Ë 100 - l ˜¯

Maximum current through the fuse : P 2260 I = = = 10.3 A V 220

...(ii)

Rh =

V 2 (120 )2 = = 60 W Ph 240

255

CURRENT ELECTRICITY

Equivalent resistance of the circuit when both the bulb and heater are on (heater and bulb are connected in parallel) : R eq 2 = R +

1 1 =6+ = 54 W 1 1 1 1 + + Rb Rh 240 60

Equivalent resistance of only combination of bulb and heater : 1 1 1 =  + R eq 3 Rb Rh R eq3 = 48 W

the

parallel

Current through the lead wires when both the bulb and heater are on : V 120 = = 2.2 A i2 = R eq 2 54 Voltage drop across the bulb when both the bulb and heater are on : R eq 3 48 V2 = = ¥ 120 = 106.7 V R eq 2 54 Decrease in voltage across the bulb when the heater is switched on : V1 - V2 = 117 - 106.7 = 10.3 V



[

[

Direction of field will be perpendicular to plane containing current element and the point of observation

helical

]

]

Magnetic effect of current & Magnetism Part-1 Se nsi

tivity

current carrying circular ring

wire can deflect a nearby magnetic compass needle.

256 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

between

Magnetic Effects of Current & Magnetism Part-2

(

> a). The magnitude of the torque on the loop about z-axis is given by : (a) (c)

2 m0 I2 a 2 pb 2 m0 I2 a3 p b2



(b) (d)

m0 I2 a 2 2p b m0 I2 a3

2p b 2 [JEE (Main) – 5 Sep. 2020 - Shift-1] Q.12. A galvanometer is used in laboratory for detecting the null point in electrical experiments. If, on passing a current of 6 mA, it produces a deflection of 2°, its figure of merit is close to : (a) 6 × 10–3 A/div (b) 3 × 10–3 A/div (c) 666° A/div (d) 333° A/div [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.13. An iron rod of volume 10–3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be : (a) 0.5 × 102 A m2 (b) 5 × 102 A m2 2 2 (c) 500 × 10 A m (d) 50 × 102 A m2 [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.14. A particle of charge q and mass m is moving with a velocity − vi( v ≠ 0) towards a large screen placed th

in the Y-Z plane at distance d. If there is magnetic   the minimum value of v for which field B = B0 k, the particle will not hit the screen is : 2 qdB0 qdB0 (a) (b) m 3m (c)

qdB0 m

(d)

qdB0 2m

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.15. An electron is moving along +x direction with a velocity of 6 × 106 m s–1. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be : (a) 5 × 10–3 T, along + z direction (b) 3 × 10–4 T, along +z direction (c) 3 × 10–4 T, along –z direction (d) 5 × 10–3 T, along –z direction [JEE (Main) – 6th Sep. 2020 - Shift-1]

PHYSICS

Q.16. A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with  speed v under the influence of magnetic field B. The magnetic moment of this moving particle is :   mυ 2 B mυ 2 B (a) (b) B2 2B2   mυ 2 B mυ 2 B − (c) − (d) 2p B2 2B2 [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.17. A square loop of side 2a and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be : (a) (c)

2 m0 I2 a 2 pb m0 I2 a 2b

2p ( a 2 + b 2 )

(b)

(d)

2 m0 I2 a 2b

p ( a2 + b2 ) m0 I2 a 2 2p b

[JEE (Main) – 6th Sep. 2020 - Shift-2] Q.18. Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by fi; The magnetic flux through the area of the circular coil area is given by f0. Which of the following option is correct? (a) fi = f0 (b) fi < f0 (c) fi > f0 (d) fi = –f0 [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.19. A particle of mass m and charge q has an initial   velocity v = v0 j. If an electric field E = E0 i and  magnetic field B = B0 i act on the particle, its speed will double after a time : 3mv0 3mv0 (a) (b) qE0 qE0 (c)

2mv0 qE0

2mv0 (d) qE 0

[JEE – Mains - 7th Jan. 2020 - Shift-2] Q.20. Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10–27 kg) (a) 7.1 mT (b) 71 mT (c) 0.071 mT (d) 0.71 mT [JEE (Main) – 8th Jan. 2020 - Shift-1] Q.21. A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

261

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

(a)

m0 I 2p R

(c)

m0 I (p + 1) 2p R

1   p +  2 

(b)

m0 I 2R

(d)

m0 I 2p R

1   p −  2 

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.22. A galvanometer having a coil resistance 100 W gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V? (a) 8.9 kW (b) 10 kW (c) 9.9 kW (d) 7.9 kW [JEE (Main) – 8th Jan. 2020 - Shift-2] Q.23. A charged particle of mass ‘m’ and charge ‘q’ moving under the influence of uniform electric   field Ei and a uniform magnetic field Bk follows a trajectory from point P to Q as shown in figure.  The velocities at P and Q are respectively, vi and −2v j. Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale)

Q.24. A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at a distance and 2a, respectively from the axis of 3 the wire is : 3 (a) (b) 2 2 1 2 (c) (d) 2 3 [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.25. An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns/ length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed v. If the electron does not hit the surface of the solenoid, maximum possible value of v is (all symbols have their standard meaning) :

(a)

2 e m0 nIR m

(b)

e m0 nIR 2m

e m0 nIR (d) m [JEE (Main) - 9th Jan. 2020 - Shift-2] Q.26. A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 N m–1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 W and air drag negligible, N will be close to : (c)

e m0 nIR 4m

XXXX XXXX B

(A) E =

3  mv 2    4  qa 

(B) Rate of work done by the electric field at P is 3  mv 3    4  a  (C) Rate of work done by both the fields at Q is zero (D) The difference between the magnitude of angular momentum of the particle at P and Q is 2 mav. (a) (A), (C), (D) (b) (A), (B), (C) (d) (A), (B), (C), (D) (d) (B), (C), (D) [JEE (Main) – 9th Jan. 2020 - Shift-1]

10cm XXXX XXXX

(a) 1000 (c) 5000

(b) 50000 (d) 10000 [JEE (Main) – 8th April 2019 - Shift-1] Q.27. A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field Bi^. The torque on the coil due to the magnetic field is : (a)

Br 2 I pN

(b) Bpr2I N

(c)

Bpr 2 I N

(d) Zero



[JEE (Main) – 8th April 2019 - Shift-1]

262 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.28. Two very long, straight and insulated wires are kept at 90° angle from each other in xy-plane as shown in the figure. y I d •P d

x

I



These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be : m I (a) Zero (b) – 0 ( x + y ) 2p d (c)

+ m0 I  ( z) pd

(d)

m0 I   (x + y) 2p d

[JEE (Main) – 8th April 2019 - Shift-2] Q.29. A moving coil galvanometer has resistance 50 W and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kW resistance. The maximum voltage, that can be measured using this voltmeter, will be close to : (a) 40 V (b) 15 V (c) 20 V (d) 10 V [JEE (Main) – 9th April 2019 - Shift-1] Q.30. A rectangular coil (Dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3 A in the clockwise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the coil is : (a) 0.38 Nm (b) 0.55 Nm (c) 0.42 Nm (d) 0.27 Nm [JEE (Main) – 9th April 2019 - Shift-1] Q.31. A rigid square loop of side ‘a’ and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be : I1

Q.32. A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant 10–6 N-m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately : (a) 10–4 (b) 10–2 –1 (c) 10 (d) 10–3 [JEE (Main) – 9th April 2019 - Shift-2] Q.33. The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A? (a) 0.5 ohm (b) 0.002 ohm (c) 0.02 ohm (d) 0.2 ohm [JEE (Main) – 9th April 2019 - Shift-2] Q.34. A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 20 kW is required to convert the above galvanometer into a voltmeter of range 0 – 5 V. Therefore the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0-10 mA is : (a) 30 kW (b) 100 W (c) 200 W (d) 10 W [JEE (Main) – 10th April 2019 - Shift-1] Q.35. A proton, an electron and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let rp, re and rHe be their respective radii, then, (a) re > rp = rHe (b) re < rp = rHe (c) re < rp < rHe (d) re > rp > rHe [JEE (Main) – 10th April 2019 - Shift-1] Q.36. Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are : C A

I2

I1

a a

(a) Repulsive and equal to

m0 I1I 2 2p

(b) Attractive and equal to

m0 I1I 2 3p

(c) Repulsive and equal to

m0 I1I 2 4p

(d) Zero

[JEE (Main) – 9th April 2019 - Shift-1]

PHYSICS

B

I2

x d

Ê I ˆ I2 d (a) x = Á 1 ˜ d and x = (I1 + I 2 ) Ë I1 - I 2 ¯

Ê I2 ˆ Ê I2 ˆ d and x = Á d (b) x = Á Ë I1 + I 2 ˜¯ Ë (I1 - I 2 ) ˜¯ Ê I ˆ I2 (c) x = Á 1 ˜ d and x = d (I1 - I 2 ) Ë I1 + I 2 ¯ (d) x = ±

I1d (I1 - I 2 ) [JEE (Main) – 10th April 2019 - Shift-1]

263

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Q.37. A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is : 2 Ê 4 q2 ˆ Á ˜ m Ë 15 4p ε 0 a ¯

1/2

(a)

Ê 2 q2 ˆ Á ˜ Ë 15 4p ε 0 a ¯

1/2

(c)

2 m

(b)

2 Ê 1 q2 ˆ Á ˜ m Ë 5 4p ε 0 a ¯



(d)

2 Ê 1 q2 ˆ Á ˜ m Ë 15 4p ε 0 a ¯

R1

R2

1/2

R1=2000  R2=8000  R3=10000  2V

20V

G R1

R2 2V

R1

10V R2

20V

8 cm

(a) 11.65 cm (c) 1.22 cm

(b) 12.87 cm (d) 2.25 cm [JEE (Main) – 12th April 2019 - Shift-2] Q.42. A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance RV to it. Then, ÊI -I ˆ Ê Ig ˆ (a) R R = G2 Á 0 g ˜ and R A = Á ˜ A V R V Ë (I0 - I g ) ¯ Ë Ig ¯

Ê Ig ˆ R (b) R A R V = G and A = Á ˜ R V Ë (I0 - I g ) ¯

10V

Ê Ig ˆ R A Ê I0 - I g ˆ =Á (c) R A R V = G Á ˜ and ˜ RV Ë Ig ¯ I I Ë 0 g¯ Ig R (d) R A R V = G2 and A = R V (I0 - I g )

[JEE (Main) – 12th April 2019 - Shift-2] Q.43. Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) (m0 = 4p × 10–7 N – A–2)

m

R1=19900  R2=9900  R3=1900 

R3 10V

P

• 5c

R2 20V

2

2

20V

G R1

2

2

R1=1900  R2=8000  R3=10000  2V

2

R3

G

(d)

2cm

m

(c)

R1=1900  R2=9900  R3=19900 

R3

P

S

5c

(b)

10V

Q

d

R3

G

(electron’s charge = 1.6 × 10–19C, mass of electron = 9.1 × 10–31 kg)

1/2



[JEE (Main) – 10th April 2019 - Shift-1] Q.38. The magnitude of the magnetic field at the center of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is : (Take m0 = 4p × 10–7 NA–2] (a) 18 mT (b) 9 mT (c) 3 mT (d) 1 mT [JEE (Main) – 10th April 2019 - Shift-2] Q.39. A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 p rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to (m0 = 4p × 10–7 N/A2). (a) 2 × 10–6 C (b) 3 × 10–5 C –5 (c) 4 × 10 C (d) 7 × 10–6 C [JEE (Main) – 12th April 2019 - Shift-1] Q.40. A galvanometer of resistance 100 W has 50 divisions on its scale and has sensitivity of 20 mA/division. It is to be converted to a voltmeter with three ranges, of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so is : (a)



2V

[JEE (Main) – 12th April 2019 - Shift-1] Q.41. An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field  ^ at S (See figure). The field B = (1.5 × 10–3T)k extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is :

•

•

A

B

6 cm

(a) 2.0 × 10–5 T (b) 1.5 × 10–5 T (c) 3.0 × 10–5 T (d) 2.5 × 10–5 T [JEE (Main) – 12th April 2019 - Shift-2]

264 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.44. A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to : O

45°

=

m 3c

3c m

•

Q

R

m

2c

m

2c S

P i = 10 A

(a) 1.0 × 10–7 T (b) 1.5 x 10–7 T –5 (c) 1.5 × 10 T (d) 1.0 × 10–5 T [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.45. An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d  a). If the loop applies a force F on the wire then :

d

(a) F = 0

Ê aˆ (b) F ∝ Á ˜ Ë d¯

Ê a2 ˆ (c) F ∝ Á 3 ˜ Ëd ¯

Ê aˆ (d) F ∝ Á ˜ Ë d¯ th

2

[JEE (Main) – 9 Jan. 2019 - Shift-1] Q.46. A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 × 10–19C) (a) 9.1 × 10–31 kg (b) 1.6 × 10–27 kg –19 (c) 1.6 × 10 kg (d) 2.0 × 10–24 kg [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.47. One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop B (BL) to that at the centre of the coil (BC), i.e. L BC will be :

(a) N

(b)

(c) N2

(d)

PHYSICS

1 N 1 N2

[JEE (Main) – 9th Jan. 2019 - Shift-2] Q.48. A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is : (a) 12 mV (b) 6 mV (c) 1 mV (d) 2 mV [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.49. In an experiment, electrons are accelerated from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. (Charge of the electron = 1.6 × 10–19 C, Mass of the electron=9.1 × 10–31 kg) : (a) 7.5 × 10–3 m (b) 7.5 m (c) 7.5 × 10–2 m (d) 7.5 × 10–4 m [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.50. A galvanometer having a resistance of 20 W and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is : (a) 100 W (b) 120 W (c) 80 W (d) 125 W [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.51. The region between y = 0 and y = d contains a  magnetic field B = B z . A particle of mass m and  charge q enters the region with a velocity v = vi. mv , the acceleration of the charged particle If d = 2 qB at the point of its emergence at the other side is : Ê1  3 Á2 i- 2 Ë Ê qvB - j + i ˆ (c) Á ˜ m Ë 2 ¯ (a)

qvB m

jˆ ˜ ¯

(b)

qvB Ê 3  1  ˆ i + j˜ m ÁË 2 2 ¯

(d)

qvB m

Ê i + j ˆ Á ˜ Ë 2¯

[JEE (Main) – 11th Jan. 2019 - Shift-2] Q.52. A particle of mass m and charge q is in an electric and magnetic field given by    E = 2i + 3 j ; B = 4 j + 6 k. The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is : (a) (0.35)q (b) 5q (c) (2.5)q (d) (0.15)q [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.53. As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4 cm, and the magnitude

265

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

of the magnetic field at O is 10–4T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (m0 = 4p × 10–7 NA–2) : y S O

L

P

•

p = v cosθ ×

M x N

(a) 20 A, perpendicular out of the page (b) 40 A, perpendicular out of the page (c) 20 A, perpendicular into the page (d) 40 A, perpendicular into the page [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.54. A proton and an a-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : ra of the circular paths described by them will be : (a) 1 : 2 (c) 1 : 3

(b) 1 : 2 (d) 1 : 3 [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.55. A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10–4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of : (a) 250 ohm (b) 200 ohm (c) 6200 ohm (d) 6250 ohm [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (c) 2. (c) 5. (c) 6. (d) 9. (b) 10. (a) 13. (b) 14. (c) 17. (b) 18. (d) 21. (a) 22. (c) 25. (b) 26. (c) 29. (c) 30. (d) 33. (d) 34. (a) 37. (c) 38. (a) 41. (b) 42. (b) 45. (d) 46. (d) 49. (d) 50. (c) 53. (c) 54. (a)

3. (a) 7. (a) 11. (a) 15. (d) 19. (a) 23. (b) 27. (b) 31. (c) 35. (c) 39. (b) 43. (b) 47. (d) 51. (b) 55. (b)

is q = 60°, mass of proton is m = 1.67 × 10–27 kg, charge on proton is e = 1.6 × 10–19 C. To find : p, the pitch of the helical path followed by the protons. Let T be the time period of proton’s motion in the magnetic field. p = v cosθ × T

4. (c) 8. (c) 12. (b) 16. (c) 20. (d) 24. (c) 28. (a) 32. (d) 36. (d) 40. (c) 44. (d) 48. (a) 52. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : Speed of beam of protons is v = 4 × 105 m/s, the protons enter a uniform magnetic field with magnitude B = 0.3 T, angle between v and B

2p m = 4 cm eB

Ans. 2. Option (c) is correct. Given : Length of a region with uniform magnetic field is l, magnitude of magnetic field in the region is B = 0.3 T, speed of beam of protons entering the region is v = 4 ×105 m/s, angle between v and B is q = 60°, number of revolutions completed by a proton in the region is n = 10, mass of proton is m = 1.67 ×10–27 kg, charge on proton is e = 1.6 ×10–19 C. To find : The value of l. Let T be the time period of proton’s motion in the magnetic field. l = n × pitch l = 10 × v cosθ × T l = 10 × v cosθ ×

2p m = 0.44 m eB

Ans. 3. Option (a) is correct. Given : A current carrying wire is bent in a shape as shown below,

Current through the wire is I, dimensions of rectangles ABCD and ADEF are a × b.  To find : M, the magnetic moment of loop ABCDEFA. Magnetic moment of loop ABCD :  M = abI k ...( i ) 1

Magnetic moment of loop ADEF :  M 2 = abI j

...(ii)

From equations (i) and (ii) :     j k  M = M1 + M 2 = 2 abI  +   2 2   Ans. 4. Option (c) is correct. Given : Edge length of a hexagon shaped coil is l = 10 cm, number of turns in the coil is n = 50, current through the coil is I, the magnetic field at m I the centre of the coil is B = p o . p To find : The value of p.

266 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Ans. 7. Option (a) is correct. Given : Two arcs of wire A and B carry currents I1 = 2A and I2 = 3A respectively, radius of arc A is r1 = 2 cm, radius of arc B is r2 = 4 cm. Let each hexagonal section of wire be at distance d from the centre of the hexagon. From construction: x tan 30° = ; d d =

x 5 × 10 −2 = = 5 3 cm tan 30° tan 30°

Magnetic field at centre due to six wires of the hexagon, equidistant from the centre: m In B = 6 × o (sin 30° + sin 30°) 4p d B =

6 × 50 4 × 5 3 × 10 −2

B = 500 3 ×

×

moI p

moI p

p = 500 3  Ans. 5. Option (c) is correct. Given : Charge on a particle is q = 1 mC, velocity of the  particle is v = ( 2i + 3 j + 4 k ) m / s, external magnetic  field in the region is B = ( 5i + 3 j − 6 k ) × 10 −3 T , force   −9 on the particle is f = F × 10 N.  To find : the vector F. Force on a particle moving in an external magnetic field :    f = q( v × B)  f = 10 −6 ( 2i + 3 j + 4 k ) × ( 5i + 3 j − 6 k )10 −3  f = 10 −9 ( 2i + 3 j + 4 k ) × ( 5i + 3 j − 6 k )  f = ( −30i + 32 j − 9 k ) × 10 −9 N  F = ( −30i + 32 j − 9 k ) Ans. 6. Option (d) is correct. Given : Magnitude of an external magnetic field is B = 0.06 T, angle the magnetic field makes with the axis of a small bar magnet is q = 30°, torque experienced by the bar magnet is t = 0.018 Nm. To find : DU, the minimum work required to rotate the bar magnet from its stable to unstable equilibrium position in the external magnetic field. Let M be the magnetic moment of the bar magnet. τ = MBsin θ M =

τ 0.018 = = 0.6 Am 2 B sin θ 0.06 × sin 30

∆U = MB cos 0 − MB cos 180° ∆U = 2 MB = 2 × 0.6 × 0.06 = 7.2 × 10 −2 J

To find :

B1 ,the ratio of magnetic fields due to arcs B2

A and B at their common centre O. Let q1 and q2 be the angle subtended at O by arc A and arc B respectively. Magnetic field due to arc A at O : m I B1 = o 1 θ1 4p r1 B1 =

m o I1  p 2p −  4p r1  2

Magnetic field due to arc B at O : m I B2 = o 2 θ 2 4p r2 B2 =

mo I2  p 2p −  4p r2  3

p  2p −  B1 I1 r2  2 = × × p B2 I 2 r1   2p − 3    B1 2 4 3 3 6 = × × × = B2 3 2 2 5 5 Ans. 8. Option (c) is correct. Given : A small bar magnet moves through a coil of wire at a constant speed. To find : The deflection in the galvanometer as the magnet, enters the coil, is completely inside the coil and exits the coil. a. As the magnet enters the coil, the galvanometer shows deflection due to flow of induced current due to change in flux linked to the coil. b. When the magnet is completely inside the coil, the galvanometer shows no deflection as the flux linked to the coil is constant. c. As the magnet exits the coil, the galvanometer shows deflection due to flow of induced current due to change in flux linked to the coil. Ans. 9. Option (b) is correct. Given : Moment of inertia of a circular coil about any of its diameter is I = 0.8 kg-m2, magnetic moment of the coil is M = 20 A-m2, initially the coil is standing vertical, the direction of external magnetic field B = 4T is also vertical, the coil rotates about its horizontal diameter in the external magnetic field.

267

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

To find : w, the angular speed of the coil after it has rotated q = 60° from its initial position. By law of conservation of energy: 1 1 2 Iω = − MB cos 60 + MB cos 0 = MB 2 2

ω=

20 × 4 MB = = 10 rad s−1 I 0.8

Ans. 10. Option (a) is correct. Given : Resistance of a galvanometer is G, Case 1 : Range of voltmeter formed by connecting G in series with resistance R1 is 0 – 1 V, Case 2 : Range of voltmeter formed by connecting G, R1 and R2 in series is 0 – 2 V. To find : The value of R2. Let the current through the galvanometer be I. For case 1 : I(G+ R1 ) = 1 For case 2 : I(G+ R1 + R 2 ) = 2 (G+ R1 + R 2 ) = 2 G+ R1

To find : M, the magnetic moment of the iron rod. Let N be total number of turns in the solenoid and A be the area of cross section of the iron rod. Magnetic moment of the iron rod : M = ( mr − 1)NIA M = ( mr − 1)NI M = ( mr − 1)

V  l

N IV l

M = ( mr − 1)nIV M = (1000 − 1) ×

10 10 −2

× 0.5 × 10 −3

M ª 500 A m 2 Ans. 14. Option (c) is correct. Given : Charge on particle is q, mass of particle is  m, initial velocity of particle is v = − vi, the distance between the particle and the screen placed in the YZ   plane is d, magnetic field in the region is B = B k. o

To find : The minimum value of v, for which the particle does not reach the screen.

G+ R1 + R 2 = 2G + 2 R1 R 2 = G+ R1 Ans. 11. Option (a) is correct. Given : Edge length of a square loop kept in XZ plane with its centre at origin is 2a, current through the loop is I, current through a wire kept parallel to z axis at (0, b, 0) is also I, b >> a. To find : t, the magnitude of torque on the square loop about the z axis. As b >> a, the distance between the wire and the square loop is b. Magnitude of magnetic field due to the wire at the location of square loop : m I B = o 2p b Magnetic moment of the square loop : M = I( 2 a )2 Torque on the square loop

τ = MB sin θ = I( 2 a )2 τ =

moI sin 90 2p b

2 mo I2 a2 pb

Ans. 12. Option (b) is correct. Given : Current through the galvanometer is I = 6 mA, corresponding deflection in the galvanometer is q = 2°. To find : The figure of merit of the galvanometer. Figureofmerit =

I 6 × 10 −3 = = 3 × 10 −3 A div −1 θ 2

Ans. 13. Option (b) is correct. Given : Volume of an iron rod is V = 10–3 m3, relative permeability of the rod is mr = 1000, the rod is placed as core of a solenoid with n = 10 turns cm–1, current through the solenoid is I = 0.5 A.

Radius of the circular path followed by a charged particle in external magnetic field : mv d = qBo v =

qdBo m

Ans. 15. Option (d) is correct. Given : Electric field in a region of space is  −1 E = 300 jV cm −1 = 30000 j V m , velocity of an electron with which it enters the region is  v = ( 6 × 10 6 )i m / s.  To find : B, the magnitude and direction of magnetic field in the region so that the electron keeps moving along +x direction. For an electron to go un-deflected, the forces acting on it due to electric and magnetic field should balance :   FE = FB    eE = e ( v × B )  E  30000 B =  = = 5 × 10 −3 T v 6 × 10 6     As, E = ( v × B); B = ( 5 × 10 −3 )( − k )T

268 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 16. Option (c) is correct. Given : Charge on particle is q, mass of particle is m, speed of particle travelling under influence of magnetic field B is v. To find : M, magnetic moment of the particle. A charged particle moving under influence of magnetic field will trace a circle with radius, mv r= ; circumference 2pr and time period of its qB motion will be : 2pr . T = v Current due to motion of charge q : q qv = i = T 2p r

Ans. 18. Option (d) is correct. Given : Current through a circular coil of wire is I, flux through the plane that contains the circular coil excluding the coil area is fi, flux through the area enclosed by the coil is f0. fi fo

To find : Relation between fi and fo. Magnetic field lines associated with the current through the circular coil will form closed loops. So, number of magnetic lines entering the area enclosed by the coil will be same as number of magnetic lines leaving the infinite plane. Therefore, fi = –fo.

Magnetic moment of charge q : qv M = i (p r 2 ) = × pr2 2p r M =

PHYSICS

qvr qv mv = × 2 2 qB

mv 2 2B   Direction of M will be that of opposite of B.  mv 2  M = − 2B 2B M =

Ans. 17. Option (b) is correct. Given : Edge length of a square loop kept in XZ plane with its centre at origin is 2a, current through the loop is I, current through a wire kept parallel to z axis at (0, b, 0) is also I, b >> a. To find : t, the magnitude of torque on the square loop about the z axis.

Ans. 19. Option (a) is correct. Given : Mass of the particle is m, charge on the particle is q, the initial velocity of the particle is   vi = vo j , an electric field Eo i and magnetic field B i act on the particle. o

To find : t, the time in which the speed of the particle doubles its initial speed. As applied electric and magnetic field are in x direction, only the x component of the particle’s velocity will change with time. So, the final velocity of particle will be :  v = v j + v i f

o

x

Relation between initial and final speed of particle : ( 2 vo )2 = vo2 + vx2 vx =

3 vo

...(i)

Change in speed of particle due to action of electric and magnetic field : qE vx = 0 + 0 t m Torque on sides AB and CD is zero.

3 vo =

F' BC    

t =

I

F

AD

τ = F cosθ × 2 a = τ =

2 mo I2 a2b

p ( a2 + b2 )

mo 2p

2 aI 2 2

a +b

2

×

b 2

a + b2

× 2a

qE0 t m 3mvo qE0

Ans. 20. Option (d) is correct. Given : Kinetic energy of proton is K = 1 MeV, direction of motion of proton is from north to south, applied magnetic field is from west to east, the acceleration of proton due to the field is a = 1012 m/s2, rest mass of proton is m = 1.6 × 10–27 kg. To find : B, the value of magnetic field acting on the proton.

269

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Velocity of proton : v = =

 2 × 1 × 10 × 1.6 × 10 2K =   m 1.6 × 10 −27  m 2 × 107 s 6

−19

  

By Ohm’s law : ( R + R g )I = V ( R + 100 ) × 1 × 10 −3 = 10 R + 100 = 10000 R = 9900Ω

Magnetic force acting on the proton : ma = qvB ma 1.6 × 10 −27 × 1012 = = 0.71 × 10 −3 T qv 1.6 × 10 −19 × 2 × 107 = 0.71mT

B =

Ans. 21. Option (a) is correct. Given : Current through a long wire ABDMNDC is I. M

.

R = 9.9 k Ω Ans. 23. Option (b) is correct. Given : Mass of the particle is m, charge on the particle is q, an electric field Ei and magnetic field Bk act on the particle, the particle follows a trajectory from P to Q as shown below, velocity of   particle at P is vi and velocity of particle at Q is   −2v j

O

N 45°

45° B

D

R 3 C

1 I A

To find : B, magnetic field at O. We have divided the given wire into three parts (1, 2 and 3). The net magnetic field at O, will be vector sum of contributions from all the three parts.     B = B1 + B2 + B3  m I B = o (sin 90 − sin 45)( − k ) 4p R m I m I + o (sin 90 + sin 45)( k ) + o k 4p R 2R  B =

2 moI  moI  k+ k 4p R 2R

To find : 1. The magnitude of electric field acting on the particle, 2. Rate of work done by electric field at P, 3. Rate of work done by both the fields at Q, 4. The change in angular momentum of the particle as it goes from P to Q. 1. Change in kinetic energy of the particle as it goes from P to Q : 1 1 K = m( 2 v )2 − mv 2 2 2

 m I  1  + p  k B = o  2p R  2   m I  1  +p  B = o  2p R  2  Ans. 22. Option (c) is correct. Given : Resistance of galvanometer coil is Rg = 100 W, current through galvanometer coil at full scale deflection is I = 1 mA. To find : The value of resistance R, to be connected in series to the galvanometer to convert it into voltmeter giving full scale deflection for potential difference of V = 10 V. The required circuit will be : Rg R I Voltmeter Required voltage drop across series combination of Rg and R is V = 10 V.

K =

3 2 mv 2

...(i)



Work done on the particle by the two fields as it goes from P to Q : W = Wmag + Wele = 0 + qE( 2 a ) = qE( 2 a )



By work energy theorem : W = K qE( 2 a ) =

3 2 mv 2

E =

3 mv 2 4 qa

...(ii)

2. The rate of work done by electric field at P, is the power of electric field at P. Power = Force ×velocity = qE × v  3 mv 2  3 mv 3 = q ×v =  4 qa  4 a  

...(iii)

270 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

3. As the direction of electric field at Q is perpendicular to the particle’s velocity, the rate of work done by both the fields on the particle at Q is zero. WQ = 0

...(iv)

4. Angular momentum of particle at Q : = LQ

m = ( 2 v )( 2 a ) 4 mva

Angular momentum of particle at P : = LP

m = ( v )( a ) mva

So, ∆L = LQ − LP = 3mva

...(v)

From equations (ii), (iii), (iv) and (v) we can see the statements A, B and C are correct. So, option b is correct.

Looking at the figure above, we can see that the R maximum radius the electron’s trajectory is . 2 Balancing the forces acting on the electron :

Ans. 24. Option (c) is correct. Given : Radius of a long straight wire is a, the wire carries a uniform current I. BA ,the ratio of magnetic fields due to the BB a wire at distance and 2a from the axis of the wire. 3 To find :

mv 2 R 2

evB =

eBR e m onIR = 2m 2m

= v

Ans. 26. Option (c) is correct. Given : Length of strip is l = 10 cm, resistance of the U shaped wire is negligible, spring constant of spring connected to the strip is k = 0.5 N/m, magnitude of applied magnetic field is B = 0.1 T, mass of strip is m = 50 g, resistance of strip is R = 10 W, number of oscillations performed by the strip before its amplitude decreases by a factor of e is N when pulled from its equilibrium position and released. XXXX XXXX B

10cm XXXX XXXX

kx, FB

BA

a moI    3  = moI = 6p a 2p a 2

= BB

moI moI = 2p ( 2 a ) 4p a

ma

To find : Value of N. ...(i) ...(ii)

From equations (i) and (ii) : BA 2 = BB 3

When the strip is pulled from its equilibrium position, there are two forces acting on the strip which try to bring it back to its equilibrium position, as shown in the above diagram. - kx - FB = ma

So,

(FB = – Bil, is the force due to magnetic field) - kx - Bil = ma Blv Ê is the current through the strip i= Á R Á Ë and v is its velocity

Ans. 25. Option (b) is correct. Given : Radius of a long solenoid is R, current through the solenoid is I, the solenoid has n turns/ length, an electron travels along the radius of the solenoid with velocity v. To find : The maximum value of v, such that the electron does not touch the surface of the solenoid.

- kx Put v =

B 2l 2 v = ma R

dx d 2x , a = 2 in the equation above. dt dt

ˆ ˜ ˜ ¯

271

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

- kx -

Ê d 2x ˆ B2l 2 Ê dx ˆ = m Á 2˜ R ËÁ dt ¯˜ Ë dt ¯

Ê d 2 x ˆ B 2l 2 Ê dx ˆ mÁ 2˜ + + kx = 0 R ÁË dt ˜¯ Ë dt ¯

Ans. 29. Option (c) is correct.

Ê B 2l 2 ˆ A(t ) = A0 exp Á t˜ Ë 2 Rm ¯ A0 Put A(t ) = in the above equation. e Ê B 2l 2 ˆ A0 t˜ = A 0 exp Á e Ë 2 Rm ¯ 2 2

B l t = 1 2 Rm 2 Rm B 2l 2

Put the given values : t =

-3

-2 2

( 0.1) ¥ (10 ¥ 10 )

= 10000 s

Time period of oscillation of spring : T = 2p

Given : Resistance of a moving coil galvanometer is Rg = 50 W, the galvanometer shows a full deflection at Ig = 4 mA, galvanometer is converted into a voltmeter using an addition resistance R = 5 kW. To find : E, the maximum voltage that can be measured using this voltmeter. Maximum current drawn by a galvanometer is given as : Ig =

E R + Rg

E = Ig (R + R g ) = 4 ¥ 10 -3 ¥ ( 5 ¥ 10 3 + 50 ) ª 20 V Ans. 30. Option (d) is correct.

2 ¥ 10 ¥ 50 ¥ 10 2

So, net magnetic field at point P : Bnet = B1 + B2 = 0

Above equation is equation of a damped simple harmonic motion, with amplitude :

t =

We can see that B1 and B2 are same in magnitude and opposite in direction.

m 50 ¥ 10 -3 = 2p ª 2 s k 0.5

So, total number of oscillations in time t : t 10000 N = = = 5000 T 2 Ans. 27. Option (b) is correct. Given : Number of turns in a coil is N, radius of the coil is r, current through the coil is I, the coil is located in the XZ plane in a magnetic field Bi^. To find : t, the torque acting on the coil. Torque acting on the coil : τ = m¥B (m = NIA is the magnetic moment of the coil)

τ = NIAB sin 90 = NIAB = NIpr 2 B (A = p r2 is area of cross section of coil)



Ans. 28. Option (a) is correct. Given : An arrangement of two very long, straight wires in xy plane, current through both the wire is I. y 1 I d •P d  I 2

Given : Dimensions of a rectangular coil are A = 5 cm × 2.5 cm, number of turns in the coil is N = 100, current through the coil is I = 3 A, direction of current is clockwise, the coil rests in XZ plane in an applied magnetic field of B = 1 T ^ i. To find : t, the torque on the coil if it is tilted through 45° about the z-axis. Torque on the coil :

τ = m¥B (m = NIA is the magnetic moment of the coil and A is area of cross section of the coil)

τ = NIAB sin 45 Put the given values in the above equation.

τ = 100 ¥ 3 ¥ 5 ¥ 10 -2 ¥ 2.5 ¥ 10 -2 ¥ 1 ¥

2

= 0.27 N m Ans. 31. Option (c) is correct. Given : Side of a square loop is a, a current I2 flows through the loop in counter clockwise direction, the loop lies at a distance a from a long straight wire, the current through the straight wire is I1. To find : Fnet, the net force on the loop due to the wire.

x I1

I2

A

 To find : Magnetic field at point P.

1

a

D

B

 C

a

272 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Force on section AD of the loop due to the straight wire : m II FAD = 0 1 2 a 2p a Here a in denominator is the distance between section AD and the straight wire and a in the numerator is the length of section AD. As I1 and I2 in section AD are in opposite direction the force will be repulsive and will act away from the wire as shown in the above diagram. Force on the section BC of the loop due to the straight wire : m0 I1I 2 a FBC = 2p ( 2 a ) Here 2a in denominator is the distance between section BC and the straight wire and a in the numerator is the length of section BC. As I1 and I2 in section BC are in same direction the force will be attractive and will act towards the wire as shown in the above diagram. Force on the sections AB and DC of the loop due to the straight wire will be zero as both the sections are lying perpendicular to the straight wire. FAB = FDC = 0 That gives, net force on the square loop : m II m II Fnet = FAD - FBc = 0 1 2 a - 0 1 2 a 2p a 2p ( 2 a ) As both the forces are acting in opposite directions we subtract the terms. m II Fnet = 0 1 2 4p and repulsive.

for a current of i = 1 mA. Torque on the coil due to magnetic field : τ = NIBA

...(i)

Restoring torque on the coil due to torsion band : τ = kθ ...(ii) Equating the above two equations. NIBA = kθ B =

Ans. 33. Option (d) is correct. Given : Resistance of a galvanometer is Rg = 50W, maximum current that can pass through the galvanometer is ig = 0.002 A. To find : R, the value of resistance that must be added to the galvanometer in order to convert it into an ammeter of range 0 – 0.5 A. To convert a galvanometer into an ammeter we have to connect a resistance in parallel to the galvanometer. R

i

kθ NIA

p 180 = ª 10 -3 T 175 ¥ 1 ¥ 10 -3 ¥ 1 ¥ 10 -4

G

0.002 ¥ 50 = ( 0.5 - 0.002 ) ¥ R R = 0.2 W Ans. 34. Option (a) is correct. Given : Maximum current through a galvanometer is I = 10–4 A, a series resistance of R = 20 kW is connected to the galvanometer to convert it into a voltmeter, the maximum voltage read by the voltmeter is V = 5 V. To find : The value of the shunt resistance required to convert the galvanometer into an ammeter of range of 0.1 mA. Rg

R Voltmeter

From the above diagram : I( R g + R ) = V (Rg is the resistance of the galvanometer) V 5 Rg = - R = -4 - 20 ¥ 10 3 = 30 kW I 10 Ans. 35. Option (c) is correct. Given : Energy of proton = energy of electron = energy of a helium nucleus, all three are subjected to a magnetic field directed perpendicular to the plane of their motion, rp, re, rHe are the respective radii of the orbit of proton, electron and the helium nucleus. To find : Relation between rp, re, rHe. Radius of the proton’s orbit in a magnetic field : rp =

10 -6 ¥

i – ig ig

In the above circuit diagram, i = 0.5 A is the maximum current we want the ammeter to measure. So, i g R g = (i - i g )R

I

Ans. 32. Option (d) is correct. Given : number of turns in a coil of a moving coil galvanometer is N = 175, area of cross section of the coil is A = 1 cm2, torsion constant of the torsion band is k = 10–6 N-m/rad. To find : The value of magnetic field B for which the coil deflects by angle p θ = 1∞ = rad, 180

PHYSICS

mp v p eB

=

2 mp E eB

...(i) Ê mv ˆ ÁË Use r = qB ˜¯

273

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Radius of the electron’s orbit in a magnetic field : 2 me E mv re = e e = eB eB

...(ii)

Radius of the orbit of helium nucleus in a magnetic field : rHe =

mHe vHe = 2 eB

2mHe E 2 eB

...(iii)

Energy of the three particles is same. Compare equations (i), (ii) and (iii), re : rp : rHe = m1e / 2 : m1p / 2 :

/2 m1He 2

As me < mp < mHe, we get : re < rp < rHe Ans. 36. Option (d) is correct. Given : Current through wire A is I1, current through wire B is I2, separation between the wires is d, current through wire C is I, distance of wire C from wire A is x. C

A

x = -

I1d I1 - I 2

...(ii)

From equations (i) and (ii), x = ±

I1d I1 - I 2

Ans. 37. Option (c) is correct. Given : Total charge on ring placed in the xy plane is q, radius of ring is R = 3a, the ring is centred at the origin, charge q is moving towards the ring along the z axis with speed v at z = 4a. To find : Minimum value of v such that the charge crosses the origin. q R

•q

z=4a

B

Potential due to the charged ring at z = 4a : I1

V =

I2

x d

=

To find : Value of x, such that the net force acting on wire C is zero. If direction of current in C is parallel to I1, there will be force of attraction between A and C and force of repulsion between B and C. Both the forces will act in same direction. If direction of current in C is parallel to I2, there will be force of attraction between B and C and force of repulsion between A and C. Again, both the forces will act in same direction. So, net force on C placed between A and B at a distance x from A will be : m II m0 I2 I Fnet = 0 1 + =0 2p x 2p (d - x ) I1 I = - 2 x d-x I1d I1 + I 2

I1d + I1x = I 2 x

=

kq 2

( 3 a ) + ( 4 a )2

ka 5a

...(i)

Potential due to the charged ring at z = 0 : VC = =

kq 2

R +z

2

=

kq ( 3 a )2 + ( 0 )2

ka 3a

...(ii)

Loss of potential energy of charge q as it travels from z = 4a to z = 0 towards the ring : DU = q( Vc - V ) =

2 kq 2  15 a

The lost potential energy must be less than or equal to the change in kinetic energy of the particle q as it travels from z = 4a to z = 0 towards the ring, so it can cross the origin.

...(i)

Now if we place wire C to the left of wire A. For both possible directions of current in C, the force from A and B will act in opposite directions. m II m0 I2 I Fnet = 0 1 =0 2p x 2p ( d + x ) I1 I2 = x d+x

R +z

2

1 2 2 kq 2 mv ≥ 2 15 a

I1d - I1x = - I 2 x x =

kq 2

v =

2 ¥ m

Ê 1 ˆ ÁË k = 4pe ˜¯ 0 2q2 4pe 0 a ¥ 15

Ans. 38. Option (a) is correct. Given : A current loop is of the shape of an equilateral triangle, side of the triangular loop is l = 1 m, current through the triangular loop is I = 10 A. To find : Magnitude of magnetic field at the centre of the triangular loop.

274 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) A

s = 20

mA . division

To find : Select the appropriate circuit for the converting the galvanometer into a voltmeter with three ranges 0 – 2 V, 0 – 10 V, 0 – 20 V. Current through galvanometer :

O

•

60° 60° i r B

Ig = n ¥ s

C

Using Biot-Savart’s law magnetic field at O due to segment BC will be : mi BBC = 0 (sin 60 + sin 60 ) 4p r =

m0i sin 60 2p r

...(i)

= 50 divisions ¥ 20

A series resistance required to convert the galvanometer into a voltmeter of range 0 – 2V : 2 = I g ( R1 + R g )

Bnet = BBc + B AB + BCA = 3 ¥ BBC m0i sin 60 2p r

2 2 - Rg = - 100 = 1900 W Ig 1 ¥ 10 -3

R1 = ...(ii)

From the diagram we can see :

A series resistance required to convert the galvanometer into a voltmeter of range 0 – 10 V : 10 = I g ( R1 + R 2 + R g )

1 2 = 1 m r = tan 60 2 3

...(iii)

10 - R g - R1 Ig

R2 =

Put value of r from equation (iii) in equation (ii). Bnet = 3 ¥

4p ¥ 10 -7 ¥ 10 ¥ 2p ¥

1 2 3

3 2 = 18 m T

10

=

1 ¥ 10 -3

- 100 - 1900 = 8000 W

A series resistance required to convert the galvanometer into a voltmeter of range 0 – 20 V :

Ans. 39. Option (b) is correct. Given : Radius of a thin ring is r = 10 cm, angular speed of the ring is w = 40p rad/s, the ring rotates about its axis perpendicular to its plane, magnetic field at the centre of the ring is B = 3.8 × 10–9 T. To find : q, charge on the ring. Magnetic field at the centre of the ring carrying current I is given by : m I B = 0 ...(i) 2r Current through the ring rotating with angular speed of w = 40p rad/s : 40p q qω I = = =q¥ = 20 q ...(ii) t 2p 2p Substitute in equation (i) from equation (ii). m ¥ 20 q B = 0 2r q =

mA division

= 1mA

Due to symmetry of the problem :

Bnet = 3 ¥

PHYSICS

-2

20 = I g ( R1 + R 2 + R 3 + R g )

Ans. 40. Option (c) is correct. Given : Resistance of a galvanometer is Rg = 100 W, number of divisions on the galvanometer scale is n = 50, sensitivity of the galvanometer is

1 ¥ 10 -3

- 100 - 1900 - 8000

= 10000W Ans. 41. Option (b) is correct. Given : Initial energy of electron is E = 100 eV,   the electron initial velocity of electron is v = vi, enters a region of magnetic field B = 1.5 ¥ 10 -3 T k , the magnetic field extends from x = 0 to x = 2 cm, distance SP = 8 cm. To find : Distance PQ (d = h + y).

•

2rB 2 ¥ 10 ¥ 10 ¥ 3.8 ¥ 10  = 20 m0 20 ¥ 4p ¥ 10 -7

= 3 ¥ 10 C

20

=

-9

-5

20 - R g - R1 - R 2 Ig

R3 =



Q R h

R

 X

6 cm y

S

• x=0

x=2

P

275

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Radius of circular orbit of an electron in a magnetic field : r =

=

mv = qB

2mE qB

IgG Ig I 2g RA = ¥ = RV I0 - I g G( Io - I g ) ( I0 - I g )2 Ans. 43. Option (b) is correct. Given : Length of wire AB is l = 6 cm, current through the wire is i = 5 A. To find : The magnetic field at point P shown in the diagram below.

2 ¥ 9.1 ¥ 10 -31 ¥ 100 ¥ 1.6 ¥ 10 -19 1.6 ¥ 10 -19 ¥ 1.5 ¥ 10 -3

= 2.25 cm From the above diagram : x 2 sin θ = = ; θ = 62.734∞ R 2.25

P

h = 6 ¥ tan θ = 11.642 cm

• m

5c

m

1 2

That gives, d = h + y = 11.642 + 1.22 ª 12.87 cm

d

Ans. 42. Option (b) is correct. Given : resistance of a moving coil galvanometer is G, maximum current through the galvanometer is Ig, the galvanometer can be converted into an ammeter of range (0 – I0) (I0 > Ig) by connecting a shunt resistance RA, the galvanometer can also be converted into an voltmeter of range (0 – V) (V = GI0) by connecting a series resistance RV. To find : RARV and RA/RV Circuit for converting galvanometer into ammeter : RA

•

Using Biot-Savart’s law, magnetic field at P due to wire segment AB is given as : mi B = 0 (sin θ1 + sin θ 2 ) 4p d From the figure shown above : 3 sin θ1 = sin θ 2 = , 5 d =

( 5)2 - ( 3)2 = 4 cm

B=

4p ¥ 10 -7 ¥ 5 Ê 3ˆ Á 2 ¥ 5 ˜¯ 4p ¥ 4 ¥ 10 -2 Ë

That gives, G

= 1.5 ¥ 10 -5 T

IgG

...(i)

I0 - I g

Circuit for converting galvanometer into voltmeter  : RV

Ans. 44. Option (d) is correct. Given : current through the loop of wire formed by two circular arcs is i = 10 A. To find : The magnetic field at point O.

G

O

•

Also,

V = GI0

3c (given)

G( I0 - I g ) GI0 -G= Ig Ig

From equations (i) and (ii), IgG G( I0 - I g ) RARV = ¥ = G2 I0 - I g Ig

R

m 2c ...(ii)

m

RV =

Q

2c

GI0 = I g (G + R V )

m

V = I g (G + R V )

3c

m

Voltmeter

45°

Ig

B

6 cm

( I0 - I g ) R A = I g G RA =

•

A

=

i0

i0 – ig ig

5c

Also, using the left triangle : y = R - R cosθ = 2.25 (1 - 0.458 ) = 1.22 cm

S

P i = 10 A

Magnetic field at O due to wire PQ = magnetic field at O due to wire RS = 0, as O lies on the axial position of the current carrying wire.

276 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Magnetic field at O due to wire QR : θ Ê m 0i ˆ B1 = 2p ËÁ 2 a ¯˜

...(i)

In equation (i), a is the distance of point O from the current carrying wire and in this case a = 3 cm. p From the above figure θ = 45∞ = . Put this value 4 in equation (i). B1 =

4p ¥ 10 -7 ¥ 10 p 2p ¥ 4 2 ¥ 3 ¥ 10 -2

= 2.6 ¥ 10 T Similarly, Magnetic field at O due to wire PS : B2 =

As an additional electric field is switched on, the particle moves in a straight line, that means both the forces balance out. FB = Fe qvB = qE vB = E Put value of v from equation (i). qrB B= E m m=

-4

=

-7

4p ¥ 10 ¥ 10 p 2p ¥ 4 2 ¥ 5 ¥ 10 -2

PHYSICS

qrB2 E 1.6 ¥ 10 -19 ¥ 0.5 ¥ 10 -2 ¥ ( 0.5)2 100

= 2 ¥ 10 -24 kg

= 1.56 ¥ 10 -4 T  Direction of both the field will be opposite to each other. So, net magnetic field at O due to the given current loop will be : Bnet = B1 - B 2 = 1 ¥ 10 -5 T Ans. 45. Option (d) is correct. Given : radius of a small current carrying loop is a, the loop lies a distance d from a long current carrying wire, d  a. To find : F, the force exerted by the loop on the wire. Let the current through the small loop of wire be I. Then, the dipole moment of the current carrying loop will be :

Ans. 47. Option (d) is correct. Given : Length of wire-1 = length of wire-2 = L, wire-1 is bent in to form a circular loop and wire-2 is bent in to form a circular coil with N number of turns, current through both the wires is I. B To find : L , the ratio of magnetic field at the BC centre of the loop to that at the centre of the coil. Radius of the circular loop formed by wire-1 : L 2p R L = L ; R L = 2p Magnetic field at the centre of the circular loop formed by wire-1 : m I m0 I pm I ...(i) BL = 0 = = 0 2 R L 2 L / 2p  L

m = Ip a 2 ; m ∝ a 2

...(i)

dB d (d )

...(ii)

Radius of the circular coil formed by wire-2 : L 2 Np R C = L ; R C = 2 Np

...(iii)

Magnetic field at the centre of the circular coil formed by wire-2 : m I m0 I BC = N 0 = N 2 RC 2 L / 2 Np 

F= m

B is magnetic field of the wire : m I¢ dB 1 B= 0 ; ∝ 2 2p d d ( d ) d From equations (i), (ii) and (iii), F∝

a2 d2

= N2

Ans. 46. Option (d) is correct. Given : charge on particle is q = 1.6 × 10–19 C, radius of circular path traced by the particle is r = 0.5 cm, applied magnetic field is B = 0.5 T. To find : m, mass of the particle if it moves in a straight path under influence of an additional electric field of E = 100 V/m. Under influence of magnetic field only, the radius of the particle will be : mv qrB r = ; v= ...(i) qB m The magnetic force on particle is FB = qvB

...(ii)

pm0 I L

...(ii)

From equations (i) and (ii), BL 1 = 2 BC N Ans. 48. Option (a) is correct. Given : Edge length of a solid metal cube is a = 2 cm, speed of cube is v = 6 m/s along the y-axis, magnetic field is along +z-axis, magnitude of magnetic field is B = 0.1 T. To find : V, the potential difference between the two faces of the cube perpendicular to the x-axis. Potential difference between the two opposite faces of the cube = emf induced

V = Bav = 0.1 ¥ 2 ¥ 10 -2 ¥ 6 = 12 mV 

277

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Ans. 49. Option (d) is correct.

Y centre of circular path v cos  Y=2d r=2d  vf  Y=d v sin 

Given : Potential difference through which the electrons are accelerated is V = 500 V. To find : r, radius of the electron’s path after magnetic field B = 100 mT, is applied. Radius of the circular path of the electron when it is moving through a magnetic field is given as : mv r = = eB

2mE = eB

2meV = eB

Y=0  v

2mV 1 ¥ e B

x

(E = eV is energy of electron) Radius of circular path of particle of charge q in magnetic field B : mv r = ...(ii) qB

Put given values in the above equation. 2 ¥ 9.1 ¥ 10 -31 ¥ 500

r =

1.6 ¥ 10 -19

¥

1 100 ¥ 10 -3

= 7.5 ¥ 10 -4 m

From equation (i) : r = 2d

Ans. 50. Option (c) is correct. Given : Resistance of a galvanometer is Rg = 20 W, number of divisions on either side of zero of galvanometer are n = 30, figure of merit for the galvanometer is FM = 0.005

Ampere . division

To find : The value of series resistance R that should be connected to the galvanometer such that it can be used as a voltmeter up to V = 15 V. Maximum current that can be measured by the given galvanometer : I g = FM ¥ n I g = 0.005 ¥ 30 = 0.15 A So, maximum current through the circuit composed of a series combination of Rg and R is Ig = 0.15 A. The required voltage drop is V = 15 V. That implies, V = I g R eq = I g ( R + R g ) Put given values in the above equation. 15 = 0.15 ( R + 20 ) R = 80W Ans. 51. Option (b) is correct.

  exists in a region Given : Magnetic field B = Bk, from y = 0 to y = d, mass of particle is m, charge on particle is q, the particle enters the magnetic field   region with velocity v = vi. To find : Acceleration of the charged particle at the point of its emergence from the magnetic field region if d =

mv . 2 qB

...(i)

As depicted in the diagram : d 1 sin θ = = ; θ = 60∞ 2d 2 The final velocity vector of the particle when it emerges from the magnetic field region :  Ê 3  1 ˆ v f = v (sin θ i + cosθ j ) = v Á i + j˜ 2 ¯ Ë 2

...(iii)

Using equation (ii), the acceleration of the particle when it emerges from the magnetic field region :        dv f d Ê qBrf ˆ qB drf qB  v f ...((iv) = = af = Á ˜= dt dt ÁË m ˜¯ m dt m  Put value of v f   qvB af = m

from equation (iii) in equation (iv). Ê 3  1 ˆ Á 2 i + 2 j˜ Ë ¯

Ans. 52. Option (b) is correct. Given : Mass of particle is m, charge on particle is q, applied electric field is E = 2i + 3 j ,N/C, applied magnetic field is B = 4 j + 6 k .T. To find : Work done on particle when it is shifted from origin to point P(x = 1, y = 1) along a straight line. Net force on the particle due to the electric and magnetic field : Fnet = qE + qv ¥ B As the particle is moving along the x-axis v × B = 0. F = qE = 2 qi + 3q j ...(i) net

Work done on the particle : work = force ¥ displacement W =

x =1

y =1

ÚdW = Úx = 0 2qdx + Úy = 0 3qdy

= 2 q + 3q = 5q

278 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 53. Option (c) is correct. Given : Two infinitely long identical wires are bent by 90° and placed as shown in the figure below, OP = OQ = 4 cm, magnitude of net magnetic field at O is Bnet = 10–4 T, the two wires carry equal current I. y

To find : The ratio of their radii, rp : ra. Radius of the circular path of a charged particle q when it is moving through a magnetic field B is given as : r =

mv = qB

S

P

2mE = qB

2mqV

...(iii)

qB

(E = qV is energy of charged particle) For proton :

O Q

L

PHYSICS

•

M x N

rp = 

2 mp q p V qp B

=

2 mp V qp

¥

1 B

...(iv)

For a particle : To find : Value of I and direction of B at O. Magnetic field at O due to wire segment LP = magnetic field at O due to wire segment QM = 0, as O lies on the axial position of the current carrying wire. Magnetic field at O due to wire segment PS : m I ...(i) B1 = 0 4pr Magnetic field at O due to wire segment MN : m I B2 = 0 ...(ii) 4pr Using thumb rule we can see that both B1 and B2 are into the plane of paper. m I Bnet = B1 + B2 =  0 2p r I =

2p rBnet 2p ¥ 4 ¥ 10 -2 ¥ 10 -4 = = 20 A m0 4p 10 -7

Ans. 54. Option (a) is correct. Given : Mass of a proton and a particle are in ratio mp 1 = . ...(i) mα 4 the charge on proton and charge on a particle are in ratio qp 1 = , ...(ii) qα 2 both the particles are accelerated from rest through a potential difference V, applied magnetic field is perpendicular to their velocities.

rα = 

2mα qα V qα B

=

2mα V 1 ¥ qα B

...(v)

From equations (i), (ii), (iv) and (v), rp : rα =

mp qα mα q p

Ê 1ˆ Ê 2ˆ = Á ˜ ¥Á ˜ = 1: 2 Ë 4¯ Ë 1¯

Ans. 55. Option (b) is correct. Given : Resistance of a galvanometer is Rg = 50 W, number of divisions on the galvanometer are n = 25, when current through the galvanometer is 4 × 10–4 A its needle deflects by one division. To find : The value of series resistance R that should be connected to the galvanometer such that it can be used as a voltmeter up to V = 2.5 V. Maximum current that can be measured by the given galvanometer : I g = 4 ¥ 10 -4 ¥ n I g = 4 ¥ 10 -4 ¥ 25 = 10 -2 A So, maximum current through the circuit composed of a series combination of Rg and R is Ig = 10–2 A. The required voltage drop is V = 2.5 V. That implies, V = I g R eq = I g ( R + R g ) Put given values in the above equation. 2.5 = 10 -2 ( R + 50 ) R = 200W

279

Magnetic Effects of Current and Magnetism

Topic-2

Magnetism Concept Revision (Video Based) Para, Dia and Ferro Magnetic Substances

Magnetic Dipole Moment     

Part -1

 

Part - 2

      

Part -1

 

    Part - 2

Magnetic Permeability

Hysteresis               

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following, which property best matches for the type of magnet required ? (a) T : Large retentivity, large coercivity (b) P : Large retentivity, large coercivity (c) T : Large retentivity, small coercivity (c) T : Small retentivity, large coercivity [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic substance. The whole system is  placed in a uniform magnetic field B. Then the field inside the paramagnetic substance is :

Q.4.



The figure gives experimentally measured B vs. H variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of the material are : (a) 1.5T, 50 A/m and 1.0 T (b) 1.0 T, 50 A/m and 1.5 T (c) 1.5 T,50 A/m and 1.0 T

  (a) much large than B but opposite to B (b) Zero

  (c) much large than B but parallel to B  (d) B [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.3. A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external magnetic field of 0.4 T at a temperature of 4K. When the sample is placed in an external magnetic field of 0.3 T at a temperature of 24K, then the magnetisation will be : (a) 4 A/m (b) 0.75 A/m (c) 2.25 A/m (d) 1 A/m [JEE (Main) – 4th Sep. 2020 - Shift-2]

(d) 150 A/m, 1.0 T and 1.5 T [JEE (Main) – 7th April 2019 - Shift-2] Q.5. Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their midpoint P, at angle q = 45° with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant ? (d is much larger than the dimensions of the dipole) d N

S X (M)

P



S N Y (2 M)

280 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a) Th = Tc (c) Th = 1.5Tc

Ê m0 ˆ M ¥ qv (a) ÁË ˜¯ 4p ( d / 2 )3 (b) 0 (c)

Êm ˆ M 2 Á 0˜ ¥ qv Ë 4p ¯ ( d / 2 )3

Ê m0 ˆ 2 M ¥ qv (d) ÁË ˜¯ 4p ( d / 2 )3 [JEE (Main) – 8th April 2019 - Shift-2] Q.6. A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be : (a)

m p

2m (c) p

PHYSICS

(b)

3m p

4m (d) p

[JEE (Main) – 10th April 2019 - Shift-2] Q.7. A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45° and 40 times per minute where the dip is 30°. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is best given by : (a) 1.8 (b) 0.7 (c) 3.6 (d) 2.2 [JEE (Main) – 12th April 2019 - Shift-1] Q.8. A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns and carrying a current of 5.2 A. The coercivity of the bar magnet is : (a) 285 A/m (b) 2600 A/m (c) 520 A/m (d) 1200 A/m [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.9. An insulating thin rod of length l has a linear x on it. The rod is rotated l about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is : p 3 (a) p nrl3 (b) nrl 3 p 3 (c) nrl (d) nrl3 4 charge density r(x) = r0

[JEE (Main) – 10th Jan. 2019 - Shift-1] Q.10. A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc. respectively, then :

(b) Th = 2Tc (d) Th = 0.5Tc [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.11. A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 × 10–6 J/T when a magnetic intensity of 60 × 103 A/m is applied. Its magnetic susceptibility is : (a) 3.3 × 10–2 (b) 4.3 × 10–2 –2 (c) 2.3 × 10 (d) 3.3 × 10–4 [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.12. A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 × 10–4. Its susceptibility at 300 K is : (a) 3.267 × 10–4 (b) 3.672 × 10–4 –4 (c) 3.726 × 10 (d) 2.672 × 10–4 [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (b) 5. (b) 9. (c)

2. (b) 6. (d) 10. (a)

3. (b) 7. (b) 11. (d)

4. (b) 8. (b) 12. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. Given : A permanent magnet P and a transformer magnet T. To find : The correct property of the two given magnets. As permanent magnets should retain their magnetic fields even in the absence of external magnetic field. They should have large retentivity and large coercivity. Transformer magnets are magnetized and demagnetized multiple numbers of times. So they should have small retentivity and small coercivity. Ans. 2. Option (b) is correct. Given : A diamagnetic sphere has a small spherical cavity at its centre, the cavity is filled with a paramagnetic material and the sphere is kept in a  uniform external magnetic field B. To find : the magnetic field in the spherical cavity. The diamagnetic sphere will tend to repel the magnetic field lines. As a result the magnetic field lines will not penetrate the sphere and magnetic field inside the spherical cavity will be zero. Ans. 3. Option (b) is correct. Given : Case 1 : For external magnetic field B1 = 0.4 T and temperature T1 = 4K, magnetization A of a paramagnetic sample is M1 = 6 , m Case 2 : for external magnetic field B2 = 0.3 T and temperature T2 = 24 K, magnetization of a paramagnetic sample is M2. To find : The value of M2. Let C be the Curie’s constant. Magnetization of a paramagnetic sample : CB1 M1 = ...(i) T1

281

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

M2 =

CB2 T2

...( ii )

From equations (i) and (ii) : M1 B T = 1× 2 M2 B 2 T1

M 2 = M1 × M2 = 6 ×

B2 =

0.3 4 A × = 0.75 0.4 24 m

=

Ans. 4. Option (b) is correct. Given : B versus H graph for a ferromagnetic material,

To find : The retentivity, co ercivity and saturation field for the ferromagnet. From the given graph we can see: Retentivity (1) = 1.0 T, Co-ercivity (2) = 50 A/m and Saturation field (3) = 1.5 T.

Ans. 5. Option (b) is correct. Given : Distance between two magnetic dipoles X and Y is d angle between the axes of two dipoles is 90°, dipole moment of Y is twice that of X. To find : The force on a charged particle q passing through the midpoint P with its velocity making an angle of q = 45° with the horizontal line connecting the two dipoles. d N

S X (M)

P



Bnet B1

S N Y (2 M)

Let M be dipole moment of X. That gives; dipole moment of Y will be 2M. The distance of separation of P from the centre of dipole X and centre of dipole d Y is . 2 So, magnetic field at axial point P due to dipole X (d  dipole dimension) : B1 =

m0 2 M 4p Ê d ˆ 3 ÁË 2 ˜¯

...(i)

Similarly, magnetic field at equatorial point P due to dipole Y (d  dipole dimension) :

B2 T1 × B1 T2

B2

=

2 ¥ dipole moment m0 4p ( distanceofseparation )3

dipole moment m0 4p (distanceofseparation )3 m0 2 M 4p Ê d ˆ 3 ÁË 2 ˜¯

...(ii)

The direction of B1 is along the horizontal line joining the two dipoles and direction of B2 is perpendicular to the horizontal line joining the two dipoles, as shown in the figure. That gives,    Bnet = B1 + B 2 ...(iii)    As magnitude of B1 and B2 are same, Bnet will make an angle of q = 45° with the horizontal line connecting the two dipoles.  Force on charge q due to Bnet :   F = qv ¥ Bnet   As angle between Bnet and v is zero, F = 0. Ans. 6. Option (d) is correct. Given : Current through a square loop is I, magnetic dipole moment of the square loop is m. To find : M, dipole moment of the circular loop carrying current I, formed by deforming the square. Let the edge length of given square be a. Then, m = I ¥ Areaof  square = Ia 2

...(i)

The same square is deformed to form a circle of radius r. So, 2a 2p r = 4 a ; r = p Hence, magnetic moment of the circular loop will be : M = I ¥ Areaof  circle = Ipr 2 2

4 a2 I 4 Ê 2a ˆ = Ip Á ˜ = = m Ëp¯ p p Ans. 7. Option (b) is correct. Given : Magnetic field of earth at point A is B1, the frequency of oscillation of magnetic compass 30 , dip at A is q1 = 45°, needle at A is ω1 = minute magnetic field of earth at point B is B2 the frequency of oscillation of magnetic compass needle at B is 40 , dip at B is q2 = 30°. ω2 = minute To find : The ratio

B1 . B2

282 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Time period of oscillation of the magnetic compass needle at A : 1 60 T1 = ...(i) = = 2 s ω1 30

y

x

Time period of oscillation of the magnetic compass needle at B : 1 60 3 T2 = ...(ii) = = s ω2 40 2 Horizontal component of earth’s magnetic field at A: B BH1 = B1 cosθ1 = B1 cos 45 = 1 ...(iii) 2 Horizontal component of earth’s magnetic field at B: BH2 = B 2 cosθ 2 = B2 cos 30 = We know, T ∝

1 BH

3B2 ...(iv) 2

O

BH 2 = B H1

x

dx

Consider a thin length segment dx at distance x from axis of rotation. Let charge on this length segment be dq. So, magnetic moment of this length segment will be : dm = dq ¥ frequency ¥ Areacovered bytherodwhilerotation = nA ( dq ) Put dq = r(dx) and A = px2 dm = np x 2 ρ ( dx )

.

So, from equations (i), (ii), (iii) and (iv), T1 = T2

PHYSICS

3B2 2 = 2 B1 3 2 2

Squaring both sides of the above equation : 3 B2 16 = 2 B1 9

Substitute from equation (i) in equation (ii), x dm = np x 2 ρo ( dx ) l

Magnetic field intensity required to reduce the magnetisation of the given bar magnet to zero : B N 100 A H = = I= ¥ 5.2 = 2600 . l 0.2 m m0 Ans. 9. Option (c) is correct. Given : Linear charge density on an insulating thin rod is x ρ ( x ) = ρ0 , ...(i) l length of the rod is l, the rod lies along the x axis and rotates about a perpendicular axis passing through x = 0, frequency of rotation of rod is n. To find : m, time averaged magnetic moment of the rod.



...(iii)

Integrate equation (iii) to get time averaged magnetic moment of the rod : l x m = Údm = Ú np x 2 ρ0 ( dx ) 0 l

B1 9 3 = ¥ = 0.7  B2 16 2 Ans. 8. Option (b) is correct. Given : Length of solenoid is l = 0.2 m, number of turns in the solenoid is N = 100, current through the solenoid is I = 5.2 A. To find : The coercivity of the bar magnet that is demagnetised by inserting it inside the given solenoid. Magnetic field at centre of a solenoid : N B = m0 I ...(i) l

...(ii)

At

=

np ρ0 l

=

np ρ0l 3 4

l 3

Ú0 x

dx =

np ρ0 l 4 ¥ l 4

x = l , ρ = ρ0 . m =

np ρ l 3 4



Ans. 10. Option (a) is correct. Given : Radius of hoop = radius of solid cylinder = r, mass of hoop = mass of cylinder = m, magnetic moment of both the bodies lie parallel to their respective axes, magnetic moment of the two are related as Mh = 2Mc, the bodies are placed in a magnetic field in such a manner that their magnetic moments make a small angle with the field. To find : Relation between the oscillation periods of the hoop and the cylinder, Th and Tc. Let the applied magnetic field be B, Ih = mr2 is 1 moment of inertia of the hoop and I c = mr 2 is 2 moment of inertia of the cylinder. Time period of oscillation of hoop : Th = 2p = 2p

Ih m = 2p r MhB MhB m r 2Mc B

...( i )

283

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Time period of oscillation of hoop : Tc = 2p = 2p

Ih Mc B m r 2Mc B

...(ii)

From equations (i) and (ii), Th = Tc Ans. 11. Option (d) is correct. Given : Magnetic dipole moment of a paramagnetic substance is m = 20 × 10–6 J/T, the paramagnetic substance is in the form of a cube with edge length a = 1 cm. To find : The magnetic susceptibility of the substance when a magnetic intensity of H = 60 × 103 A/m is applied. Magnetisation in the paramagnetic substance : m m M = = Volume a3 M =

20 ¥ 10

-6

(10 -2 )3

= 20 A / m

Susceptibility of the paramagnetic substance : M 20 χ = = = 3.3 ¥ 10 -4 H 60 ¥ 10 3

Sol. Given : Number of turns in a galvanometer coil is N = 500, area of each turn is A = 3 × 10–4 m2, current through the coil is I = 0.5 A, torque on the coil due to external magnetic field B is t = 1.5 Nm,   p angle between A and B is θ = . 2

To find : The magnitude of B. Torque on the galvanometer coil:   τ = M × B = NIABsin θ B = B =

τ NIA 1.5 500 × 0.5 × 3 × 10 −4

= 20 T

Q.2. A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region  is B = (3i + 4 k )T . The quantity of flux through the loop ABCDEFA (in Wb) is [JEE (Main) – 7th Jan. 2020 - Shift-1] Sol. A loop ABCDEFA as shown below, the magnetic field in the region shown is  B = ( 3i + 4 k )T ,

Ans. 12. Option (a) is correct. Given : Number density of atoms in a paramagnetic atoms material is n = 10 28 , magnetic susceptibility m3 at T1 = 350 K is c1 = 2.8 × 10–4. To find : c2, magnetic susceptibility at T2 = 300 K. Using curie’s law for paramagnetic materials 1ˆ Ê ÁË χ ∝ T ˜¯ , we get :

χ1 ∝

1 T1

...(i)

χ2 ∝

1 T2

...(ii)

From equations (i) and (ii), χ2 T = 1 χ1 T2

χ2

ÊT ˆ = χ1 Á 1 ˜ = 2.8 ¥ 10 -4 ËT ¯ 2

To find : f, the magnetic flux through the loop ABCDEFA. Magnetic flux through the loop ABCDEFA :  φ = B ⋅ ( areavectorforloopABCD + areavectorforloopADEF)

φ = ( 3i + 4 k ) ⋅ ( 25k + 25i) = 75 + 100 = 175 Wb Ê 350 ˆ ÁË 300 ˜¯

= 3.267 ¥ 10 -4

Subjective Questions (Chapter Based) Q.1. A galvanometer coil has 500 turns and each turn has an average area of 3 × 10–4 m2. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is [JEE (Main) – 3rd Sep. 2020 - Shift-2]

Q.3. Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, B then what will be value of ratio A ? BB Sol. Given : Length of wire A = length of wire B = l, wire A is bent to form a circle and wire B is bent to form a square, radius of circle is R, edge length of square is a, magnetic field at the centre of circle is BA, magnetic field at the centre of square is BB. B To find : The ratio A . BB

284 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

Radius of circle formed by bending a wire of length l :

R =





l 2p

Magnetic field at the centre of circle : BA =





m0 I m0 I pm I = 0 = 2R l Ê l ˆ 2Á ˜ Ë 2p ¯

...(i)

Magnitude of magnetic field due to arc BC : 1 m 0i B2 = 2 2R2

Magnetic field at the centre of a square : BB = 4 ¥ magneticfielddueto singlearmofthesquare

...(ii)

Direction of B1, by thumb rule will be out of the paper and direction of B2, by thumb rule will be into the paper. 1 m 0i 1 m 0i So, Bnet = B2 - B1 = 2 2 R 2 2 2 R1

=

Edge length of square formed by bending a wire of length l : l a = 4

BB = 4 ¥



PHYSICS

m 0i Ê 1 1ˆ 4 ÁË R 2 R1 ˜¯

Q.5. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k is coming out of the plane of the paper. What will be the magnetic moment of the current loop? y

m0 I (sin 45 + sin 45) 4pd

I a

x a

a=1/4

Sol. Given : A current carrying loop is made of 4 semicircles and a square, diameter of circle = edge length of the square = a, current through the loop is I. To find : The magnetic moment of the current loop. Area of the given loop :

45° 45° d

BB = 4 ¥

=

m0 I (sin 45 + sin 45) l 4p  8

8 2 m0 I pL

2

...(ii)

From equation (i) and (ii), 2

BA p = BB 8 2

Q.4. Find the magnetic field at point O, as shown in the figure below. i



Ê aˆ pÁ ˜ Ë 2¯ Êp ˆ A = 4¥ + a 2 = a 2 Á + 1˜ Ë2 ¯ 2 Magnetic moment of the loop : Êp ˆ m = IA = Ia 2 Á + 1˜ k Ë2 ¯

Q.6. Find magnetic field at point B due to the current loop ABC. A

i

R2

•

R1

O A B C D Sol. Given : Current through the loop is i, radius of the inner circle is R2, radius of the outer circle is R1. To find : The magnetic field at point O. Magnetic field due to wire segment AB = magnetic field due to wire segment CD = 0, as point O lies on the axial position of the current carrying wire. Magnitude of magnetic field due to arc AD : 1 m 0i B1 = ...( i ) 2 2 R1

6

B

8

10

C

Sol. Given : A current loop ABC is in shape of right angled triangle, current through the loop is i. To find : The magnitude of magnetic field at point B. Magnetic field due to wire segment AB = magnetic field due to wire segment BC = 0, as point B lies on the axial position of the current carrying wire.

285

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM



Magnetic field due to wire segment AC :



A 53°

To find : The resultant horizontal magnetic induction at the midpoint O. N S S

i

•

6 10 d

37°



37° 8

C

m0 I 0.07 (sin 53 + sin 37 ) = m0 I p 4p d

Q.7. Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. What is the value of the resultant horizontal magnetic induction at the mid - point O of the line joining their centre? (Horizontal component of earth’s magnetic induction is 3.6 × 10–5 Wb/m2) Sol. Given : length of bar magnet 1 = length of bar magnet 2 = 1 cm, magnetic moment of bar magnet 1 is m1 = 1.20 Am2, magnetic moment of bar magnet 2 is m2 = 1.00 Am2, the arrangement is as shown in the figure below, the separation between the magnets is d = 20.0 cm = 0.2 m, horizontal component of earth’s magnetic induction at the midpoint O of the line joining their centres is



d = 6 cos 37 = 4.8 B =

•

N

53° B

•

O

BH = 3.6 ¥ 10 -5 Wb m -2 .





W N

1

2

S

Magnetic induction due to bar magnet 1 at O : m m1 B1 = 0 ...(i) 4p Ê d ˆ 3 ÁË 2 ˜¯ Magnetic induction due to bar magnet 2 at O : m m2 B1 = 0 ...(ii) 4p Ê d ˆ 3 ÁË 2 ˜¯ Earth’s magnetic induction at O : BH = 3.6 ¥ 10 -5 Wb m -2



From equations (i), (ii) and (iii),



Bnet = B1 + B 2 + B H =

Bnet =

E

m0 Ê dˆ 4p Á ˜ Ë 2¯ 10 -7 ( 0.1)3

3

...(iii)

( m1 + m2 ) + BH

(1.2 + 1) + 3.6 ¥ 10 -5

= 2.56 ¥ 10 -4 Wb m -2



2-

e uc nd d n i co EMF ting rota

d uc i n a t or

1- Whenever magnetic flux through an area bounded by a closed conducting loop changes, an emf is produced in the loop

Electromagnetic Induction & Alternating Currents Part-1

286 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

–

The hindrance offered by inductor or capacitor or both to the flow of AC ( )

. .

Electromagnetic Induction & Alternating Currents Part-2

.

..

Transfomer

Electromagnetic Induction and Alternating Currents

287

Electromagnetic Induction and Alternating Currents

Chapter 14 Syllabus

Electromagnetic induction; Faraday’s law, induced emf and current; Lenz’s Law, Eddy currents'; Self and mutual inductance; Alternating currents, peak and rms value of alternating current/voltage; reactance and impedance; LCR series circuit, resonance; Quality factor, power in AC circuits, wattless current. AC generator and transformer. RC, LR and LC circuits with DC and AC sources.

Topic-1

LIST OF TOPICS : Topic-1 : Electromagnetic Induction

Electromagnetic Induction

.... P. 288

Topic-2 :  RC, LR, LC and LCR Circuits .... P. 295

Concept Revision (Video Based) Faraday’s Law      

 

         

   

Part -1    Part - 2

     

  

    

Part -1    Part - 2           

Lenz’s Law

Part -1      Part - 2

Eddy Currents    

 

Self and Mutual Inductance

  

Part -1 Part - 2 AC Transformers

          Part -1             Part - 2   

Reactance and Impedance   

     

            Part -1     Part - 2

Peak and RMS Value of Alternating Current/Voltage

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. An elliptical loop having resistance R, of semi major axis a and semi minor axis b is placed in magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency w, the average power loss in the loop due to Joule heating is :

289

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

p 2 a 2 b 2 B2ω 2 R

(a)

p abBω R

(b)

(c)

p 2 a 2 b 2 B2ω 2 2R

(d) Zero

[JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.2. A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field dB changes with time at a steady rate = 0.032 dt T s–1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23 × 10–8 W-m) (a) 0.61 A (b) 0.34 A (c) 0.43 A (d) 0.53 A [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.3. An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown in the figure. The connector has length l and resistance R. It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is :

v

I

R

l

r

(a)

m0 Ivl p Rr

(b)

2 m0 Ivl p Rr

(c)

m0 Ivl 4p Rr

(d)

m0 Ivl 2p Rr

[JEE (Main) – 5th Sep. 2020 - Shift-2] Q.4. A long solenoid of radius R carries a time (t) dependent current I(t) = I0t(1 – t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 £ t £ 1, the induced current (IR) and the induced EMF(VR) in the ring change as : (a) Direction of IR remains unchanged and VR is zero at t = 0.25 (b) Direction of IR remains unchanged and VR is maximum at t = 0.5 (c) At t = 0.25 direction of IR reverses and VR is maximum (d) At t = 0.5 direction of IR reverses and VR is zero [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.5. A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

(a) 2.5 s and 7.5 s (b) 5.0 s and 7.5 s (c) 5.0 s and 10.0 s (d) 2.5 s and 5.0 s [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.6. At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is :

(a) 36 mV (c) 56 mV

(b) 48 mV (d) 28 mV [JEE (Main) – 8th Jan. 2020 - Shift-1] Q.7. An alternating voltage v(t) = 220 sin 100pt volt is applied to a purely resistive load of 50 W. The time taken for the current to rise from half of the peak value to the peak value is : (a) 5 ms (b) 2.2 ms (c) 7.2 ms (d) 3.3 ms [JEE (Main) – 8th April 2019 - Shift-1] Q.8. The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to  : (a) L (b) L2 (c) 1/L2 (d) 1/L [JEE (Main) – 9th April 2019 - Shift-1] Q.9. A very long solenoid of radius R is carrying current I(t) = kte–at (k > 0), as a function of time (t ³ 0). Counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by :

(a)

I t=0

(b)

I t=0

(c)

I t=0

(d)

t=0

I



t

t

t

t

[JEE (Main) – 9th April 2019 - Shift-2]

290 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.10. Two coils ‘P’ and ‘Q’ are separated by some distance. When a current of 3 A flows through coil’P’, a magnetic flux of 10–3 Wb passes through ‘Q. No current is passed through ‘Q’. When no current passes through ‘P’ and a current of 2 A passes through ‘Q’, the flux through ‘P’ is : (a) 6.67 × 10–4 Wb (b) 3.67 × 10–3 Wb –3 (c) 6.67 × 10 Wb (d) 3.67 × 10­–4 Wb [JEE (Main) – 9th April 2019 - Shift-2] Q.11. A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are : (a) 220 V and 20 A (b) 440 V and 20 A (c) 440 V and 5 A (d) 220 V and 10 A [JEE (Main) – 10th April 2019 - Shift-1] Q.12. The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s–1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 W, the current in the loop at that instant will be close to :

B

. . . . .

. . . . .

. . . . .

. . . . .

v=1 cm s–1 L 1 A 1

B

D

2 C 2

5 cm

(a) 60 mA (c) 150 mA

(b) 170 mA (d) 115 mA [JEE (Main) – 12th April 2019 - Shift-1] Q.13. A conducting circular loop made of a thin wire, has area 3.5 × 10–3 m2 and resistance 10 W. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4 T) sin (50 pt). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to : (a) 0.14 mC (b) 7 mC (c) 21 mC (d) 6 mC [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.14. A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be : (a) 50 A (b) 45 A (c) 35 A (d) 25 A [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.15. A magnet of total magnetic moment 10–2 ^ i A-m2 is placed in a time varying magnetic field, B ^ i (coswt) where B = 1 Tesla and w = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is :

PHYSICS

(a) 0.01 J (b) 0.007 J (c) 0.028 J (d) 0.014 J [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.16. The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is : (a) 740 J (b) 437.5 J (c) 540 J (d) 637.5 J [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.17. At some location on earth the horizontal component of earth’s magnetic field is 18 × 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 A-m is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is : (a) 3.6 × 10–5 N (b) 1.8 × 10–5 N –5 (c) 1.3 × 10 N (d) 6.5 × 10–5 N [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.18. There are two long co-axial solenoids of same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is : n n r (a) 2 . 1 (b) 1 n n1 r2 2 (c)

n2 r22 . n1 r12

(d)

n2 n1

[JEE (Main) – 11th Jan. 2019 - Shift-1] Q.19. A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil : (a) decreases by a factor of 9 (b) increases by a factor of 27 (c) increases by a factor of 3 (d) decreases by a factor of 9 3 [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.20. A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, of 0.3 × 10–4 Wb/m2. The value of the induced emf in wire is : (a) 1.5 × 10–3 V (b) 1.1 × 10–3 V –3 (c) 2.5 × 10 V (d) 0.3 × 10–3 V [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY 1. (c) 5. (d) 9. (a) 13. (a) 17. (d)

2. (a) 6. (c) 10. (a) 14. (b) 18. (d)

3. (d) 7. (d) 11. (c) 15. (d) 19. (b)

4. (d) 8. (d) 12. (b) 16. (b) 20. (a)

291

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : resistance of an elliptical loop lying in an xy plane is R, semi major axis of the loop is a and semi minor axis of the loop is b, angular frequency of the loop about the x axis is w, external magnetic field in the region is Bk . To find : Pav, average power loss in the loop due to Joule heating. Flux linked to the elliptical loop : φ = (p ab ) B cos ωt E.m.f induced in the loop : dφ ∈ = = p ω abBsin ωt dt Power loss in the loop due to joule heating : P =

ε 2 p 2ω 2 a 2 b 2 B2 sin 2 ωt = R R

Average power loss in the loop due to joule heating  : Pav =

To find : I, the induced current in the square loop. Flux linked to the square loop : l φ = B  4

2

E.m.f. induced in the square loop : dφ l 2 dB =  dt 16 dt

Let R be the resistance and A be the area of cross section of wire. ρl ρl 4 ρl R = = = A pd2 pd2 4 Induced current through the loop :

ε pd 2l 2 dB I =  = R 4 ρ l × 16 dt I =

v

I

R

p × ( 4 × 10 −3 )2 × ( 30 × 10 −2 )2 64 × 1.23 × 10 −8 × 30 × 10 −2 × 0.032 = 0.61A

Ans. 3. Option (d) is correct. Given : Current through an infinitely long wire is I, the wire lies in the same plane as a rectangular loop with of length l, I || l, the resistance of the loop is R,

l

r

To find : i, the induced current in the loop as function of separation r. Magnetic field at r due to a current carrying wire : m I B( r ) = o 2p r Current induced in the loop =

p 2ω 2 a 2 b 2 B 2 2R

Ans. 2. Option (a) is correct. Given : A wire of length l = 30 cm and diameter d = 4 mm is turned to make a square loop, resistivity of the wire is r = 1.23 × 10–8 W-m, direction of magnetic field is perpendicular to the plane of the square, the rate of change of magnetic field in the region is dB = 0.032 T s−1 dt

ε =

at some moment the distance between the wire and far edge of the loop is r, the width of the rectangular loop is increasing at the rate of v.

i =

induced e.m.f. Resistance Bvl m o I vl = × R 2p r R

Ans. 4. Option (d) is correct. Given : Radius of a long solenoid is R, current through the solenoid is given as I(t ) = I ot(1 − t ), ...(i) radius of the ring placed coaxially near the middle of the solenoid is 2R. To find : The profile of inducted current IR and induced emf VR in the ring for the time interval 0 £ t £ 1. Flux through the ring :

φ = B.A = m onI(t ) × p ( 2 R )2 Emf induced in the ring due to changing current through the solenoid : dφ d VR = − = − [ m onI(t ) × p ( 2 R )2 ] dt dt d = −4pm onR 2 [ I(t )] dt d [ I0t(1 − t )] dt = −4pm onR 2 I0 (1 − 2t )

VR = −4pm onR 2

From equation (ii) : 1 dI(t ) At = t = s,VR 0 , =0 2 dt

...(ii)

...(iii)

So, the current through the solenoid has maximum 1 value at t = s. Therefore, the induced current 2 through the ring IR will reverse its direction at 1 t = s. 2 Ans. 5. Option (d) is correct. Given : The period of rotation of a planar loop of wire in a uniform magnetic field is T = 10 s.

292 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

To find : t¢, the time taken for the current to rise from half of the peak value to the peak value. Current through the AC circuit at time t : V 220 I(t ) = = sin (100p t ) R 50 I(t ) =

To find : The time when the induced emf through the loop will be maximum tmax and minimum tmin. Frequency of rotation of the loop : 2p 2p p ω = = = T 10 5 The induced emf will be maximum for p ωt = 2 p p t = 5 2 tmax=

5 = 2.5 s 2

The induced emf will be minimum for wt = p p t = p 5

To find : V, the emf induced in the loop. Emf induced in the loop due to changing magnetic field : ∆φ ∆B.A ( B1 − B2 )A V = = = ∆t ∆t ∆t (1000 − 500 ) × 10 −4    1   −4   × × − × × × × 16 4 4 2 2 10    2    V =  5 V =

500 × 56 × 10 5

−8

= 56 × 10 −6 V

V = 56µ V Ans. 7. Option (d) is correct. Given : Load resistance is R = 50 W, applied voltage is V(t ) = 220 sin (100p t ). ...(i)

...(ii)

Time period for one complete cycle of current : 2p 2p 1 T = = = s ...(iii) ω 100p 50 Time period to reach the peak value of current : 1 T tpeak = = s ...(iv) 4 200 Let the peak value of current through the circuit be Ipeak. So, half the peak value of current will be :

I peak

. 2 Let time taken by the current to reach half its peak value be t1. So from equation (i), I peak = I peak sin (100p t1 ) 2 100p t1 =

tmin = 5 s Ans. 6. Option (c) is correct. Given : Magnetic field passing perpendicularly through a loop at t = 0 is B1 = 1000 gauss, the field reduces to B2 = 500 gauss in time Dt = 5 s, the loop is as shown below,

22 sin (100p t ) 5

t1 =

p 6 1 600

...(v)

From equation (iv) and (v), time taken for the current to rise from half of the peak value to the peak value : 1 1 1 t¢ = = = 3.3 ms 200 600 300 Ans. 8. Option (d) is correct. Given : Number of turns in a solenoid is a constant = N, cross sectional area of the solenoid is a constant = A, length L of the solenoid is a variable. To find : The induction of the solenoid. Self-induction of a solenoid : Lsolenoid =

m0 N 2 A L

...(i)

As N and A are constant we can see from equation (i), 1 Lsolenoid ∝ . L Ans. 9. Option (a) is correct. Given : Radius of solenoid is R, current through the solenoid is I(t ) = kte -α t ,

...( i )

k > 0, radius of circular conducting coil placed in the equatorial plane of the solenoid and concentric with the solenoid is 2R. To find : The current induced in the outer coil as function of time.

293

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

Magnetic field of a solenoid : B = m0 nI Magnetic flux associated with the outer coil : φ = m0 nI ¥ p ( 2 R )2 = 4pm0 nR 2 I ...(ii) Induced emf in the outer coil : - dφ d ε = = - ( 4pm0 nR 2 I) dt dt d ε = -4pm0 nR 2 ( kte -α t ) dt

To find : Vin, input voltage and ip, current in the primary coil. Output power : Pout = is Vout ; Vout = Np

Also :

Ns

= - 4pm0 nR 2 ( ke -α t - α kte -α t )

ε = -4pm0 nkR 2 (1 - α t )e -α t

Vin =

...(iii)

Induced current in the outer coil : ε i = Resistance 2

-4pm0 nkR (1 - α t )e r From equation (iv) : =

-α t

i ∝ (α t - 1) e -α t ,

...(iv)

...(v)

t>

N PφPQ iQ

=

NQφQP iP

For coils P and Q, number of turns : N P = NQ = 1 φPQ =

φQP iP

¥ iQ =

10 -3 ¥ 2 3

B

. . . . .

. . . . .

Ans. 11. Option (c) is correct. Given : Number of turns in primary coil of a transformer is Np = 300, number of turns in secondary coil of a transformer is Ns = 150, output power of transformer is Pout = 2.2 kW, current through secondary coil is is = 10 A.

Ns

¥ Vout =

300 ¥ 220 = 440V 150

is Vout 10 ¥ 220 = = 5A Vin 440

. . . . .

. . . . .

v=1 cm/sec L

B 1 A 1

3 D

2 C 2

5 cm Given network of resistors is a balanced Wheatstone bridge. Equivalent resistance of the Wheatstone bridge : 1 ˆ Ê 1 RW = Á + Ë 1 + 1 2 + 2 ¯˜

-1

4 = W 3

Equivalent resistance of the entire circuit : 4 R eq = R W + R = + 1.7 ª 3 W 3 Induced emf in the square loop : ε = BLv So, induced current in the square loop will be : ε BLv i = = R eq R eq Put given values in the above equation.

-4

= 6.67 ¥ 10 Wb 

Np

Ans. 12. Option (b) is correct. Given : Side of a square loop shown in the figure below is L = 5 cm, the loop is connected to a network of resistors and is moving out of the region of magnetic field B = 1 T, velocity of whole set up is v = 1 cm/s, resistance of square loop is R = 1.7 W To find : i, the current through the square loop when it is partially into the region of magnetic field.

1 ,i = positive α

The above qualities of i are well represented in graph a. Ans. 10. Option (a) is correct. Given : Two coils P and Q are separated by a distance, Case a : Current through P is iP = 3 A, magnetic flux linked to coil Q is fQP = 10–3 Wb, current through Q is 0, Case b : Current through Q is iQ = 2 A current through P is 0, To find : fPQ, flux through P for case b. Coefficient of mutual induction for coils P and Q : M PQ = MQP

Vin ; Vout

For a transformer : i p Vin = is Vout ip =

as rest of the quantities are constant. From equation (v) : at t = 0, i =negative 1 At t = , i = 0 α At

=

Pout 2.2 ¥ 10 3 = = 220 V is 10

i =

1 ¥ 5 ¥ 10 -2 ¥ 10 -2 ª 170 m A 3

Ans. 13. Option (a) is correct. Given : Area of a conducting circular loop is A = 3.5 × 10–3 m2, resistance of the loop is R = 10 W, the loop is placed in a perpendicular time dependent magnetic field.

294 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) B(t ) = 0.4 T sin ( 50p t ).

...(i)

To find : dq, the net charge flowing through the loop during t = 0 to t = 10 ms. 1 dφ Induced charge : dq = idt, where i = is the R dt induced current. A 1 dφ 1 dq = dt = - dφ = ( Bt =10 ms - Bt = 0 ) R dt R R

dq =

3.5 ¥ 10 -3 ¥ 0.4 [sin ( 0.5p ) - sin 0] 10

= 1.4 ¥ 10 -4 C = 0.14 mC Ans. 14. Option (b) is correct. Given : For a step down transformer Vin = 2300 V, number of turns in primary windings is Np = 4000, Vout = 230 V, ip =5 A, efficiency of transformer is h = 90%. To find : is, the output current for the transformer. Input power to the step down transformer : Pin = Vin i p ...( i ) Output power from the step down transformer : Pout = Vout is ...(ii) From equations (i) and (ii) : P η = out ¥ 100 Pin

η =

Voutis Vini p

...( iii )

Put given values in the above equation. 230 ¥ is 90 = 100 2300 ¥ 5 is = 45A Ans. 15. Option (d) is correct. Given : Total magnetic moment of a magnet is  i A m2, the magnet is placed in a magnetic m = 10–2 ^  field B(t) = B0 ^ i cos(wt), B0 = 1 T, w = 0.125 rad/s. To find : The work done in reversing the direction of magnetic moment at t = 1 s. At t = 1 s : 180 ˆ Ê B(t ) = 1 ¥ cos Á 0.125 ¥ ¥ 1˜ Ë ¯ p = cos(7∞) = 0.9925 Work done in reversing the direction of magnetic moment : W = 2mB (t = 1s) = 2 ¥ 10 -2 ¥ 0.9925 = 0.0198 J 0.014 J is the nearest value. Ans. 16. Option (b) is correct. Given : The self-induced emf of a coil is e = 25 V, current through the coil changes at a uniform rate from I1 = 10 A to I2 = 25 A in time dt = 1 s. To find : DE, the corresponding change in the energy of the inductance.

PHYSICS

Induced emf in the coil :

ε = L L =

dI L ( I 2 - I1 ) = dt dt

ε dt 25 ¥ 1 5 = = H  I 2 - I1 25 - 10 3

Energy stored in an inductor : 1 E = LI 2 2 Change in energy stored in the inductor when current changes from I1 = 10 A to I2 = 25 A : 1 DE = L ( I 22 - I12 ) 2 =

1 5 ¥ ¥ [( 25)2 - (10 )2 ] = 437.5 J 2 3

Ans. 17. Option (d) is correct. Given : Horizontal component of earth’s magnetic field at some location is BH = 18 × 10–6 T, a magnetic needle is suspended from its mid-point using a thread at this location, length of the needle is l = 0.12 m, pole strength of the needle is m = 1.8 A m, when suspended the needle makes an angle of q = 45° with BH. To find : The vertical force that should be applied at one of the ends of the needle to keep the needle horizontal. As the needle makes an angle of q = 45° with BH at this location, the angle of dip, qdip = 45°. The vertical component of earth’s magnetic field : BV = BH tan θ dip = BH

...(i)

Let an external force F, be applied to the needle to keep it in horizontal position. mBv

mBH

mBH

• F mBv

F = 2mB V = 2 ¥ 1.8 ¥ 18 ¥ 10 -6 = 6.5 ¥ 10 -5 N Ans. 18. Option (d) is correct. Given : Length of solenoid-1 = length of solenoid-2 = l, radius of solenoid-1 = r1, radius of solenoid-2 = r2, r1 < r2, number of turns per unit length of solenoid-1 = n1, number of turns per unit length of solenoid-2 = n2. To find : M/L, ratio of mutual inductance to the selfinductance of solenoid-1. Mutual inductance for solenoid-1 : M = m0 n1n2p r12l

...(i)

Self-inductance for solenoid-1 : L = m0 n12p r12l

...(ii)

295

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

From equations (i) and (ii) : n M = 2 n1 L

Self-inductance of the triangular coil, when dimension of each arm of triangle is 3a :

Ans. 19. Option (b) is correct. Given : Shape of a coil is in the form of an equilateral triangle. To find : The change in the coil’s self-inductance if dimension of each arm of the triangle is increased by a factor of 3, keeping the number of turns per unit length of the coil the same. Let length of each arm of equilateral triangle be a. That gives total length of coil, l = 3 × a. Area of equilateral triangle : A =

3 2 a 4

Self-inductance of the triangular coil dimension of each arm of triangle is a :

when

Ê 3 2ˆ L1 = m0 n 2 Al = m0 n 2 Á a ˜ (3 ¥ a) Ë 4 ¯ 3 3 = m0n2 a3 4

Ê 3 ˆ L2 = m0 n 2 Al = m0 n 2 Á ( 3 a )2 ˜ ( 3 ¥ 3 a ) Ë 4 ¯ =

3 3 m0 n 2 a3 ¥ 27 4

...(ii)

From equations (i) and (ii), we can say that the self-inductance of the triangular coil increases by a factor of 27 when dimension of each arm of triangle is increased by a factor of 3. Ans. 20. Option (a) is correct. Given : Length of horizontal wire is l = 10 m, the speed of wire towards earth is v = 5.0 m/s, horizontal component of earth’s magnetic field is BH = 0.3 ¥ 10 -4 Wb m -2 , angle between v and BH is q = 90°. To find : e, the value of induced emf in the wire. Induced emf in the wire :

ε = BHlv = 0.3 ¥ 10 -4 ¥ 10 ¥ 5 ...(i)

= 1.5 ¥ 10 -3 V

Topic-2

RC, LR, LC and LCR Circuits Concept Revision (Video Based) RC Circuit      

          

 

   

Part -1   Part - 2

     

Part -1    Part - 2

  

LR Circuit

Part -1      Part - 2

LC Circuit    

 

LCR Circuit     

  

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. An inductance coil has a reactance of 100W. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is : (a) 5.5 × 10–5 H (b) 1.1 × 10–2 H (c) 6.7 × 10–7 H (d) 1.1 × 10–1 H [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. A 750 Hz, 20 V (rms) source is connected to a resistance of 100 W, an inductance of 0.1803 H and a capacitance of 10 mF all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to : (a) 245 s (b) 365 s (c) 418 s (d) 348 s [JEE (Main) – 3rd Sep. 2020 - Shift-1]

Q.3. A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor  1 reaches   times of its maximum value, is :  n (a)

L  n  ln    R  n − 1 

(b)

L  n  ln    R  n + 1 

(d)

L  n −1 ln   R  n 

L  n +1 ln   R  n − 1  [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.4. An AC circuit has R = 100W, C = 2mF and L = 80 mH, connected in series. The quality factor of the circuit is : (c)

296 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a) 400

(b) 0.5

(c) 2

(d) 20

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.5. For the given input voltage waveform Vin(t), the output voltage waveform V0(t), across the capacitor is correctly depicted by :

(a)

Q.8.



As shown in the figure, a battery of emf e is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit) is : εL εL  1 (a) 2  1 −  (b) R 2 e R  (c)

εR 2

eL

(d)



εL eR 2

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.9. In LC circuit the inductance L = 40 mH and capacitance C = 100 mF. If a voltage V(t) = 10sin (314 t) is applied to the circuit, the current in the circuit is given as : (a) 10 cos 314 t (b) 0.52 cos 314 t (c) 5.2 cos 314 t (d) 0.52 sin 314 t [JEE (Main) – 9th Jan. 2020 - Shift-2] Q.10. A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is :

(b)

(c)

i 10 

(d)

E— –

(a) [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.6. A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical springmass damped oscillator having damping constant ‘b’, the correct equivalence would be : (a) L ´ k, C ´ b, R ´ m (b) L ´ m, C ´ k, R ´ b (c) L ´ m, C ´ (d) L ´

PHYSICS

1 ,R´b k

1 1 1 ,C´ , R´ m b k

[JEE (Main) – 7th Jan. 2020 - Shift-1] Q.7. An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 W resistor. The ratio of the currents at time t = ∞ and at t = 40 s is close to : [Take e2 = 7.389] (a) 1.06

(b) 1.15

(c) 0.84

(d) 1.46 [JEE (Main) – 7th Jan. 2020 - Shift-2]

2 ln 2

(c) 2 ln 2

20 H

(b)

1 ln 2 2

(d) ln 2 [JEE (Main) – 8th April 2019 - Shift-1] Q.11. A circuit connected to an ac source of emf e = e0 sin (100 t) with t in seconds, gives a phase difference p of between the emf e and current i. Which of 4 the following circuits will exhibit this? (a) RL circuit with R = 1 kW and L = 10 mH (b) RL circuit with R = 1 kW and L = 1 mH (c) RC circuit with R = 1 kW and C = 1 mF (d) RC circuit with R= 1 kW and C = 10 mF [JEE (Main) – 8th April 2019 - Shift-2] Q.12. A coil of self inductance 10 mH and resistance 0.1 W is connected through a switch to a battery of internal resistance 0.9 W. After the switch is closed, the time taken for the current to attain 80% of the saturation value is : (take ln 5 = 1.6) (a) 0.324 s (b) 0.103 s (c) 0.002 s (d) 0.016 s [JEE (Main) – 10th April 2019 - Shift-2] Q.13. Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of

297

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

charge that passes through the battery between t = 0 and t = L/R is : L i

R

S

(a) (c)

R2 7.3 EL R2

— –

R

(b)



(d)



2.7 R 2

(a) 5.5 A (c) 3 A

EL 7.3 R 2 Q.17.



S2

— –

S1



the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time(t0), the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time ‘t’ is given by :

(b) 7.5 A (d) 6 A [JEE (Main) – 12th Jan. 2019 - Shift-1] I2

— —

L

R

R

EL

[JEE (Main) – 12th April 2019 - Shift-2] Q.14. A series AC circuit containing an inductor (20 mH), a capacitor (120 mF) and a resistor (60 W) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is : (a) 5.65 × 102 J (b) 2.26 × 103 J 2 (c) 5.17 × 10 ] (d) 3.39 × 103 J [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.15. In the circuit shown,



L

S

E

2.7 EL

15 V is connected in the circuit. What will be the current through the battery long after the switch is closed ?



C

R2

L

R1

In the above circuit, C =

I1

3 mF, R2 = 20W, 2

3 H and R1 = 10 W. Current in L-R1 path 10 is I1 and in C-R2 path it is I2. The voltage of A.C. source is given by, V = 200 2 sin (100 t) volts. The phase difference between I1 and I2 is : (a) 60° (b) 30° (c) 90° (d) 0° [JEE (Main) – 12th Jan. 2019 - Shift-2] L=

ANSWER – KEY

I

(a) t0

t

I



1. (b) 5. (d) 9. (b) 13. (b) 17. (b)

2. (d) 6. (c) 10. (c) 14. (c)

3. (a) 7. (a) 11. (d) 15. (c)

4. (c) 8. (d) 12. (d) 16. (d)

ANSWERS WITH EXPLANATIONS

(b) t0

t

I

Ans. 1. Option (b) is correct. Given : Reactance of an inductance coil is X 2L + R 2 = 100 Ω , frequency of the AC signal

(c) t0

t

I

(d) t0 t [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.16. In the figure shown, a circuit contains two identical resistors with resistance R = 5 W and an inductance with L = 2 mH. An ideal battery of

applied to the coil is f = 1000 Hz, the applied voltage leads the current in the coil by q = 45°. To find : L, the self inductance of the coil. As the voltage leads the current by q = 45° : X tan θ = L = 1;X L = R R Reactance of the inductance coil :  X 2L + R 2 = 100 X L = R = 50 2 Ω X L = Lω = 50 2

298 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) L =

50 2 2p f

L =

50 2 = 1.1 × 10 −2 H 2p × 1000

t =

Ans. 2. Option (d) is correct. Given : In a series LCR circuit, L = 0.1803 H, C = 10 mF, R = 100 W, the rms voltage supplied by the source is V = 20 V at frequency f = 750 Hz, heat capacity of the resistor is C = 2 J/°C. To find : t, time required to raise the temperature of the resistor by DT = 10°C. Total impedance of the circuit :  1  R 2 +  2p fL −  2 p fC  

Z =

PHYSICS

L  n  ln    R  n − 1 

Ans. 4. Option (c) is correct. Given : In an AC series LCR circuit, L = 80 mH, C = 2 mF, R = 100 W. To find : Q, quality factor of the circuit. Q =

1 L 1 80 × 10 −3 = =2 R C 100 2 × 10 −6

Ans. 5. Option (d) is correct. Given : R = 1 kW, C = 10 nF, waveform Vin(t) as shown,

input voltage

2

2

  2p × 750 × 0.1803  2  (100 ) +  1  −   2p × 750 × 10 × 10 −6   = 834 Ω

Z =

Power loss in AC circuit : P = Vrms Irms cosφ

To find : Vo(t) across the capacitor. Time constant of the circuit : RC = 10 −5 s Voltage across the capacitor at the end of the first charging pulse (t = 5 ms) : V1 = Vo (1 − e − t / RC ) = 5 × (1 − e −5×10

V R P = Vrms × rms × Z Z = P

2 Vrms R = 0.0575 W 2 Z

Time taken to heat the resistor : Pt = C∆T t =

C∆T 2 × 10 = = 348 s P 0.0575

Ans. 3. Option (a) is correct. Given : In a series LR circuit the emf of the source is V, the circuit is switched on at t = 0. To find : t, time in which the energy stored in the 1 inductor reaches of its maximum value. n At time t, let the energy stored in the inductor be U and current through the circuit be I. 1 U = LI 2 ...(i) 2 Let the maximum energy stored in the inductor be Uo and current through the circuit then be Io. 1 U o = LI 2o ...(ii) 2 From equations (i) and (ii) : U = Uo Also,

I2 1 = I 2o n

I = 1 − e − Rt / L Io 1 n

= 1 − e −Rt / L

−6

/ 10 −5

)

V1 = 5 × (1 − e −0.5 ) ≈ 2 V The capacitor will discharge when no pulse is applied for the next 5 ms. Voltage across the capacitor at t¢ = 10 ms. V2 = V1e −t / RC = 1.2 V Voltage across the capacitor at the end of the second charging pulse (t≤ = 15 ms) applied for = 5 ms : V3 = V2 + Vo (1 − e −t / RC ) ≈ 3 V Ans. 6. Option (c) is correct. Given : An LCR circuit behaves like a damped harmonic oscillator. To find : Compare the LCR circuit with the physical spring mass damped oscillator with damping constant b. For physical spring-mass damped oscillator : ma = − kx − bv m

d 2x dt 2

+b

dx + kx = 0 dt

...(i)

For a damped LCR circuit : dI q − IR − L − = 0 dt C L

d 2q dt 2

+R

dq 1 + q = 0 dt C

...( ii )

On comparing equations (i) and (ii), we get : 1 L ↔ m,R ↔ b ,C ↔ k Ans. 7. Option (a) is correct. Given : A circuit consists of a series combination of

299

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

L = 10 mH inductor and R = 5 W resistor, an emf of V = 20 V is applied to the circuit for all times t ≥ 0, e2 = 7.389. To find :

I( • ) , the ratio of currents at time t = ∞ I( 40 )

Z =

and t = 40 s through the circuit. Current through the LR circuit at some time t :

Z =

I(t ) = I o [1 − e( − R / L )t ] I(t ) =

...(i)

I( ∞ ) = 4

...(ii)

)

    1 1 I( 40 ) = 4  1 − 2 10000  = 4  1 − 10000      (e ) (7.389 )     I( 40 ) ª 4

V I(t ) = = Z

p . 2

p  10 sin  314t +  2  = 0.52 cos( 314t ) 19.24

Ans. 10. Option (c) is correct. Given : For an LR circuit shown below, L = 20 H, R = 10 W. To find : t¢, the time at which rate of dissipation of energy across resistance is equal to the rate at which magnetic energy is stored in the inductor.

Put t = 40 in equation (i) : I ( 40 ) = 4(1 − e

1 1 − ωL = − 314 × 40 × 10 −3 Cω 100 × 10 −6 × 314

As XC > XL, the current leads the voltage by

Put t = ∞ in equation (i) :

−500×40

( R )2 + ( X C − X L )2 = X c − X L

Z = 19.24 Ω

V 20 [1 − e( − R / L )t ] = [1 − e( −5 / 0.01)t ] R 5

I(t ) = 4(1 − e −500t )

Ans. 9. Option (b) is correct. Given : For an LC circuit L = 40 mH, C = 100 mF, V(t) = 10 sin(314t). To find : I(t), the current through the circuit. Impedance of the LC circuit :

...(iii)

i 10 

So, from equations (ii) and (iii) :

E— –

I(∞) ≈ 1 I ( 40 ) and slightly more than 1.

20 H

Current through the LR circuit : E i = 1 - e - Rt / L R

(

Ans. 8. Option (d) is correct. Given : An LR circuit as shown in the figure below, the switch is closed at t = 0.

)

...(i)

Energy stored in the inductor : 1 2 UL = Li 2 Rate at which energy is stored in the inductor : dU L 1 di di = L ( 2i ) = Li ...(ii) dt 2 dt dt

To find : The total charge that flows from the battery between t = 0 and t = tc, tc is the time constant of the circuit.

Rate at which energy is dissipated by the resistor : dU R = i 2R ...(iii) dt

Current through an LR circuit at some time t :

At t = t¢ :

ε I ( t ) = (1 − e − t / t c ) R

dU L dU R = dt dt

...(i)

From equation (i), the charge flowing from the L battery between t = 0 and = will be : t t= c R tc

I(t )dt = ∫

tc

ε (1 − e −t / tc )dt R

q =

∫0

q =

ε tc ε (1 − e −t / tc ) dt = [t + tc e −t / tc ]t0c R ∫0 R

q =

t ε  tc − 0 + c − tc   e R 

ε tc ε L q = = R e eR 2

0

Li

di = i 2R dt di R = i dt L

...(iv)

Put value of i from equation (i). d ÈE R E ˘ 1 - e - Rt / L ˙ = ¥ 1 - e - Rt / L dt ÍÎ R L R ˚

(

)

(

Put t = t¢.

(

R - Rt ¢ / L R e = 1 - e - Rt ¢ / L L L

(

e - Rt ¢ / L = 1 - e - Rt ¢ / L

)

)

)

300 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) e - Rt ¢ / L =

1 2

R = 1kW = 10 3 W XC = 1 R

Rt¢ = ln 2 L

So, option (d) is the correct circuit.

L t¢ = ln 2 R t¢ =

20 ln 2 = 2 ln 2 10

Ans. 11. Option (d) is correct. Given : An AC source of emf e = e0 sin (100t )

...(i)

is connected to a circuit, the circuit gives a phase p difference of φ = between e and current i. 4 To find : Which among the given circuits will exhibit the above property. For RL circuit : X tan φ = L R X Êpˆ tan Á ˜ = 1 = L Ë 4¯ R

...(ii)

XC R

X Êpˆ tan Á ˜ = 1 = C Ë 4¯ R

Ans. 12. Option (d) is correct. Given : For an LR circuit with dc source, L = 10 mH, R = 0.1 W, internal resistance of the battery is r = 0.9 W. To find : t¢, time taken by the current to attain 80% of its saturation value. Equivalent resistance of the circuit : R eq = R + r = 1W Current through the LR circuit :

(

i = i0 1 - e

Either equation (ii) or equation (iii) should be satisfied by the required circuit. From equation (i) : ω = 100 rad / s

)

So, equation (i) becomes : 4 - R t¢ / L i0 = i0 1 - e eq 5

(

- R eq t ¢ / L

e ...(iii)

- R eq t / L

...(i)

In equation (i), i0 is the peak current. At t =t¢, 80 4 i = i0 = i0 100 5

(1 - e

For RC circuit : tan φ =

PHYSICS

)

- R eq t ¢ / L

R eq t¢ L

=

4 5

=

1 5

)

= ln 5

t¢ =

10 ¥ 10 -3 L ln 5 = ¥ 1.6 1 R eq

= 0.016 s X L = Lω = 10 ¥ 10 -3 ¥ 100 = 1W R = 1kW XL π 1 R

Option b :

Ans. 13. Option (b) is correct. Given : An LR circuit as shown in the figure below. L i

XL π 1 R Option c : XC

1 1 = = = 10 4 W Cω 1 ¥ 10 -6 ¥ 1 ¥ 10 2

R = 1kW XC π 1 R Option d : XC =

1 1 = = 10 3 W 6 Cω 10 ¥ 10 ¥ 1 ¥ 10 2

S

E

X L = Lω = 1 ¥ 10 -3 ¥ 100 = 0.1W R = 1kW

R

— –

Option a :

To find : q, the charge that passes through the battery between t = 0 and t = L/R. Current through the LR circuit : E i = 1 - e - Rt / L ...(i) R

(

)

Charge passing through the battery in time dt : E dq = idt = 1 - e - Rt / L dt ...(ii) R

)

(

Integrate equation (ii) to get the answer. L/R E q = Údq = Ú 1 - e - Rt / L dt 0 R

(

q =

)

L / R - Rt / L ˆ E Ê L/R dt ˜ ÁË Ú0 dt - Ú0 e ¯ R

301

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

=

EL R

2

+

So, equivalent resistance of the circuit will be : 5¥5 5 R eq = = W 5+5 2

EL Ê 1 ˆ EL  Á - 1˜ = R 2 Ë e ¯ 2.7 R 2

Ans. 14. Option (c) is correct. Given : A series LCR circuit has components L = 20 mH, C = 120 mF, R = 60 W, the rms value of ac voltage supply is Vrms =24 V, frequency of the voltage supply is f = 50 Hz. To find : E, the energy dissipated in the circuit in t = 60 s. Impedance of the LCR circuit : Z =

1 ˆ Ê R 2 + ( X L - X C ) 2 = R 2 + Á Lω Ë Cω ˜¯

And current through the circuit will be : E 15 i = = = 6A 5 R eq 2 Ans. 17. Option (b) is correct.

W, L =

(60 )

2

= 63.3 W Rms value of current in the LCR circuit : V 24 Irms = rms = = 0.379 A Z 63.3 Energy dissipated in t = 60 s : 2 E = Irms Rt = ( 0.379 )2 ¥ 60 ¥ 60 = 5.17 ¥ 10 2 J

Ans. 15. Option (c) is correct. Given : An RL circuit with two switches and a battery, switch S1 is closed at t = 0, from t = 0 and t = t0 S1 is closed but S2 is open, at t > t0 S2 is closed but S1 is open. To find : The behaviour of current I as a function of time t. From t = 0 and t = t0 the inductor is charging and current through it will be : V I1 = 1 - e - Rt / L ...(i) R

)

(

After t > t0, the inductor is discharging and current through it will be : V - R / L(t - t0 ) I2 = e ...(ii) R From equations (i) and (ii), option c seem to depict the correct behaviour of current I as a function of time t. Ans. 16. Option (d) is correct. Given : For the circuit shown below, R = 5 W, L = 2 mH, E = 15 V. To find : i, current through the battery a long time after the switch was closed. S E

L R R

After sufficiently long time the resistance offered by the inductor will be zero.

I2

— —

Z =

1 ˆ Ê + Á 20 ¥ 10 -3 ¥ 2p 50 ˜ -6 Ë 120 ¥ 10 ¥ 2p 50 ¯

mF, R2 = 20

3 H, R1 = 10 W, V = 200 2 sin (100t). 10 To find : Df, the phase difference between I1 and I2.

2

Put the given values : 2

3 2

Given : For the circuit below, C =

C

R2

L

R1 V

I1

Phase difference between I2 and V in the C – R2 arm of the circuit : X 1 1 tan φ1 = C = ¥ R 2 Cω R 2 =

1 3 ¥ 10 -6 ¥ 100 2

φ1 ª 90∞

¥

1 10 3 = ; 20 3 ...(i)

Phase difference between I1 and V in the L – R1 arm of the circuit : tan φ2 =

X L Lω 3 1 = = ¥ 100 ¥ = 3; R1 R1 10 10

φ2 ª 60∞

...(ii)

From equations (i) and (ii) the phase difference between I1 and I2 will be : Dφ = φ1 - φ2 = 30∞

Subjective Questions (Chapter Based) Q.1. A circular coil of radius 10 cm is placed in a uniform magnetic field of 3.0 × 10–5 T with its plane perpendicular to the field initially. It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in 0.2 s. The maximum value of EMF induced (in mV) in the coil will be close to the integer... [JEE (Main) – 2nd Sep. 2020 - Shift-1] Sol. Given : Radius of circular coil is R = 10 cm magnitude of uniform magnetic field in the region is B = 3.0 × 10–5 T, initial angle between the plane  p of the coil and B is θ = , the coil rotates along 2 one of its diameters perpendicular with respect to  B and completes half a rotation in t = 0.2 s.

302 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) To find : e0, maximum value of induced e.m.f. in the coil. Let A be the area of the coil and w be its angular speed. Flux through the coil : φ = BA cos ωt E.m.f. induced in the coil : dφ ε = = BAω sin ωt dt Maximum e.m.f induced in the coil : ε o = BAω  p  ε o = B(p r 2 )   = 15 m V  0.2  Q.2. Two concentric circular coils, C1 and C2, are placed in the XY plane. C1 has 500 turns, and a radius of 1 cm. C2 has 200 turns and radius of 20 cm. C2 carries a time dependent current I(t) = (5t2 – 2t + 3) A where t is in s. The emf induced in C1 (in mV), at 4 the instant t 1s is . The value of x is ....... x [JEE (Main) – 5th Sep. 2020 - Shift-1] Sol. Given : C1 and C2 are two concentric coils in XY plane, number of turns in C1 is n1 = 500, radius of C1 is r1 = 1 cm, number of turns in C2 is n2 = 200, radius of C2 is r2 = 20 cm, current through C2 is I(t) = (5t2 – 2t + 3)A, induced e.m.f. in C1 at t = 1 is 4 ε = mV. x To find : The value of x. Magnetic field produced by current in C2 at location of C1 : m In B = o 2 2r2 Flux linked to C1 : m In φ = o 2 × n1p r12 2r2 E.m.f. induced in C1 : dφ n1n2 m op r12 dI n1n2 m op r12 ∈ = = = 2r2 2r2 dt dt × (10t − 2 ) E.m.f. induced in C1 at t = 1 s : 4 ε ≈ 0.8 V = V 5

Sol. Given : For the above circuit, L = 50 mH, R = 2 W, V = 30 V, I = 1 A, dI/dt =102 A/s. To find : VP – VQ. L dI VP − − 30 + RI = VQ dt VP − VQ = L

P

Q 30 V th [JEE (Main) – 6 Sep. 2020 - Shift-1]

dI + 30 − RI = 33V dt



Q.4 In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series  n  to the L and R. Taking the value of C as   mF,  3p  then value of n is ___________ [JEE (Main) – 6th Sep. 2020 - Shift-2] Sol. Given : Case 1 : Peak voltage supplied by a source in a series LR circuit is V = 250 V at f = 50 Hz, power dissipated from the source is P = 400 W, the power factor of the circuit is cos f = 0.8. Case 2 : For the same LR circuit as in case 1, a n m F, capacitor is added in series to L and R, C = 3p new power factor is cos f¢ = 1. To find : The value of n. For case 1 : Let I be the peak current in the circuit Power dissipated from the source : P = VI cosφ I =

P 400 = = 2 A V cosφ 250 × 0.8

Value of resistor R in the circuit : P 400 R = 2 = = 100Ω 4 I Power factor : cosφ =

0.8 =

R 2

R + X 2L 100 100 2 + X 2L

X L = 75Ω For case 2 : As cosφ ′ = 1,X L = XC XC =

x = 5 Q.3. A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A and it is decreasing at a rate of 102A s–1. The value of the potential difference Vp – VQ, (in volts) at that instant is : L = 50 mH R= 2W I

PHYSICS

C =

1 = 75 Ω Cω 1 n = m F 75 × 2p f 3p

n ª 400 Q.5. In a fluorescent lamp choke (a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _________ . [JEE (Main) – 9th Jan. 2020 - Shift-1]

303

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

Sol. Given : The choke current changes from I1 = 0.25 A to I2 = 0 A in time Dt = 0.025 ms, the reverse voltage produced in this process is V = 100 V. To find : L, the self-inductance of the choke. Using equation : dI V = L dt I −I 100 = L 1 2 ∆t L =

−3

R

2L I2

I1

— –

I S1



V

S2

Sol. Given : An arrangement of two inductors, 2 resistors, 2 switches and 2 batteries, resistance of both the resistors is R, emf of both the batteries is V, the value of inductance is as indicated in the circuit above, the switches S1 and S2 are closed at t = 0. To find : The time t in which the current though the middle wire reaches its maximum value and the value of Imax. Variation of current in the left loop : V I1 = (1 - e -tR / L ) ...( i ) R

Variation of current in the right loop : V I2 = (1 - e - tR / 2 L ) R

At t = t, I = Imax,



...(ii)

Variation of current in the middle wire : I = I1 - I 2 =

V - tR / 2 L - tR / L (e -e ) R dImax dt

= 0

d È V - tR / 2 L - tR / L ˘ (e -e )˙ = 0 dt ÍÎ R ˚ -

τ

R -τ R / 2 L R -τ R / L e + e = 0 2L L

Substituting t = τ =

= 10 −2 H

Q.6. In the figure below, the switches S1 and S2 are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude Imax at time t = t. Find value of Imax and t?

— V–

τR = ln 2 2L







= e -τ R / L

eτ R / 2 L = 2





L = 10mH

L





100 × 0.025 × 10 0.25

R

e- τ R / 2L 2

Imax =

=

2L ln 2  R

2L ln 2 in equation (iii) : R

V -τ R / 2 L -τ R / L V (e -e )= R 4R

Q.7. The instantaneous voltages at three terminals marked X, Y and Z are given by VX = V0 sin ω t , 2p ˆ Ê VY = V0 sin Á ω t + ˜ and Ë 3 ¯ 4p ˆ Ê VZ = V0 sin Á ω t + ˜ . Ë 3 ¯

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. What will be the reading(s) of the voltmeter? Sol. Given : Instantaneous voltage at terminal X is VX = V0 sin ωt, ...( i ) Instantaneous voltage at terminal Y is 2p ˆ Ê VY = V0 sin Á ωt + , ...(ii) Ë 3 ˜¯ Instantaneous voltage at terminal Z is 4p ˆ Ê VZ = V0 sin Á ωt + . ...(iii) Ë 3 ˜¯ To find : Reading of voltmeter when connected between terminals X and Y, Vrms|XY and Y and Z, Vrms|YZ. VXY = VY - VX 2p ˆ Ê = V0 sin Á ωt + - V0 sin ωt Ë 3 ˜¯

È Ê ˘ 2p ˆ = V0 Ísin Á ωt + - sin ωt ˙ ˜ 3 ¯ Î Ë ˚



2p ˆ Ê 2ωt + Á 3 ˜ sin Ê 2p ˆ = 2 V0 cos Á ÁË 3 ˜¯ ˜ 2 Á ˜ Ë ¯

VXY

...( iii )



=

pˆ Ê 3V0 cos Á ωt + ˜ Ë 3¯



4p ˆ 2p ˆ Ê Ê VYZ = V0 sin Á ωt + ˜ - V0 sin Á ωt + ˜ Ë Ë 3 ¯ 3¯



È Ê 4p ˆ 2p ˆ ˘ Ê = V0 Ísin Á ωt + ˜¯ - sin ÁË ωt + 3 ˜¯ ˙ Ë 3 Î ˚

304 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) VYZ

z

6p ˆ Ê 2ωt + Á 3 ˜ sin Ê 2p ˆ = 2 V0 cos Á ÁË 3 ¯˜ ˜ 2 Á ˜ Ë ¯

45°

a

3V0 cos(ωt + p )

=

3R

So, rms value as read by the voltmeter will be : Vpeak 3 Vrms XY = Vrms YZ = = V0 2 2

Q.8. A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed w on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, what will be the e.m.f. induced across the ends of the rod ?

R

Sol. Given : Length of metallic rod is l, length of string tied to the rod is 2l, angular speed of rod is w, direction of magnetic field B is perpendicular to the rod. To find : e, the emf induced across the ends of the rod. V l dx

2l x



Consider a thin length element dx as shown in the above diagram. That gives : dε = (ω x ) Bdx ...(i) Integrate the above equation :

ε =



=

3l

Ú2l (ωx ) Bdx

x

Sol. Given : Radius of circular loop shown in the figure above is R, edge length of the square loop is a, the square loop is centred at z = 3 R, number of turns in the square loop is N = 2, the plane of the square loop makes an angle of q = 45° with the z-axis, the mutual inductance between the two loops is

of p?

m0 a 2

loop at z = 3R ) ¥ ( Areaof

square

loop ) ¥ cosθ φ = 2¥



m0 a2

2p/2 R

, then what is the value

...(i)

2p / 2 R

To find : The value of p. Let the current through the circular loop be i. Then flux through the square loop will be : φ = N ¥ ( Magnetic fieldof

circular



φ =

m 0iR 2

2( R 2 + ( 3R )2 )3 / 2

¥ a 2 cos 45

m 0i a 2 m ia 2 = ( 70/ 2 ) 8R 2 2 R

Mutual inductance : M =

m a2 φ = 70/ 2 i 2 R

...(ii)

On comparing equations (i) and (ii) we get : p = 7.

Q.10. In the given circuit, the AC source has w = 100 rad/s. Considering the inductor and capacitor to be ideal, calculate the current I through the circuit and the voltage drop across the 50 W and the 100 W resistor. I1 100 F

Q.9. A circular wire loop of a radius R is placed in the x-y plane centered at the origin O. A square loop of side a (a  R) having two turns is placed with

loops is given by

M =

5l 2ω B ωB (9l 2 - 4 l 2 ) = 2 2

its centre at z = 3 R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45° with respect to the z-axis. If the mutual inductance between the

y O





PHYSICS

0.5H I



I2

~

100  50 

20 V

Sol. Given : An ac source (w = 100 rad/s) connected across a capacitor C = 100 mF, inductor L = 0.5 H and 2 resistors R1 = 100 W, R2 = 50 W as shown in the circuit above. To find : The current I and the voltage drop across the two resistors.

305

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

I1 = =

=

V V = Z1 Ê 1 ˆ 2 2 ÁË cω ˜¯ + R1



20 Ê ˆ 1 + 100 2 ˜ Á Ë (100 ¥ 10 -6 ¥ 100 )2 ¯

=

5 2

A



φ2 = 45∞



φ1 = 45∞

-1 Ê



As it is the capacitive arm of the circuit the current

I2 = =



V V = Z 2 ( Lω )2 + R 22 20 [( 0.5 ¥ 100 )2 + 50 2 ]



=

20 50 2

As it is the inductive arm of the circuit the voltage From equations (i), (ii), (iii) and (iv), I1 and I2 have mutual phase difference of 90° :

...(ii)

leads the voltage by 45°.

...(iv)

leads the current by 45°



...(iii)



...(i)

100 ˆ φ1 = cos Á ˜ = cos Á ; Ë 100 2 ˜¯ Ë Z1 ¯



100 2

2 A 5

ÊR ˆ Ê 50 ˆ φ2 = cos-1 Á 2 ˜ = cos-1 Á ; Ë 50 2 ˜¯ Ë Z2 ¯



1

-1 Ê R1 ˆ



20

=



I =

I12 + I 22 = 0.32 A

From equation (i), voltage drop across R1 : V100 = R1 ¥ I1 =

100 5 2

= 10 2 V

From equation (ii), voltage drop across R2 : V50 = R 2 ¥ I 2 =

50 ¥ 2 = 10 2 V 5



Hi

sto ry

result into

they

306 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Chapter 15

Electromagnetic Waves

Syllabus Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, Xrays, gamma rays). Applications of e.m. waves.

Concept Revision (Video Based) Electromagnetic Waves

  

        

     

Part -1     Part - 2

Characteristics of Electromagnetic Waves

  

     

            Part -1     Part - 2

Intensity of Electromagnetic Waves

Electromagnetic Specturm  

 

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. A plane electromagnetic wave, has frequency of 2.0 × 1010 Hz and its energy density is 1.02 × 10–8 J/m3 in vacuum. The amplitude of the magnetic Nm 2 1 = 9 × 10 9 field of the wave is close to ( C2 4pe 0 and speed of light = 3 × 108 m s–1) : (a) 160 nT (b) 180 nT (c) 190 nT (d) 150 nT [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. In plane electromagnetic wave, the directions of electric field and magnetic field are represented by k and 2i − 2 j , respectively. What is the unit vector along direction of propagation of the wave. 1  1   (i + 2 j ) ( j + k) (a) (b) 5 2 (c)

1 5

( 2i + j )

(d)

1   (i + j ) 2

[JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. The magnetic field of a plane electromagnetic  wave is B = 3×10 −8 sin[ 200p ( y + ct )]i T where c = 3 ×108 ms–1 is the speed of light. The corresponding electric field is :

 V (a) E = −9 sin[200p ( y + ct )] k m  V (b) E = 9 sin[200p ( y + ct )] k m

 V (c) E = −10 −6 sin[200p ( y + ct )] k m  V (d) E = 3 × 10 −8 sin[200p ( y + ct )] k m

[JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.4. The electric field of a plane electromagnetic wave propagating along the x-direction in vacuum is   E = E j cos(wt – kx). The magnetic field B, at the 0

moment t = 0 is :  E0 cos( kx ) k (a) B = m0 ε 0  (b) B = E0 m0 ε 0 cos( kx ) k  (c) B = E0 m0 ε 0 cos( kx ) j  (d) B =

E0 m0 ε 0

cos( kx ) j [JEE (Main) – 3rd Sep. 2020 - Shift-2]

308 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.5. Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum : (a) lradio waves > lmicrowaves > lvisible > lx-rays (b) lvisible < lmicro waves < lradio waves < lx-rays (c) lvisible > lx-rays > lradio waves > lmicrowaves (d) lx-rays < lmicrowaves < lradio waves < lvisible [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.6. The electric field of a plane electromagnetic wave   )sin( kz − ω t ). Its magnetic is given by E = E ( x + y 0

field will be given by : E0   ( x + y )sin( kz − ωt ) (a) c (b)

E0   ( x − y )sin( kz − ωt ) c

(c)

E0   ( x − y )cos( kz − ωt ) c

(d)

E0   ( − x + y )sin( kz − ωt ) c

[JEE (Main) – 4th Sep. 2020 - Shift-2l] Q.7. An electron is constrained to move along the y-axis with a speed of 0.1c (c is the speed of light) in the presence of electromagnetic wave, whose electric  V field is E = 30 j sin(1.5 × 107 t – 5 × 10–2x) . m The maximum magnetic force experience by the electron will be : (given c = 3 × 10­8 ms–1 and electron charge = 1.6 × 10–19C) (a) 2.4 × 10–18 N (b) 1.6 × 10–19 N –19 (c) 4.8 × 10 N (d) 3.2 × 10–18 N [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.8. The correct match between the entries in column I and column Il are : I II Radiation Wavelength (A) Microwave (i) 100 m (B) Gamma rays (ii) 10–15 m (C) A.M. radio (iii) 10–10 m (D) X-rays (iv) 10–3 m (a) (A) – (iv), (B) – (ii), (C) – (i), (D) – (iii) (b) (A) – (iii), (B) – (ii), (C) – (i), (D) – (iv) (c) (A) – (i), (B) – (iii), (C) – (iv), (D) – (ii) (d) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii) [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.9. For a plane electromagnetic wave, the magnetic field at a point x and time t is :  B( x , t ) = [1.2 × 10 −7 sin(0.5 × 10 3 x + 1.5 × 10 11 t )k ]T .  The instantaneous electric field E corresponding  to B is :  V (a) E.( x , t ) = [36 sin(1 × 10 3 x + 0.5 × 1011 t ) j] m  V 3 11 (b) E.( x , t ) = [36 sin( 0.5 × 10 x + 1.5 × 10 t )k ] m

PHYSICS

 V 3 11 (c) E.( x , t ) = [36 sin(1 × 10 x + 1.5 × 10 t )i ] m  V 3 11 (d) E.( x , t ) = [ −36 sin( 0.5 × 10 x + 1.5 × 10 t ) j] m JEE (Main) – 6th Sep. 2020 - Shift-2) Q.10. If the magnetic field in a plane electromagnetic wave is given by  B = 3 × 10 −8 sin(1.6 × 10 3 x + 48 × 10 10 t ) j T , then what will be expression for electric field?   V (a) E =  9 sin(1.6 × 10 3 x + 48 × 1010 t )k m    (b) E =  60 sin(1.6 × 10 3 x + 48 × 1010 t )k V  m    V (c) E =  3 × 10 −8 sin(1.6 × 10 3 x + 48 × 1010 t )i  m    V −8 3 10 (d) E =  3 × 10 sin(1.6 × 10 x + 48 × 10 t ) j  m   [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.11. The electric field of a plane electromagnetic wave is given by i + j  cos( kz + ω t ) E = E0 2 At t = 0, a positively charged particle is at the point p  ( x , y , z) =  0 , 0 ,  . If its instantaneous velocity 0   at (t = 0) is v k , the force acting on it due to the wave is :

(a) parallel to

0

i + j 2

(c) antiparallel to

(b) zero

i + j 2



(d) parallel to k

[JEE (Main) –7th Jan. 2020 - Shift-2] Q.12. A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time,  the magnetic field is given by B = 5 × 10 −8 j T. The  corresponding electric field E is (speed of light c 8 –1 = 3 × 10 ms ) V V (a) −1.66 × 10 −16 i (b) 1.66 × 10 −16 i m m V (c) −15i m

V (d) 15i m

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.13. The electric fields of two plane electromagnetic plane waves in vacuum are given by  E1 = E0 j cos(ω t − kx ) and  E 2 = E0 k cos(ω t − ky ) At t = 0, a particle of charge q is at origin with  a velocity v = 0.8 c j (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :

309

ELECTROMAGNETIC WAVES

(a) E0 q ( 0.8i − j + 0.4 k )

(b) E0 q ( 0.4i − 3 j + 0.8 k )

   (c) E0 q ( −0.8i + j + k ) (d) E0 q ( 0.8i + j + 0.2 k ) [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.14. A plane electromagnetic wave is propagating i + j along the direction , with its polarization 2 along the direction k . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant) : i + j   i − j cos  ωt − k (a) B0   2 2   j − i  i + j  cos  ωt + k  (b) B0  2 2    i + j i + j  cos  ωt − k  (c) B0  2 2    i + j   (d) B0 k cos  ωt − k 2   [JEE (Main) – 9th Jan. 2020 - Shift-2] Q.15. A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 Vm–1 along y-direction. Its corresponding magnetic field component, B would be : (a) 2 × 10–8 T along z-direction (b) 6 × 10–8 T along x-direction (c) 6 × 10–8 T along z-direction (d) 2 × 10–8 T along y-direction [JEE (Main) – 8th April 2019 - Shift-1] Q.16. The magnetic field of an electromagnetic wave is given by :  Wb B = 1.6 ¥ 10 -6 cos( 2 ¥ 107 z + 6 ¥ 10 15 t )( 2i + j ) 2 m The associated electric field will be:  (a) E = 4.8 ¥ 10 2 cos( 2 ¥ 107 z - 6 ¥ 1015 t )( 2i + j )V/ m  (b) E = 4.8 ¥ 10 2 cos( 2 ¥ 107 z - 6 ¥ 1015 t )( -2 j + k )V/ m  (c) E = 4.8 ¥ 10 2 cos( 2 ¥ 107 z + 6 ¥ 1015 t )( -i + 2 j )V/ m  (d) E = 4.8 ¥ 10 2 cos( 2 ¥ 107 z + 6 ¥ 1015 t )(i - 2 j )V/ m [JEE (Main) – 8th April 2019 - Shift-2] Q.17. The magnetic field of a plane electromagnetic wave is given by :  B = B0 i [cos( kz - w t )] + B1 j cos( kz + ωt ) where B0 = 3 × 10–5 T and B1 = 2 × 10–6 T. The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to : (a) 0.6 N (b) 0.1 N (c) 0.9 N (d) 3 × 10–2N [JEE (Main) – 9th April 2019 - Shift-1] Q.18. The electric field of a plane electromagnetic wave is given by  E = E0 i cos( kz ) cos(ωt )



 The corresponding magnetic field B is then given by :  E (a) B = 0 j sin( kz ) sin(ωt ) C  E (b) B = 0 j sin( kz ) cos(ωt ) C  E (c) B = 0 j cos( kz ) sin(ωt ) C  E (d) B = 0 k sin( kz ) cos(ωt ) C

[JEE (Main) – 10th April 2019 - Shift-1] Q.19. An electromagnetic wave is represented by the   sin[ω t + (6 y - 8 z )]. Taking electric field E = E n 0

unit vectors in x, y and z directions to be i , j , k , the direction of propogation s , is :

3i - 4 j (a) s = 5

   -4 k + 3 j (b) s = 5

   Ê -3 j + 4 k ˆ (c) s = Á ˜ 5 Ë ¯

Ê 4 j - 3k ˆ (d) s = Á ˜ Ë 5 ¯

[JEE (Main) – 12th April 2019 - Shift-1] Q.20. A plane electromagnetic wave having a frequency n = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave ?  (a) B = 2 ¥ 107 sin ( 0.5 ¥ 10 3 z + 1.5 ¥ 1011 t ) i  (b) B = 2 ¥ 10 -7 sin ( 0.5 ¥ 10 3 z - 1.5 ¥ 1011 t ) i  (c) B = 60 sin ( 0.5 ¥ 10 3 x + 1.5 ¥ 1011 t ) k  (d) B = 2 ¥ 10 -7 sin (1.5 ¥ 10 2 x + 0.5 ¥ 1011 t ) k [JEE (Main) – 12th April 2019 - Shift-2] Q.21. A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time,  E = 6.3 j V / m. The corresponding magnetic field  B, at that point will be : (a) 18.9 × 10–8 k T (b) 2.1 × 10–8 k T (c) 6.3 × 10–8 k T (d) 18.9 × 108 k T [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.22. The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then : U (a) U E = B (b) UE > UB 2 (c) UE < UB (d) UE = UB [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.23. If the magnetic field of a plane electromagnetic wave is given by (The speed of light = 3 × 108 m/s)

È xˆ˘ Ê B = 100 ¥ 10 -6 sin Í2p ¥ 2 ¥ 10 15 Á t - ˜ ˙ Ë c¯˚ Î

310 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

then the maximum electric field associated with it is : (a) 6 × 104 N/C (b) 3 × 104 N/C 4 (c) 4 × 10 N/C (d) 4.5 × 104 N/C [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.24. The electric field of a plane polarized electromagnetic wave in free space at time t =0 is given by an expression  E( x , y ) = 10 j cos[(6 x + 8 z )]  The magnetic field B( x , z , t ) is given by : (c is the velocity of light) 1 (a) ( 6 k + 8i ) cos [( 6 x - 8 z + 10 ct )] c (b)

1  ( 6 k - 8i ) cos [( 6 x + 8 z - 10 ct )] c

(c)

1  ( 6 k + 8i ) cos [( 6 x + 8 z - 10 ct )] c

(d)

1  ( 6 k - 8i ) cos [( 6 x + 8 z + 10 ct )] c

Ê 1 ˆ , n˜ (a) Á Ë n ¯

1 ˆ Ê 1 , (b) Á ˜ Ë n n¯

1 ˆ Ê (c) Á n , ˜ Ë n¯

(d)

(

n, n

ANSWER – KEY 1. (a) 5. (a) 9. (d) 13. (d) 17. (a) 21. (b) 25. (c)

2. (d) 6. (d) 10. (a) 14. (a) 18. (a) 22. (d) 26. (d)

)

[JEE (Main) – 11th Jan. 2019 - Shift-1] Q.26. A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by : (Given permittivity of space e0 = 9 × 10–12 SI units, Speed of light c = 3 × 108 m/s) (a) 2 kV/m (b) 0.7 kV/m (c) 1 kV/m (d) 1.4 kV/m [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.27. A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propogating in the glass medium will be : (a) 30 V/m (b) 10 V/m (c) 24 V/m (d) 6 V/m [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.28. The mean intensity of radiation on the surface of the Sun is about 108 W/m2. The rms value of the corresponding magnetic field is closest to : (a) 1T (b) 102 T –2 (c) 10 T (d) 10–4 T [JEE (Main) – 12th Jan. 2019 - Shift-2]

3. (a) 7. (c) 11. (c) 15. (a) 19. (c) 23. (b) 27. (c)

4. (b) 8. (a) 12. (d) 16. (d) 20. (b) 24. (b) 28. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Given : Frequency of electromagnetic wave is f = 2.0 × 1010 Hz, energy density of the wave is Œ = 1.02 × 10–8 J/m3. To find : B, the amplitude of magnetic field associated with the wave. Energy density associated with an electromagnetic wave in vacuum :

ε =

[JEE (Main) – 10th Jan. 2019 - Shift-2] Q.25. An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index ’n’ without any loss. The ratio of the magnitudes of electric fields and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by :

PHYSICS

1 B2 2 µo

B =

2m o ε

B =

2 × 4p × 10 −7 × 1.02 × 10 −8

B = 160nT Ans. 2. Option (d) is correct. Given : Direction of electric field in a plane electromagnetic wave is k , direction of magnetic field in a plane electromagnetic wave is 2i − 2 j. To find : v , unit vector in the direction of propagation of wave.   E × B k × ( 2i − 2 j ) i + j v = = = E B 2 2 2 Ans. 3. Option (a) is correct. Given : Magnetic field of a plane electromagnetic wave is  i T , B = 3 × 10 −8 sin{200p ( y + ct )} ...(i) speed of light is c = 3 × 10 8 m s−1  To find : E, electric field associated with the wave.  Direction of E :  From equation (i), B is along the x axis and direction of propagation is along the y axis that  means direction of E associated with the wave will be along the ( y × x ) = − z axis.  Magnitude of E : E = cB E = 3 × 10 8 × 3 × 10 −8 = 9 From equations (i) and (ii) :  E = −9 sin{200p ( y + ct )}k V m −1

...(ii)

311

ELECTROMAGNETIC WAVES

Ans. 4. Option (b) is correct. Given : Electric field of a plane electromagnetic field is  E = Eo j cos(ωt − kx ), ...(i) direction of propagation of wave is i.  To find : B( 0 ), magnetic field associated with the wave at t = 0.  Direction of B :  From equation (i), E is along the y axis and direction of propagation of wave is along the x axis,  that means direction of B associated with the wave will be along the ( x × y ) = z axis.  Magnitude of B : E Bo = o = Eo m o ε o ...(ii) c From equations (i) and (ii) :  B = E m ε k cos(ωt − kx ) o

o o

At t = 0 :  B( 0 ) = Eo m oε o k cos( kx ) Ans. 5. Option (a) is correct. Given : Various electromagnetic waves; like radio waves, micro waves, visible light and x-rays. To find : Compare the wavelengths of the given electromagnetic waves.

λradio  waves > λmicro  waves > λvisible > λx − rays Ans. 6. Option (d) is correct. Given : Electric field of a plane electromagnetic field is  E = E ( x + y )sin( kz − ωt ). ...(i) o

 To find : B, magnetic field associated with the wave.  Direction of B :  From equation (i), E is along the ( x + y ) axis and direction of propagation of wave is along the z axis,  that means direction of B associated with the wave will be along the ( z × ( x + y )) = ( − x + y ) axis.  Magnitude of B : E ...(ii) Bo = o c From equations (i) and (ii) :  E B = o ( − x + y )sin( kz − ωt ) c Ans. 7. Option (c) is correct. Given : Direction of motion of an electron is along the y axis, speed of electron is v = 0.1c, electric field associated with the electromagnetic wave in the region is  V E = 30 j sin(1.5 × 107 t − 5 × 10 −2 x ) . m

To find : Fmax, the maximum magnetic force experienced by the electron. Magnitude of magnetic field associated with the electromagnetic wave in the region : E 30 Bo = o = T c 3 × 10 8 Direction of magnetic field associated with the electromagnetic wave in the region : ( −i × j ) = − k Angle between the direction of motion of electron and the magnetic field : θ = 90° Fmax = evB o sin 90 Fmax = 1.6 × 10 −19 × 0.1 × 3 × 10 8 ×

30 3 × 10 8

Fmax = 4.8 × 10 −19 N Ans. 8. Option (a) is correct. Given : Various electromagnetic waves; like radio waves, micro waves, gamma rays and x-rays. To find : The wavelengths of the given electromagnetic waves.

λmicro waves ª 10 −3 m λgamma rays ª 10 −15 m λradio waves ª 100 m λx − rays  ª 10 −10 m Ans. 9. Option (d) is correct. Given : Magnetic field at point x and time t in a plane electromagnetic wave is  B( x ,) t = 1.2 × 10 −7 sin( 0.5 × 10 3 x + 1.5 × 1011 t )kT ...(i)  t the instantaneous electric field To find : E( x ,), associated with the wave.  Direction of E :  From equation (i), B is along the z axis and direction of propagation is along the x axis that  means direction of E associated with the wave will be along the ( x × z ) = − y axis.  Magnitude of E : E0 = c B0 E0 = 3 × 10 8 × 1.2 × 10 −7 = 36 V m −1

...(ii)

From equations (i) and (ii) :  E( x ,) t = −36 sin(0.5 × 10 3 x + 1.5 × 1011 t ) j V m −1 Ans. 10. Option (a) is correct. Given : Magnetic field in a plane electromagnetic wave is :  B = 3 × 10 −8 sin(1.6 × 10 3 x + 48 × 1010 t ) j T ...(i)

312 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)  To find : E the electric field associated with the wave.  Direction of E :  From equation (i), B is along the y axis and direction of propagation is along the x axis that  means direction of E associated with the wave will be along the ( x × y ) = z axis.  Magnitude of E : E0 = c B0 E0 = 3 × 10 8 × 3 × 10 −8 = 9 V m −1

...(ii)

From equations (i) and (ii) :  V E = 9 sin(1.6 × 10 3 x + 48 × 1010 t )k m Ans. 11. Option (c) is correct. Given : The electric field in a plane electromagnetic wave is : i + j  E = Eo cos( kz + ωt ), ...(i) 2 at t = 0 position of a positively charged particle q is p  ( x ,y ,) z =  0 ,0 ,  and its velocity is v = vo k . k  To find : The direction of net force acting on the particle due to the wave. At t = 0, electric field due to the wave at the position of the particle : i + j i + j  p   cos  k × + ω × 0  = − Eo ...(ii) E = Eo k 2 2   Force acting on the particle due to the electric field of the wave :    i + j  ...(iii) FE = qE = qEo  −   2    Direction of B :  i + j From equation (i), E is along the direction 2 and direction of propagation is along the z axis that  means direction of B associated with the wave will i − j −i + j be either along or . 2 2 Force acting on the particle due to the magnetic field of the wave :   i − j      FB = q( v × B) = qvo B o  k ×     2     From equation (iv),   FB  || E

...(iv)

Ans. 12. Option (d) is correct. Given : Frequency of a plane electromagnetic wave is f = 25 GHz, direction of propagation of the wave is along the z axis, magnetic field associated with the wave at certain point in space and time is  m B = 5 × 10 −8 j T ,c = 3 × 10 8 . s  To find : E , the corresponding electric field associated with the electromagnetic wave.  Direction of E :  As B is along the y axis and direction of propagation  is along the z axis that means direction of E associated with the wave will be along the x axis.  Magnitude of E : E0 = c B0 E0 = 3 × 10 8 × 5 × 10 −8 = 15 V m −1 That gives,  V E = 15i m Ans. 13. Option (d) is correct. Given : The electric field electromagnetic waves is :  E = E j cos(ωt − kx ), 1

From equations (iii) and (v) the direction of net force acting on the particle due to the wave will be i + j . antiparallel to 2

of

two

plane ...(i)

o

 E2 = Eo k cos(ωt − ky ),

...(ii)

at t = 0 a charged particle q is at origin and its velocity is v = 0.8 c j.  To find : F, the instantaneous force experienced by the particle due to the two waves.  Force on the particle due to E1 at t = 0 :   F = q E = qE j cos(ω × 0 − k × 0 ) = qE j ...(iii) E1

1

o

o

 Force on the particle due to E2 at t = 0 :   FE2 = qE2 = qEo k cos(ω × 0 − k × 0 ) = qEo k ...(iv)  Direction of B1 , magnetic field of wave 1 :  From equation (i), E1 is along the y direction and direction of propagation is along the x axis that  means direction of B1 associated with the wave will be along z direction.  Force on the particle due to B1 at t = 0 :    E  FB1 = q( v × B1 ) = q0.8 cBo ( j × k ) = q0.8 c  o  i  c   = q0.8 E i ...(v) 0

...(v)

PHYSICS

 Direction of B2 , magnetic field of wave 2 :  From equation (ii), E2 is along the z direction and direction of propagation is along the y axis that  means direction of B2 associated with the wave will be along x direction.

313

ELECTROMAGNETIC WAVES

 Force on the particle due to B2 at t = 0 :    E  FB2 = q( v × B2 ) = q0.8 cBo ( j × i) = −q 0.8 c  o  k  c   = − q0.8E k ...(vi) 0

From equations (iii), (iv), (v) and (vi) :      F = FE1 + FE2 + FB1 + FB2  F = qEo ( 0.8i + j + 0.2 k ) Ans. 14. Option (a) is correct. Given : Direction of propagation of a plane i + j , the electric field of electromagnetic wave is 2 the wave is along direction k .  To find : B,correct expression for magnetic field of the wave.  Direction of B : As the direction of propagation, direction of electric field and direction of magnetic field associated with the wave should form three mutually perpendicular  i − j vectors, direction of B will be along . 2 i − j i + j    cos  ωt − k B = Bo   2 2   Ans. 15. Option (a) is correct. Given : Direction of propagation of an electromagnetic wave is along the x-axis, the electric field component of the wave at a particular point of space and time is  V E = 6 , j . m  To find : B, the magnetic field component  corresponding to the given E. For an EM wave :  E  = c ; B  B =

 E c

=

6 3 ¥ 10 8

 To find : E,the corresponding electric field. Magnitude of electric field :  E  = c B   E = c B = 3 ¥ 10 8 ¥ 1.6 ¥ 10 -6 = 4.8 ¥ 10 2 V/ m

Ans. 16. Option (d) is correct. Given : The magnetic field of an EM wave is  B = 1.6 ¥ 10 -6 cos( 2 ¥ 107 z + 6 ¥ 1015 t ) Wb ( 2i + j ) 2 ...(i) m

...( ii )

Also, for an EM wave :   E.B = 0

 That gives direction of E : n = i - 2 j n = -i + 2 j

or

...(iii) ...(iv)

From equation (i), the propagation of the given e.m. wave is along the k direction. We know, direction of propagation of an EM wave = direction   of E × direction of B k = n ¥ ( 2i + j ) ...(v) Equation (v) is satisfied for n = i - 2 j . So, from equations (i), (ii), (iii) and (v) :  E = 4.8 ¥ 10 2 cos ( 2 ¥ 107 z + 6 ¥ 1015 t ) (i - 2 j )V/ m Ans. 17. Option (a) is correct. Given : The magnetic field of an e.m. wave is  B = B0 ÎÈcos( kz - ωt )˚˘ i + B1[cos( kz - ωt )]j , B0 = 3 ¥ 10 -5 T ,B1 = 2 ¥ 10 -6 T

...( i )

To find : The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0.  Components of E associated with the e.m. wave will be : E0 = cB0 = 3 ¥ 10 8 ¥ 3 ¥ 10 -5 = 9 ¥ 10 3 = 9000 V / m

= 2 ¥ 10 -8 T

Also, direction of propagation of an EM wave =   direction of E × direction of B, i = j ×direction  of B  That gives, direction of B is along the z-axis. So,  B = 2 ¥ 10 -8 k T



...( ii )

E1 = cB1 = 3 ¥ 10 8 ¥ 2 ¥ 10 -6 = 600 V / m

...(iii)

From equation (i), propagation of the given e.m. wave is along the k direction. From equations (ii), (iii) and the two required     properties of e.m. wave, E.B = 0 , k = E ¥ B :   E = -9000 j ,E = 600 i 0

1

That gives,  E = 600 ÈÎcos ( kz - ωt )˘˚ i - 9000

[cos( kz - ωt )] j V / m ...(iv)

314 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Maximum force experienced by the stationary charge Q :    F = Q E0 + E1 = 10 -4 -9000 j + 600 i

(

)

(

)

Substitute from equations (iv) and (v) in equation (iii). kx x + k y y + kz z = 8 z - 6 y ;

= -0.9 j + 0.06i   | F| =

(0.9 )

2

So,

+ (0.06 ) = 0.9 N 2

Ans. 18. Option (a) is correct. Given : The electric field of a plane e.m. wave is  E = E i cos ( kz )cos(ωt ) ...(i) 0

 To find : B, the corresponding magnetic field. From equation (i) direction of propagation of wave is along z-axis and the electric field vector is along the x-axis. That implies, direction of magnetic field vector will be along the y-axis. From equation (i) and Maxwell’s equation, ∂B — ¥ E = - we get : ∂t

...(vii)

k =

2p 2pν 2p ¥ 23.9 ¥ 10 9 = = λ c 3 ¥ 10 8

= 0.5 ¥ 10 3 m -1

...(i)

Angular frequency : = 1.5 ¥ 1011 Hz

...(iii)

Integrating equation (iii) : E0 B = k sin( kz ) sin(ωt ) ω Ê kˆ = E0 Á ˜ sin( kz ) sin(ωt ) Ëω¯ E0 sin( kz ) sin(ωt ) c

...(iv)

E0  j sin( kz ) sin(ωt ) c

Ans. 19. Option (c) is correct. Given : The electric field of a plane EM wave is  E = E0 n sin ÎÈωt + ( 6 y - 8 z )˚˘ ...(i) To find : s , the direction of propagation of the EM wave. Standard expression for electric field of an EM wave :   E = E0 n Èsin ωt - k . r ˘ ...(ii) Î ˚

(

 k = -6 j + 8 k

ω = 2pν = 2p ¥ 23.9 ¥ 10 9

∂B - E0 k sin( kz )cos(ωt ) = ∂t

 B =

...(vi)

Ans. 20. Option (b) is correct. Given : Frequency of plane e.m. wave is n = 23.9 GHz, direction of propagation of e.m. wave is along the z-axis, the peak value of electric field associated with the e.m. wave is E0 = 60 V/m. To find : The corresponding magnetic field vector associated with the e.m. wave. Wave number associated with the given e.m. wave:

∂E ∂B = ∂z ∂t

So,

k z = 8 ,k y = -6

That gives, direction of propagation of EM wave :  k -6 j + 8 k -3 j + 4 k = s =  = 5 k 62 + 82

Rms value of force :   | F| 0.9 Frms = = = 0.6 N 2 2

=

PHYSICS

)

Comparing equations (i) and (ii),  k . r = 8 z - 6 y

...(iii)

 k = k x i + k y j + k z k

...(iv)

r = xi + y j + xk

...(v)

Peak value of magnetic field : E 60 B0 = 0 = = 2 ¥ 10 -7 T c 3 ¥ 10 8

...(ii)

...(iii)

As the given wave is propagating along the z-axis, we can take the direction of magnetic field vector  B, either along the x or the y-axis, among the given options.   B = B sin ( kz - ωt ) i or B = B sin( kz - ωt ) j 0

0

From equations (i), (ii) and (iii), the most acceptable  value of B will be :  B = 2 ¥ 10 -7 È ( 0.5 ¥ 10 3 z - 1.5 ¥ 1011 t )˘ i sin Î ˚

Ans. 21. Option (b) is correct. Given : Frequency of plane e.m. wave is n = 50 MHz, direction of propagation of EM wave is along the x-axis, the electric field component of the wave at a  particular point in space and time is E = 6.3 V / m j.  To find : B, the magnetic field component  corresponding to the given E. Magnitude of magnetic field vector associated with the given e.m. wave will be :  E  6.3 B = = = 2.1 ¥ 10 -8 T c 3 ¥ 10 8 We know, direction of propagation of an e.m.wave   = direction of E × direction of B  i = j × direction of B

315

ELECTROMAGNETIC WAVES

That gives, direction of magnetic field vector is along the z-axis.  So, B = 2.1 ¥ 10 -8 k T Ans. 22. Option (d) is correct. Given : Energy of an e.m. wave in electric field is UE, energy of an e.m. wave in magnetic field is UB. To find : Relation between UE and UB. Energy density of an e.m. wave in electric field : 1 U E = ε 0 E2 ...(i) 2 Energy density of an e.m. wave in magnetic field : UB =

2

1 B 2 m0

...(ii)

6i + 8 k = j ¥ ( pi + qk ) 10 6i + 8 k = - pk + qi 10 p = -

8 6 s = - i + k 10 10

m ε E2 UE = 0 02 UB B E Use relations, = c and c = B

m 0ε 0

B( x ,,) zt = in equation

(iii). UE = 1; U E = U B UB Ans. 23. Option (b) is correct. Given : Magnetic field of a plane e.m. wave is  B = 100 ¥ 10 -6 È xˆ˘ Ê sin Í2p ¥ 2 ¥ 1015 Á t - ˜ ˙ Ë c¯˚ Î

10 Ê 8  6  ˆ Á - i + k˜ 3 ¥ 10 8 Ë 10 10 ¯ cos( 6 x + 8 z - 10 ct ) 1 ( -8i + 6 k )cos( 6 x + 8 z - 10 ct ) c

Ans. 25. Option (c) is correct. Given : Intensity of e.m. wave is I = 50 W/m2, the wave enters a medium of refractive index n and suffers no loss. E To find : Ratio of magnitudes of electric fields Em B of Bm the wave before and after entering into the medium. Velocity of wave before entering the medium : 1 c = ...(i) m 0ε 0 and ratio of magnitudes of magnetic fields

...( i )

To find : Emax, the maximum electric field associated with the wave. Emax = Bmax ¥ c

Velocity of wave after entering the medium : 1 v = ...( ii ) mε

So, from equation (i) : Emax = 100 ¥ 10 -6 ¥ 3 ¥ 10 8

From equations (i) and (ii) :

= 3 ¥ 10 4 N / C Ans. 24. Option (b) is correct. Given : The electric field of a plane polarised e.m. wave in free space at t = 0 is E ( x ,y ) = 10 j cos ( 6 x + 8 z ) ...( i ) To find : The corresponding magnetic field B(x, z, t).  From equation (i), direction of E is along the y-axis and direction of propagation of the e.m. wave is 6i + 8 k 6i + 8 k = n = ...( ii ) 2 2 10 6 +8 Let direction of magnetic field be along s = pi + qk

...(v)

From equations (iv) and (v) :

...(iii) 1

...(iv)

Peak value of magnetic field : E 10 B0 = 0 = c 3 ¥ 10 8

B( x ,,) zt =

That gives,

8 6 ,q = 10 10

...( iii )

We know, direction of propagation of an e.m. wave = direction   of E × direction of B

c = v

mε m 0ε 0

For transparent medium, m = m0 ;

c ε = = K =n v ε0

...( iii )

Ê ε ˆ ÁË dielectric constan t = ε ˜¯ 0 Intensity of wave before entering the medium : 1 I = ε 0 E2 c ...(iv) 2 Intensity of wave after entering the medium : 1 2 1 2 v = ( Kε 0 ) E m v I¢ = ε Em 2 2 =

1 2 2 n ε 0 Em v 2

...( v )

316 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) As there were no losses:

Intensity of light after entering the glass :

I = I¢;

I¢ =

1 1 2 ε 0 E 2 c = n 2ε 0 E m v 2 2 2

E c =

2 n 2 Em v

=

(use equation (iii))

E = n Em Use

B =

...(vi)

1 Ê 1ˆ nÁ ˜ = Ë n¯ n

...(vii)

In equation (iv), dielectric constant K =

Intensity of the laser beam : P 27 ¥ 10 -3 = A 10 ¥ 10 -6

96 2 2 E0 c = n 2 Em v 100

...( i )

1 ε 0 E2 c 2

...( ii )

1 ε 0 E2 c = 2.7 ¥ 10 3 2

Given : mean intensity of radiation on the surface W of sun is I = 10 8 2 . m

Intensity of radiation on the surface of sun :

E02 =

= 0.2 ¥ 107

kV m

2 ¥ 10 8 9 ¥ 10 -12 ¥ 3 ¥ 10 8

=

2 ¥ 1012 27

V m

Peak value of magnetic field : B0 =

E0 0.27 ¥ 10 6 = = 9 ¥ 10 -4 T 8 c 3 ¥ 10

Rms value of magnetic field : Brms =

Given : Refractive index of glass slab is n = 1.5, light is incident normally on the glass slab out of which 4% of light gets reflected, amplitude of the electric V field of the incident light is E0 = 30 . m To find : Em, the amplitude of the electric field for the wave propagating in the glass. Intensity of light before entering the glass : 1 ε 0 E02 c 2

1 W ε 0 E02 c = 10 8 2 2 m

E0 = 0.27 ¥ 10 6

Ans. 27. Option (c) is correct.

I =

V  m

Ans. 28. Option (d) is correct.

Equate equations (i) and (ii) :

E = 1.4

96 E02 96 ( 30 )2 = ¥ 100 n 100 1.5

Em = 24

I =

Also intensity of an EM wave is given by :

9 ¥ 10 -12 ¥ 3 ¥ 10 8

Êc ˆ ÁË v = n˜¯

To find : Brms, the rms value of the corresponding magnetic field.

= 2.7 ¥ 10 3 W m -2

2.7 ¥ 10 3 ¥ 2

ε , and v is ε0

the velocity of light in the glass

2 Em =

To find : The magnitude of maximum electric field associated with the laser beam.

E2 =

...( iv )

96 Ê 1 1 ˆ 2 ε 0 E02 c˜ = n 2ε 0 Em v ¯ 100 ÁË 2 2

Given : Power of laser beam is P = 27 mW, cross sectional area of the beam is A = 10 mm2, e0 = 9 × 10–12 SI units.

I =

1 2 2 n ε 0 Em v 2

96 I = I¢ 100

Ans. 26. Option (d) is correct.

I =

1 2 1 2 ε E m v = ( Kε 0 ) E m v 2 2

As 4% light is lost in reflection :

E E , Bm = m . c v

E B E Ê vˆ = c = E Bm Em ËÁ c ¯˜ m v =

PHYSICS

...( i )

B0 2

= 6.4 ¥ 10 -4 T

which is closest to 10–4 T.

Subjective Questions (Chapter Based)  315  2 Q.1. Suppose that intensity of a laser is   W/m .  p  The rms electric field, in units of V/m associated with this source is close to the nearest integer is : (e0 = 8.86 × 10–12C2Nm–2; c = 3 × 108ms–1)

[JEE (Main) – 6th Sep. 2020 - Shift-1]

317

ELECTROMAGNETIC WAVES

Sol. Given : Intensity of a laser beam is I =

Q.3. A red LED emits light at 0.1 watt uniformly around it. What will be the amplitude of the electric field

2

315 W NC , ε o = 8.86 × 10 −12 2 ,  2 π m m

of the light at a distance of 1 m from the diode?

c = 3 × 10 8

m . s

To find : Erms, the rms electric field associated with the laser source. Intensity of an electromagnetic wave : 1 2 I = cε o Erms 2 Erms = Erms =

2I cε o 2 × 315 3.14 × 3 × 10 8 × 8.86 × 10 −12

c c = =2= c v 2 =



P = 0.1 W

To find : The amplitude of the electric field of light at the distance of 1 m from the LED source.



Intensity of light at radial distance r from the source  : I =

ª 275

V m

Q.2. An EM wave from air enters a medium. The  È Ê z ˆ˘ electric fields are E1 = E01 x cos Í2pn Á - t ˜ ˙ in air Ëc ¯˚ Î  and E2 = E02 x cos [ k ( 2 z - ct )] in medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. What will be the relation between er1 and er2, the relative permittivities of air and medium respectively? Sol. Given : the electric field of an EM wave in air is  È Ê z ˆ˘ E1 = E01 x cos Í2pn Á - t˜ ˙ ...(i) Ë c ¯˚ Î The electric field of an EM wave in medium is  E2 = E02 x cos ÈÎk ( 2 z - ct )˘˚ ...( ii ) To find : er1, er2, relation between the relative permittivities of air and medium respectively. From equation (i), velocity of light in air : ω  t c = ωz From equation (ii), velocity of light in medium : ωt c v = = ωz 2



Sol. Given : Power output from red LED source is

m r 2ε r 2 m 0ε r 1

mr2ε r 2 m 0ε 0 ...(iii)



= 8 ¥ 10 -3



E02 =



W m2

1 ε 0 E02 c 2

1 ε 0 E02 c = 8 ¥ 10 -3 2 16 ¥ 10 -3 9 ¥ 10 -12 ¥ 3 ¥ 10 8

ª6

E0 = 2.45 V



Q.4. The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. What will be the peak value of electric field strength? Sol. Given : Peak value of magnetic field associated with an e.m. wave is B0 = 20 nT.

To find : E0, peak value of electric field. E0 = B0 c = 20 ¥ 10 -9 ¥ 3 ¥ 10 8 = 6



V m

Q.5. The electric field of an electromagnetic wave is given by E = E0 sin(wt – kx), E0 = 107 V/m. Calculate the energy contained in a cylinder of cross sectional area 10–4 m2 and length 1 m along the x-axis. Sol. Given : Peak value of electric field in an e.m. wave is E0 = 107 V/m, cross sectional area of a cylinder is A = 10–4 m2, length of the cylinder is l = 1 m.

To find : The energy contained in the cylinder.



Energy density of an e.m. wave in electric field : UE =



Substitute in equation (iii).



εr2 = 4 ε r1

0.1 4p 12

Equate the above two equations.



εr2 ε r1

=



As the medium is non-magnetic, its magnetic permeability will be : mr2 = m0 . c = v

I =

Also,

P 4p r 2

1 ε 0 E0 2 2

...( i )

That gives, energy contained in a cylindrical volume  :



E =

...( ii )

Put given values in equation (ii). E =



1 ε 0 E0 2 Vcylinder 2

1 ¥ 9 ¥ 10 -12 ¥ 107 ¥ 2 107 ¥ 10 -4 ¥ 1 = 5.2 ¥ 10 -12 J

318 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.6. A plane electromagnetic wave propagating in z direction has a wavelength of 11 m. The electric field is in the y direction and its maximum amplitude is 53 V/m. Write equations for electric and magnetic fields associated with the electromagnetic wave. Sol. Given : Wavelength of EM wave is l = 11 m, direction of propagation of wave is along the z-axis, direction of electric field associated with the e.m. wave is along the y-axis, maximum amplitude of electric field is E0 = 53 V/m. To find : The equations of electric and magnetic fields associated with the electromagnetic wave. c λν = c ; ν = ; λ Ê cˆ Ê cˆ ω = 2p Á ˜ = 2p Á ˜ Ë 11 ¯ Ë λ¯







Also,

k =

2p 2p = λ 11

...( i ) ...(ii)

Standard expression for electric field of an e.m. wave propagating along z-axis, when the electric field is along the y-axis :  E = E0 y ÈÎsin (ωt - kz )˘˚ ...( iii ) Put given values and values from equations (i) and (ii) in equation (iii).  2p È ˘ E = 53 j Ísin ( ct - z )˙ ...(iv) 11 Î ˚





PHYSICS

Peak value of magnetic field associated with an EM wave : B0 =

E0 53 = = 1.8 ¥ 10 -7 T ...(v) c 3 ¥ 10 8



Direction of magnetic field associated with an e.m. wave is perpendicular to the direction of electric field and direction of propagation of wave. That  implies, for the given wave direction of B will be along x-axis.



Standard expression for magnetic field of an EM wave propagating along z-axis when the magnetic field is along the x-axis :  B = B0 x ÈÎsin (ωt - kz )˘˚ ...(vi)



Put given values and values from equations (i) and (ii) in equation (i).  B = 1.8 ¥ 10 -7 i



2p ˘ È Ísin 11 ( ct - z )˙ ...(vii) Î ˚



Equations (iv) and (vii) are the required equations.





are

When ray passes from optically denser to rarer medium, if incident angle (i) further increased till ( )c critical angle, entire light is then reflected back to the denser medium again, this process is called T.I.R. It is used in optical fibre.

are

are

(or pole)

Optics Part-1

[

] [For Combination of lens in Contact]

OPTICS

319

and

,

s

=

For bright fringe =n For dark fringe , .

Optics Part-2

F

are

s

Two waves superimpose to form a resultant wave of greater or lower or same amplitude.

320 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Optics

Chapter 16 Syllabus

Reflection and refraction of light at plane and spherical surfaces; mirror formula, Total internal reflection and its applications; Deviation and Dispersion of light by a prism; Lens Formula; Magnification, Power of a Lens; Combination of thin lenses in contact; Microscope and Astronomical Telescope (reflecting and refracting) and their magnifying powers. Wave optics : Wavefront and Huygens’ principle; Laws of reflection and refraction using Huygen’s principle; Interference; Young’s double slit experiment and expression for fringe width; coherent sources and sustained interference of light. Diffraction due to a single slit; width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, plane polarized light; Brewster’s law, uses of plane polarized light and Polaroids. Rectilinear propagation of light, Combinations of mirrors.

Topic-1

Rectilinear Propagation of Light

LIST OF TOPICS : Topic-1 : Rectilinear Propagation of Light .... P. 321 Topic-2 :  Wave Optics

.... P. 335

Concept Revision (Video Based) Reflection of Light at Plane and Spherical Surfaces

Total Internal Reflection

Refraction    

    

      Part -1    Focal Length

Concave Mirror

Concave and Convex Lens

Triangular Prism

 

Newton's Prism Experiment

Refractive Index of a Glass Slab

Deviation and Dispersion of Light

Part -1           Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A spherical mirror is obtained as shown in the figure from a hollow glass sphere. if an object is positioned infront of the mirror, what will be the nature and magnification of the image of the object? (Figure drawn as schematic and not to scale)

Object

20 (cm)

16

12

8

4

Part - 2

322 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a) Inverted, real and magnified (b) Erect, virtual and unmagnified (c) Inverted, real and unmagnified (d) Erect, virtual and magnified [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. Two light waves having the same wavelength l in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2. After this the phase difference between the two waves is : (a)

2p  L2 L1  −   λ  n1 n2 

2p ( n1 L1 − n2 L2 ) (c) λ

(b)

2p  L1 L2  −   λ  n1 n2 

(a)

Q.7. A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f fl, then the ratio l is closest to the integer : f (a) 9

(b) 1

(c) 5

(d) 17 [JEE (Main) – 7th Jan. 2020 - Shift-2]

Q.8. The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 4 3 and relative permeability for this wavelength, 3 will be :

2p ( n2 L1 − n1 L2 ) (d) λ

[JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.3. For a concave lens of focal length f, the relation between object and image distance m and n, respectively, from its pole can best the represented by (u = n is the reference line) :

PHYSICS

(a) 15°

(b) 30°

(c) 45°

(d) 60° [JEE (Main) – 8th Jan. 2020 - Shift-1]

Q.9. The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece?

(b)

(a) 30 cm

(b) 40 cm

(c) 10 cm

(d) 20 cm [JEE (Main) – 8th Jan. 2020 - Shift-1]

(c)

(d)

[JEE (Main) – 5th Sep. 2020 - Shift-1] Q.4. A point like object is placed at distance of 1 m in front of a convex lens of focal length 0.5 m. A plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the image formed by the system is : (a) 2.6 m from the mirror, real (b) 1 m from the mirror, virtual (c) 1 m from the mirror, real (d) 2.6 m from the mirror, virtual [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.5. A double convex lens has power P and same radii of curvature r of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is : R 3R (a) (b) 2 2 (c)

R 3

(d) 2R

[JEE (Main) – 6th Sep. 2020 - Shift-2] Q.6. If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eyepiece, should be close to : (a) 22 mm (b) 12 mm (c) 33 mm (d) 2 mm [JEE (Main) – 7th Jan. 2020 - Shift-1]

Q.10. An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification (m) versus distance of the object from the mirror (x) is correctly given by

(Graphs are drawn schematically and are not to scale) (a)

(b)

M 1

M 1 x

x f 2f

f 2f

(c)

(d)

M 1

M 1

x

x

f 2f

f 2f

(JEE (Main) – 9th Jan. 2020 - Shift -1] Q.11. A vessel of depth 2h is half filled with a liquid of refractive index 2 2 and the upper half with another liquid of refractive index 2. The liquids are immiscible. The apparent depth of the inner surface of the bottom of vessel will be : h 3 (a) (b) h 2 4 2 (c)

h 3 2



(d)

h 2( 2 + 1)

(JEE (Main) – 8th Jan. 2020 - Shift-1] Q.12. The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is

323

OPTICS

4 × 105 km. With light of wavelength of 5500 Å, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to : (a) 600 m (b) 20 m (c) 60 m (d) 200 m [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.13. There is a small source of light at some depth below the surface of water (refractive index = 4 ) in a tank of large cross sectional surface 3 area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) : [Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2prh] (a) 34% (b) 17% (c) 50% (d) 21% [JEE (Main) – 9th Jan. 2020 - Shift-1] Q14. An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be : (a) 20 cm from the convergent mirror, same size as the object (b) 40 cm from the convergent mirror, same size as the object (c) 40 cm from the convergent lens, twice the size of the object (d) 20 cm from the convergent mirror, twice the size of the object [JEE (Main) – 8th April 2019 - Shift-1] Q.15. In figure, the optical fibre is l = 2 m long and has a diameter of d = 20 mm. If a ray of light is incident on one end of the fiber at angle q1 = 40°, the number of reflections it makes before emerging from the other end is close to : (Refractive index of fiber is 1.31 and sin 40° = 0.64)

2

A

L

d

40°

(a) 55000 (c) 45000

Q.17. Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star. (a) 305 × 10–9 radian (b) 610 × 10–9 radian –9 (c) 152.5 × 10 radian (d) 457.5 × 10–9 radian [JEE (Main) – 8th April 2019 - Shift-2] Q.18. A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is : (a) 0.24 m (b) 1.60 m (c) 0.32 m (d) 0.16 m [JEE (Main) – 9th April 2019 - Shift-1] Q.19. Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. coming from a distant object, the limit of resolution of the telescope is close to : (a) 1.5 × 10–7 rad (b) 2.0 × 10–7 rad –7 (c) 3.0 × 10 rad (d) 4.5 × 10–7 rad [JEE (Main) – 9th April 2019 - Shift-2] Q.20. A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x1 and x2(x1 > x2) from the lens. The ratio of x1 and x2 is : (a) 2 : 1 (b) 3 : 1 (c) 5 : 3 (d) 4 : 3 [JEE (Main) – 9th April 2019 - Shift-2] Q.21. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index ml is put between the lens and the mirror, the pin has to be moved to A’, such that OA’ = 27 cm, to get its inverted real image at A’ itself. The value of ml will be : A'

(b) 66000 (d) 57000 [JEE (Main) – 8th April 2019 - Shift-1] Q.16. A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be : (a) 30 cm (b) 25 cm (c) 10 cm (d) 20 cm [JEE (Main) – 8th April 2019 - Shift-2]

M O

4 (a) 3 (c)

(b)

3 2

(d) 2 [JEE (Main) – 9th April 2019 - Shift-2] Q.22. One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is m1 and that of 2 is m2, then the focal length of the combination is : 3

1





2

324 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (a)

R m1 - m 2

(b)

2R m1 - m 2

(c)

R 2 ( m1 - m 2 )

(d)

R 2 - ( m1 - m 2 )



PHYSICS

(Refractive index of water = 1.33) Particle

• 5 cm

[JEE (Main) – 10th April 2019 - Shift-1] Q.23. A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is :

(a) 6.7 cm

(b) 13.4 cm

(c) 8.8 cm

(d) 11.7 cm [JEE (Main) – 12th April 2019 - Shift-1]

A 60°

a b

Vacuum glass

O

the face AB at an angle q (shown in the figure).

B

(c) 2 a +

2b 3

of refractive index m2, is immersed in a liquid of refractive index m1 (m1 < m2). A ray is incident on

30°

2b (b) 2 a + 3

2 3 + 2b (a) a

Q.27. A transparent cube of side d, made of a material

Total internal reflection takes place at point E on the face BC. B

E

C

(d) 2a + 2b





[JEE (Main) – 10th April 2019 - Shift-1] Q.24. The graph shows how the magnification m





produced by a thin lens varies with image distance v. What is the focal length of the lens used ? m

D

A

Then q must satisfy : (a) θ < sin -1

c

v a

b

(a)

b2 ac

(b)

b2c a

(c)

a c

(d)

b c

[JEE (Main) – 10th April 2019 - Shift-2] Q.25. The value of numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength 5000 Å is used, the minimum separation between

-1 (c) θ < sin

m1 m2 m 22 m12

(b) θ > sin -1

-1

(d) θ > sin -1

m 22 m12

-1

m1 m2

[JEE (Main) – 12th April 2019 - Shift-2] Q.28. A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is : (a) 1.1 cm away from the lens

two points, to be seen as distinct, will be :

(b) 0

(a) 0.24 mm

(b) 0.38 mm

(c) 0.55 cm towards the lens

(c) 0.12 mm

(d) 0.48 mm

(d) 0.55 cm away from the lens

[JEE (Main) – 12 April 2019 - Shift-1]

[JEE (Main) – 9th Jan 2019 - Shift-1]

Q.26. A concave mirror has radius of curvature of 40 cm.

Q.29. Two plane mirrors are inclined to each other such

It is at the bottom of a glass that has water filled up

that a ray of light incident on the first mirror (M1)

to 5 cm (see figure). If a small particle is floating

and parallel to the second mirror (M2) is finally

on the surface of water, its image as seen, from

reflected from the second mirror (M2) parallel to

directly above the glass, is at a distance d from the

the first mirror (M1). The angle between the two

surface of water. The value of d is close to:

mirrors will be :



th

325

OPTICS

(a) 45° (c) 75°

–

–

–

 (nm)

400 500 600 700

(c)

Dm

–

–

–

 (nm)

400 500 600 700

(d)

Dm

(a) 1 cm (c) 4.0 cm

–

–

–

(a) 1.16 × 10–3 m/s towards the lens (b) 3.22 × 10–3 m/s towards the lens (c) 0.92 × 10–3 m/s away from the lens (d) 2.26 × 10–3 m/s away from the lens [JEE (Main) – 11th Jan 2019 - Shift-1] Q.33. The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if Dm is the angle of minimum deviation ?

–

(b) 2 cm (d) 3.1 cm [JEE (Main) – 10thJan 2019 - Shift-2] Q.32. An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :

Dm

–

(a) m1 + m2 = 3 (b) 2m1 – m2 = 1 (c) 3m2 – 2m1 =1 (d) 2m2 – m1 = 1 [JEE (Main) – 10thJan 2019 - Shift-1] Q.31. The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

(b)

–

(b) 60° (d) 90° [JEE (Main) – 9th Jan 2019 - Shift-2] Q.30. A plano convex lens of refractive index m1 and focal length f1 is kept in contact with another plano concave lens of refractive index m2 and focal length f2. If the radius of curvature of their spherical faces is R each and f1 = 2f2, then m1 and m2 are related as  :

 (nm)

400 500 600 700

[JEE (Main) – 11th Jan 2019 - Shift-1] Q.34. A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is 3 , then the angle of incidence is : (a) 90° (b) 30° (c) 60° (d) 45° [JEE (Main) – 11th Jan 2019 - Shift-2] Q.35. What is the position and nature of image formed by lens combination shown in figure ? (f1, f2 are focal lengths) 2 cm A

B

O

1.535 –

n2 1.530 – •

20 cm

1.525 –

(b) 40 cm from point B at right; real

•

1.515 –

(c)

• –

–

–

–

1.510

f2= –5 cm

(a) 70 cm from point B at left; virtual

•

1.520 –

f1 = +5 cm

 (nm)

400 500 600 700

20 cm from point B at right, real 3

(d) 70 cm from point B at right; real [JEE (Main) – 12th Jan 2019 - Shift-1]

(a)

Dm

–

–

–

–

400 500 600 700

 (nm)

Q.36. A point source of light, S is placed at a distance L in front of the centre of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2L as shown below. The distance over which the man can see the image of the light source in the mirror is :

326 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

ANSWERS WITH EXPLANATIONS d

•S

L 2L

(a) d

(b) 2d

(c) 3d

(d)

d 2

[JEE (Main) – 12th Jan 2019 - Shift - I] Q.37. Formation of real image using a biconvex lens is shown below :

2f |

2f



|

f

f

|

screen

|

If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen ? (a) Image disappears

(b) Magnified image

(c) Erect real image

(d) No change

[JEE (Main) – 12th Jan 2019 - Shift-2] Q.38. A plano-convex lens (focal length f2, refractive index m2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index m1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be : (a) f1 – f2 (c)

(b)

2 f1 f2 f1 + f2

R m 2 - m1

(d) f1 + f2 [ JEE (Main) – 12th Jan 2019 - Shift-2]

ANSWER – KEY

1. (c)

2. (c)

3. (c)

4. (d)



5. (c)

6. (a)

7. (a)

8. (b)



9. (c)

10. (d)

11. (b)

12. (c)



13. (b)

14. (b)

15. (d)

16. (c)



17. (a)

18. (c)

19. (c)

20. (b)



21. (a)

22. (a)

23. (d)

24. (d)



25. (a)

26. (c)

27. (c)

28. (d)



29. (b)

30. (b)

31. (d)

32. (a)



33. (a)

34. (c)

35. (d)

36. (c)



37. (a)

38. (b)

Ans. 1. Option (c) is correct. Given : A spherical mirror is obtained from a hollow glass sphere of radius r = 8 cm, an object is placed in front of the mirror at u = 10 cm. To find : The nature of image formed by the mirror. From the construction, centre of curvature of the mirror is at C = 8 cm and focal point is at (f = 4 cm) from the pole of the mirror. As the object is kept slightly beyond the centre of curvature of the mirror (C), the image formed will be inverted, real and unmagnified. The location of the image will be between (C) and focal point (f ). Ans. 2. Option (c) is correct. Given : wavelength of two light waves is same l1 = l2 = l, light wave-1 travels a path of length L1 through a material of refractive index n1, light wave-2 travels a path of length L2 through a material of refractive index n2. To find : Df, the phase difference between the two waves when they are out of their respective mediums. Optical path travelled by light wave-1 : ∆1 = n1L1 Optical path travelled by light wave-2 : ∆ 2 = n2 L 2 Path difference between the two waves : ∆ = ∆1 − ∆ 2 = n1L1 − n2 L2  Corresponding phase difference between the two waves : 2p 2p ∆φ = ∆= ( n1L1 − n2 L2 ) λ λ Ans. 3. Option (c) is correct. Given : Focal length of a concave lens is f, distance of object from the lens is u, distance of image from the lens is v. To find : Correct graphical representation of v vs. u. Concave lens: a. For u = ∞, v = f b. For u = f, v < f c. For u < f, v < u All the above three points are satisfied by graph (c). Ans. 4. Option (d) is correct. Given : Focal length of a convex lens is f = 0.5 m, distance of object O placed in front of the lens is u = 1 m, distance between a plane mirror placed behind the lens and the lens d = 2m. To find : The position and nature of the final image.

Position of image (I1) formed by the lens:

327

OPTICS

Magnification of a compound microscope : L D M =  1 +  , D = 250 mm fo  fe 

1 1 1 + = v′ u f v′ = 1m The image 1 is formed at v¢ = 1m, behind the lens and d¢ = (d – v¢) = 1m in front of the mirror. (I1) acts as an object for the mirror. Position of image (I2) formed by the mirror: As the mirror is plane mirror, second image will be formed at d≤ = 1m behind the mirror. The second image will act as an object for the lens. Distance of this image (I2) from the lens is u¢ = d≤ + d = 3 m. Position of third image : 1 1 1 + = v″ u′ f 1 1 1 3 = − ; v″ = m v″ 0.5 3 5 3 The final image is formed v″ = m , in front of the 5 lens. Distance of final image from the mirror : x = v″ + d =

3 13 +2= = 2.6 m 5 5

Nature of the image is virtual. Ans. 5. Option (c) is correct. Given : Case 1 : Power of a double convex lens is P, radii of curvature for both the curved surfaces of the lens are R, Case 2 : Power of a plano-convex lens formed by the same material as the lens in case 1 is P¢ = 1.5 P. To find : R¢, the radius of curvature of the curved surface of the plano-convex lens in case 2. Let the refractive index of the material of the lens be n. For case 1 : 1  2 1 P = ( n − 1)  −  = ( n − 1)  R −R  R For case 2 :  1 1 1.5P = ( n − 1)  −   R′ ∞  1.5 ×

2  1 ( n − 1) = ( n − 1)   R  R′  R′ =

R 3

Ans. 6. Option (a) is correct. Given : Magnification from a compound microscope is M = 375, tube length of the microscope is L = 150 mm, focal length of the objective lens is fo = 5 mm. To find : fe, the focal length of the eye piece of the microscope.

is the least distance for distinct vision. 150  250  375 = 1 +  fe  5   375 250  = 1 +  fe  30  11.5 =

250 fe

= fe

250 = 22 mm 11.5

Ans. 7. Option (a) is correct. Given : Focal length of a thin lens of glass in air is f = 16 cm, refractive index of glass is ng = 1.5, focal length of a thin lens of glass in liquid is fl, refractive index of the liquid is nl = 1.42. f To find : l . f Focal length of a lens in air :  1  ng 1 1  − =  − 1    R R f n 2  1 

...(i)

In equation (i), n = 1 is the refractive index of air, R1 and R2 are radii of curvature of two faces of lens. Focal length of a lens in liquid :  1  ng 1 1  ...(ii) − =  − 1    R R fl n 2  1  l From equations (i) and (ii) :  ng  − 1   n fl  = nl  ( n g − n)  = 1.42  1.5 − 1  =  f 1  1.5 − 1.42   ng  n  n g − nl  − 1    nl  fl ª 9 f Ans. 8. Option (b) is correct. Given : Relative permittivity and relative permeability for a medium for a specific wavelength 4 l is er = 3 and µr = , respectively. 3 To find : qc, the critical angle for the medium for the same wavelength l. Refractive index for the medium : nm =

ε r µr = 3 ×

4 =2 3

The refractive index of air is n = 1. Critical angle for the medium :  n  −1  1 θ c = sin −1   = sin   n 2  m

θ c = 30

328 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 9. Option (c) is correct. Given : Magnifying power of a telescope is M = 5, tube length of the telescope is L = 60 cm. To find : fe, the focal length of the eye piece of the telescope. Let fo be the focal length of the objective lens of the telescope. Magnification from a telescope : fo M = = 5 fe

PHYSICS

Ans. 12. Option (c) is correct. Given : The aperture diameter of a telescope is a = 5m wavelength of incident light is l = 5500 Å, distance between moon and earth is D = 4 × 105 km. To find : d, the minimum separation between two objects on the surface of moon, so they can be resolved using this telescope.

fo = 5 fe Tube length of telescope : L = fo + fe 60 = 5 fe + fe fe = 10cm Ans. 10. Option (d) is correct. Given : An object which is gradually moving away from the focal point of a concave mirror along the axis of the mirror. To find : The correct graphical representation of the magnitude m of linear magnification by the mirror versus distance x of the object from the mirror. For a concave mirror : a. The magnification for an object placed at focus should be infinite and b. The magnification for an object placed at the centre of curvature of mirror (2f) should be 1. Both the above points are satisfied by graph d. Ans. 11. Option (b) is correct. Given : Lower half of vessel of depth h is filled with liquid of refractive index n1 = 2 2 , upper half of vessel of depth h is filled with liquid of refractive index n2 = 2 , the two liquids are immiscible. To find : h¢, the apparent depth of the bottom of the vessel.

Apparent depth of the bottom of the vessel from point A : n h 2 = h″ = h  2 = h n1 2 2 2 Let n = 1 be the refractive index of air. Apparent depth of the bottom of the vessel from point B : h  n 3  1  3 h′ =  + h  = h = h 2 2   n2 2  2  4

Minimum angle for clear resolution :

θ = 1.22

λ a

From the above diagram : d = θD d = 1.22

1.22 × 5500 × 10 −10 × 4 × 10 5 × 10 3 λ D= a 5

d ª 60m Ans. 13. Option (b) is correct. 4 , a light 3 source lies at depth h below the surface of water, reflection of light from the bottom surface and absorption of light by water is negligible. Given : Refractive index of water is n =

To find : The percentage of light that emerges out of the surface of water.

Let the angle of incidence for the light coming from source S, at the air water interface be i and qc be the critical angle for the interface. sin θ c =

1 3 7 = ,cosθ c = n 4 4

...(i)

For i £ qc, the light rays will escape into the air as shown in the figure. This portion of light will form a cone with half angle qc. For i > qc, the light rays will suffer total internal reflection.

329

OPTICS

The solid angle contained in the cone shown in the figure :  7 Ω = 2p (1 − cosθ c ) = 2p  1 −   4   = 0.677p

...(ii)

So, final image is formed at v3 = +40 cm in front of the lens, at the same position as object but real and inverted. The image is of same size as that of the object. Ans. 15. Option (d) is correct. x

Percentage of light that escapes from the liquid : Ω 0.677p × 100% = × 100% = 17% 4p 4p

2



d

40°

Ans. 14. Option (b) is correct f = 10cm

f = 10cm I2

object

40 cm

20 cm

I1

I3 40 cm

sin θ 2 =

Given : Distance of object from lens is u1 = 40 cm, focal length of the lens f1 = 20 cm, focal length of the mirror f2 = –10 cm, distance between the mirror and the lens d = 60 cm. To find : Location and magnification of the final image. Location (v1) and magnification (m1) of the image from convex lens : 1 1 1 1 1 1 = = = ,v1 = 40 cm , v1 f1 u1 20 40 40 m1

v =  1 = 1. u1

The first image is formed at v1 = 40 cm behind the lens and is of same size as object but inverted (I1), which acts as an object for concave mirror. For second image : u2 = -20 cm ,f2 = -10 cm. Location of image from concave mirror : 1 1 1 1 1 1 = = = - , v2 f2 u2 -10 -20 20 v2 = -20 cm ,m2 =

v2 = 1. u2

So, second image is formed at v2 = –20 cm in front of concave mirror, which is same size as object but inverted (I2), which acts as object for lens. For image three (final image I3), u3 = 40 cm, f1 = 20 cm. Location of final image : 1 1 1 1 1 1 = = = , v3 f1 u3 20 40 40 v3 = 40 cm , m3 =

Given : Length of optical fibre, l = 2 m, diameter of the fibre, d = 20 mm, angle of incidence at the entrance of fibre q1 = 40°, refractive index of fibre, m2 = 1.31. To find : Total number of reflections within the fibre. Snell’s law : sin θ 2 µ1 =  , ( m1 is refractive index of air) µ2 sin 40

u3 = 1. v3

1 sin 40 , 1.31

θ 2 = 29.39 θ = 90 - θ 2 = 90 - 29.39 = 60.6 (q is as labelled in the diagram) x tan θ = tan 60.6 = d (x is labelled in the diagram) x = ( 20 ¥ 10 -6 ) tan 60.6 = 35.5 mm Total number of reflections; l 2 m n = = = 56338. x 35.5 mm Ans. 16. Option (c) is correct. 30 cm O x

80 cm x

fL=20

f=?

Given : Focal length of the convex lens, fL = 20 cm, distance between the lens and the mirror d = 80 cm, distance of the object from lens u1 = 30 cm. To find : d, the maximum distance of an object from mirror for which the mirror itself can produce a virtual image. Distance of image from lens : 1 1 1 1 1 = = - ,v1 = 60 cm. v1 fL u1 20 30 The image is v1 = 60 cm behind the lens and 20 cm in front of the mirror. This image acts as an object for the mirror. So, for mirror : u2 = –20 cm. As the image formed by the mirror coincides with the image formed by the lens, that implies

330 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) u2 = R ; f =

PHYSICS

For virtual image :

R = 10 cm. 2

m = + 2 ,u= - x1 ,v = -2 x1

So, maximum distance of some object from concave mirror such that the mirror itself can make a virtual image is d = f = 10 cm. Ans. 17. Option (a) is correct. Given : Diameter of objective lens of a telescope, D = 200 cm, Wavelength of incident light, l = 500 nm. To find : Limit of resolution of the objective lens of the telescope.

1 1 1 1 1 1 -3 = + , = + = , f u v 20 - x1 -2 x1 2 x1 x1 = -30 cm. So, the ratio x1 : x 2  = 3 : 1. Ans. 21. Option (a) is correct. A'

-9

λ 500 ¥ 10 Dθ = 1.22 = 1.22 ¥ D 200 ¥ 10 -2

A

= 305 ¥ 10 -9 rad Ans. 18. Option (c) is correct. Given : Focal length of concave mirror f = –0.4 m, magnification of image is m = 5. To find : u, distance of object from mirror. Magnification : v m = - ,v = - mu = - 5u. u Mirror equation : 1 1 1 =  +, f u v 1 1 1 = + -0.4 u -5u -5 4 = 2 5u 8 u = = - . 0 32 m 25 Ans. 19. Option (c) is correct. Given : Diameter of objective lens of telescope D = 250 cm, wavelength of incident light, l = 600 nm. To find : Limit of resolution of the telescope, qR. By Rayleigh’s criteria : 1.22 λ  θR = D

θR =

1.22 ¥600 ¥ 10 250 ¥ 10 -2

-9

= 3 ¥ 10 -7 rad

Ans. 20. Option (b) is correct. Given : Focal length of convex lens is f = 20 cm. To find : x1 : x2 the ratio of two distances x1 and x2 of an object from the lens for which the magnification is m = 2, x1 > x2. Magnification : -v m = ; v = - mu. u For real image : m = -2 , u = - x 2 , v = 2 x 2 -1 1 1 1 1 1 1 = + , = + = , f u v 20 - x 2 2 x 2 2 x 2 x 2 = -10 cm.

L M O

Given : Refractive index of convex lens, m = 1.5, OA = 18 cm is such that image of the object placed at A coincides with itself, OA’ = 27 cm is such that image of the object placed at A’ coincides with itself if the volume between the lens and mirror is filled with some liquid. To find : ml, refractive index of the liquid. For the image and object to coincide at A, in the above assembly, the rays refracted from lens should fall normally on the mirror. That implies distance OA is the focal length (f1) of the lens. Again with liquid between the lens and the mirror, the equivalent focal length (feq) is 27 cm. Let the focal length of the lens formed by the liquid (concave) = f2. 1 = feq

1 1 + f1 f2

1 1 1 = f2 27 18 f2 = -54 cm. Lens maker’s formula : 1 Ê 2ˆ = ( m - 1) Á ˜ Ë R¯ f1 (1.5 - 1) 2 1 = ; R = 18 cm. 18 R (R = radius of curvature of the lens) For liquid lens : 1 1 Ê -1 ˆ = ( ml - 1) Á ˜ = Ë R ¯ -54 f2 1

( ml - 1) ÊÁË - 18 ˆ˜¯

= -

ml =

1 54

1 4 +1= . 3 3

331

OPTICS

Ans. 22. Option (a) is correct.

AOB = OA + mOB R

air 



air

Given : The radius of curvature for both the lenses = R, the index of refraction for Plano-convex lens = m1, the index of refraction for Plano-concave lens = m2. To find : The focal length of the combination of lenses shown in the figure above. Focal length of the Plano-convex lens, f1 : 1 = f1

Ê 1 1 ˆ ( m1 - 1) Á R - R ˜ Ë 1 2¯

Put R1 = ∞, R2 = –R for Plano-convex lens. R f1 = m1 - 1

3

m

c

v a

b

Lens equation : 1 1 1 = v u f 1-

Ê 1 1 ˆ ˜ R Ë 1 2¯

( m2 - 1) Á R

v v = u f m = 1-

v f

Put R1 = –R, R2 = ∞ for Plano-concave lens. -R f2 = m2 - 1

From the given graph, at v = a :

The net focal length of the combination, f : 1 1 1 m1 - 1 m 2 - 1 = + = + R f f1 f2 -R

At v = (a + b) :

m1 - m 2 1 R = ;f=  f R m1 - m 2 Ans. 23. Option (d) is correct.

vacuum glass

b

30° B

Given : Angle of incidence at glass slab = 60o, angle of refraction inside glass slab = 30o. To find : Optical path travelled by the ray AOB = OA + mOB. (m is the refractive index of the glass slab.) a OA = = 2a cos 60 OB = m =

b 2 = b cos 30  3 sin 60 = 3 sin 30

a f

m = m2 = 1 -

a+b f

f =

A 60° O

m = m1 = 1 -

Also, m2 – m1 = c Ê a + bˆ Ê aˆ ÁË 1 - f ˜¯ - ÁË 1 - f ˜¯ = c

normal a

b = 2 a + 2b

Ans. 24. Option (d) is correct. Given : Variation of magnification m produced by a thin lens, with image distance v. To find : f, the focal length of the lens.

Focal length of the Plano-concave lens, f2 : 1 = f2

2

= 2a + 3

b c

Ans. 25. Option (a) is correct. Given : Numerical aperture of an objective lens of telescope is NA = 1.25, wavelength of incident light is l = 5000 Å. To find : Minimum separation between two resolved points, d =

0.61λ 0.61 ¥ 5000 ¥ 10 -10 = = 0.24 mm NA 1.25

Ans. 26. Option (c) is correct. Given : distance of the object from the mirror is u = –5 cm, focal length of the concave mirror is R -40 = = -20 cm, refractive index of water is 2 2 m1 = 1.33. To find : d, distance of the final image from the surface of the water. Mirror equation : 1 1 1 =  + , f u v f=

332 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (v = the distance of the image from the mirror) 1 1 1 = - , v f u 1 1 1 3 =  = cm, v -20 -5 20 20 v = cm. 3 20 cm behind the mirror. So, image is formed at v = 3 The distance of the image from the surface of water is 20 35 d ' = 5 + cm= cm. 3 3 35 cm 3 below the surface of the water is refracted at the surface of water, such that the final image is seen at d. Light coming from this image formed d ¢ =

Êm ˆ d = d 'Á 2 ˜ , Ë m1 ¯

So, for total internal reflection : Ê m2 ˆ θ < sin -1 Á 22 - 1˜ Ë m1 ¯

Ans. 27. Option (c) is correct

Ê 1ˆ 1 ˆ 1 Ê 1.5 = cm. x = Á1 - ˜ t = Á1 Ë m¯ 1.5 ˜¯ 2 Ë

E

C



 D

A

Given : Dimension of cube is d, refractive index of cube is m2, refractive index of liquid = m1, m2 > m1. To find : q, the angle of incidence at the face AB of cube. Let the angle made by incident ray at E with the surface normal = a. For total internal reflection at face BC, m sin α = 1 m2 Snell’s law at point of incidence : m sin θ = 2 sin (90 - α ) m1

m m sin θ =  2 cosα =  2 (1 - sin 2 α )1 / 2 m1 m1

m Ê m2 ˆ =  2 Á 1 - 12 ˜ m1 Ë m2 ¯ Ê m2 ˆ sin θ = Á 22 - 1˜ Ë m1 ¯

95 = 10.55 cm. 9

d = v¢ - v = 10.55 - 10 = 0.55 cm

 

1 = 9.5 cm. 2

The new position of the image : 1 1 1 1 1 9 = - = = cm v¢ f u¢ 5 9.5 95 v¢ =

B

1/2

Ans. 28. Option (d) is correct. Given : The distance of the light source from the lens is u = 10 cm, distance of image from the lens is v = 10 cm, refractive index of the block m = 1.5, thickness of the block t = 1.5 cm. To find : d, the shift in position of the image after introducing the block. v = u, implies the object is placed at the centre of curvature of the lens, u = 2f; f = 5 cm. Let the shift in apparent position of the object after introducing the block due to refraction be x. The new position of the object will be u¢ = u – x

u¢ = 10 -

(m2 = 1 is the refractive index of air) 35 Ê 1 ˆ d =  = 8.77 = 8.8 cm 3 ÁË 1.33 ˜¯

PHYSICS

1/2

1/2

So, the image is shifted 0.55 cm away from the lens. Ans. 29. Option (b) is correct. Given : The ray of light incident on mirror M1 is parallel to mirror M2 and the ray of light reflected by M2 is parallel to M1. To find : q, the angle between the two mirrors. M2 A N2 C

180°–2 

 E

180°– 90°– N1   B

 D

M1

N1 is normal to mirror M1 and N2 is normal to mirror M2. After considering geometrical laws and laws of reflection we know the angles as labelled in the diagram above. In ΔBCE, 2 (90 - θ ) + 2 (180∞ - θ ) + θ  = 180∞

θ = 60∞. Ans. 30. Option (b) is correct. Given : Refractive index of Plano-convex lens is m1, focal length of Plano-convex lens is f1, refractive index of Plano-concave lens is m2, focal length of

333

OPTICS

Plano-concave lens is f2, radius of curvature of both the lenses = R, f1 = 2f2, both the lens are kept in contact. To find : Relation between m1 and m2. f1 = 2 f2 ;

1 1 = f1 2 f2

...( i )

( m1 - 1) Á R

1 dv v 2  dt

+

1 du u2  dt

;

dv v 2 du = 2 dt u dt

...(iii)

1 = f2

( m2 - 1) ÁÁ

Ê 1 Ë R1¢

-

Ê fu ˆ 1 du dv f2 du  = Á = ˜ 2 dt Ë f + u ¯ u dt ( f + u)2 dt

1 ˆ ˜ R¢ ˜¯

dv 0.3 ¥0.3 = ¥ 5 = . 1 16 ¥ 10 -3 m / s  dt ( 0.3 - 20 )2

2

n putting R1 = ∞, R2 = –R, R¢1 = –R, R¢2 = ∞, we O get, m -1 1 1 Ê 1ˆ = 1 = ( m 2 - 1) Á - ˜ ...(ii) ; Ë R¯ f1 R f2 Putting (ii) in (i);

So, the image is moving with velocity of dv = 1.16 ¥ 10 -3 m / s dt towards the lens. Ans. 33. Option (a) is correct.

m1 - 1 1 Ê 1ˆ = m 2 - 1) Á - ˜ ( Ë R¯ R 2

Given : Variation of refractive index n2 of a thin crown glass prism with wavelength of incident light.

m2 - 1 2

2 m1 - m 2 = 1. Ans. 31. Option (d) is correct. Given : Radius of curvature of eye R = 7.8 mm, the refractive index of medium inside eye, me = 1.34. To find : v, distance of the image formed by eye for an object at u = –∞. Relation between object, image and radius of curvature will be : me ma m - m a - = e  v u R (ma = 1 is the refractive index of air). 1.34 1 1.34 - 1 = v • 7.8

To find : Variation of minimum angle of deviation, Dm of refracted light through crown glass thin prism as function of wavelength of incident light. Minimum angle of deviation : D m = (n2 - 1) A

dv , the speed and direction of motion of dt

image. Lens formula 1 1 1 = - ; f v u

1 1 1 f +u fu = + = ;v= v f u fu f +u

...(i) ...( ii )

(Ais the angle of prism)

As Dm ∝ n2 and n2 decreases with increase in wavelength, we expect similar behaviour for Dm. So, option (a) is correct. Ans. 34. Option (c) is correct. Given : Refractive index of prism is n = 3 , prism angle is A = 60°. To find : d, the angle of minimum deviation. Prism formula : Ê A + δ ˆ sin Á Ë 2 ˜¯  n = Ê Aˆ sin Á ˜ Ë 2¯

1.34 ¥7.8 = 30.7 mm = 3.1cm. 0.34

Ans. 32. Option (a) is correct. Given : Distance of object from convex lens is u = 20 m, focal length of the lens is f = 0.3 m, speed of the object moving away with respect to the lens du is =5m / s. dt To find :

0 = -

2

Ê 1 1 ˆ Ë 1 R 2 ˜¯

1 = f1

v =

Differentiating the lens formula :

Putting value of v from (ii) in (iii), we get;

Lens Maker’s formula :

m1 - 1 =



Ê 60 + δ ˆ sin Á Ë 2 ¯˜ 3 =   Ê 60 ˆ sin Á ˜ Ë 2¯ 

3 Ê 60 + δ ˆ =  sin Á  Ë 2 ˜¯ 2

Ê 60 + δ ˆ sin 60 = sin Á  Ë 2 ˜¯ 60 + δ 60 =   2 δ = 60∞

334 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 35. Option (d) is correct.

From property of plane mirror the image S’ lies distance L behind the mirror. Δ S’PO and ΔS’AM are similar triangles. S ¢O PO So, = S ¢M AM

2 cm A

B

O f1 = +5 cm

20 cm

f2= –5 cm

Given : For lens A : u1 = 20 cm, f1 = +5 cm, for lens B : f2 = 5 cm, distance between two lenses is d = 2 cm. To find : The position and nature of final image from the combination of lenses. Lens formula (lens A) : 1 1 1 1 1 3 = = = v1 f1 u1 5 20 20 v1 =

20 cm 3

So, the image from A is formed at v1 =

20 cm to the 3

right of A and is inverted. For lens B : The image from A acts as an object, 20 14 u2 = -2= cm. 3 3 1 1 1 1 1 = = v2 f2 u2 5 14 3 v2 = -70 cm So, the image from lens B is formed to the right of B at a distance v2 = 70 cm from B and is real in nature. Ans. 36. Option (c) is correct. Given : Distance of the point source S from the plane mirror is L, distance of the man M from the mirror is 2L, width of the mirror PQ is d. To find : AB, the distance over which the man can see the image S’. As light travels in straight line, following will be the ray diagram. A

S'

d/2

O L L d/2 Q

That implies,

L d/2 =  3L AM  AM =

3d ; AB = 2 AM = 3d. 2

Ans. 37. Option (a) is correct. Given : The set up consisting of a biconvex lens is immersed in water. To find : What change one observes on the screen. As the lens goes in water its focal length will change and so will the image position. So the image at the screen will disappear. Ans. 38. Option (b) is correct. Given : Focal length of Plano-convex lens is f2, refractive index of Plano- convex lens is m2, focal length of Plano-concave lens is f1, refractive index of Plano- concave lens is m1, radius of curvature of both the lenses is R. To find : The focal length of the combination of lenses. Using lens maker’s formula for Plano-concave lens (R1 = –R, R2 = ∞) : Ê 1 1 1 ˆ = ( m1 - 1) Á f R ˜¯ ËR 1

?

P S

PHYSICS

M

1 = f1

1

2

1

1

( m1 - 1) ÊÁË - R - • ˆ˜¯

1 -m1 1 = f1 R Using lens maker’s formula for Plano-convex lens (R1 = ∞, R2 = –R) : 1 1 ˆ Ê1 = ( m 2 - 1) Á Ë • - R ˜¯ f2 m -1 1 = 2 f2 R Equivalent focal length would be : 1 1 1 = + feq f1 f2 1 - m1 m 2 - 1 1 =  +  fe q R R

L ?

B

feq =

R m 2 - m1

335

OPTICS

Topic-2

Wave Optics Concept Revision (Video Based) Polarization

Huygens Principle Diffraction due to single slit         

  

    

    

Part -1

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (l = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to : (a) 2.87 (b) 2 nm (c) 1.27 mm (d) 2.05 mm [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be : (a) 24 (b) 18 (c) 28 (d) 30 [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.3. In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to : (a) 0.17° (b) 0.07° (c) 0.57° (d) 1.70° [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.4. A beam of plane polarised light of large crosssectional area and uniform intensity of 3.3 Wm–2 falls normally on a polariser(cross sectional area 3 × 10–4 m2) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to : (a) 1.0 × 10–4 J (b) 5.0 × 10–4 J –5 (c) 1.0 × 10 J (d) 1.5 × 10–4 J [JEE (Main) - 4th Sep. 2020 - Shift-1] Q.5. Two coherent sources of sound, S1 and S2, produce sound waves of the same wavelength l = 1m, in

phase. S1 and S2 are placed 1.5 m apart (see fig). A listener, located at L, directly in front of S2 finds that the intensity is at a minimum when he is 2m away from S2. The listener moves away from S1, keeping the distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then d is :

(a) 3m (c) 12 m

(b) 5m (d) 2m [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.6. In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5m and the phase of P is ahead of that of Q by 90°. A, B and C are three distinct point of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :

(a) 0 : 1 : 2 (c) 0 : 1 : 4

(b) 4 : 1 : 0 (d) 2 : 1 : 0 [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.7. A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the

336 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is  : (a) 71.6° (b) 90° (c) 18.4° (d) 45° [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.8. Visible light of wavelength 6000 × 10–8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at q1, then q1 is close to : (a) 45° (b) 30° (c) 25° (d) 20° [JEE (Main) – 7th Jan. 2020 - Shift-1] Q.9. In a Young’s double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is : (a) 4.9 mm (b) 6.9 mm (c) 5.9 mm (d) 3.9 mm [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.10. In a double-slit experiment, at a certain point on the screen the path difference between the two 1 interfering waves is th of a wavelength. The 8 ratio of the intensity of light at that point to that at the centre of a bright fringe is : (a) 0.568 (b) 0.853 (c) 0.760 (d) 0.672 [JEE (Main) – 8th Jan. 2020 - Shift-2] Q.11. In an interference experiment the ratio of a 1 amplitudes of coherent waves is 1 = . The ratio a2 3 of maximum and minimum intensities of fringes will be : (a) 2 (b) 18 (c) 4 (d) 9 [JEE (Main) – 8th April 2019 - Shift-1] Q.12. The figure shows a Young’s double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index m is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light used is l, t will be :

a

= =

PHYSICS

Q.13. In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be : (a) 25 : 9 (b) 9 : 1 4 (d) ( 3 + 1) : 16 th [JEE (Main) – 10 April 2019 - Shift-2] Q.14. In a double slit experiment, when a thin film of thickness t having refractive index m is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is (l is the wavelength of the light used) : λ 2λ (a) (b) 2 ( m - 1) ( m - 1)

(c) 4 : 1

(c)

λ ( m - 1)

(d)

λ ( 2 m - 1)

[JEE (Main) – 12th April 2019 - Shift-1] Q.15. A system of three polarizers P1, P2, P3, is set up such that the pass axis of P3 is crossed with respect to that of P1. The pass axis of P2 is inclined at 60° to the pass axis of P3. When a beam of unpolarized light of intensity I0. is incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (I0/I) equals (nearly) : (a) 5.33 (b) 16.00 (c) 10.67 (d) 1.80 [JEE (Main) – 12th April 2019 - Shift-2] Q.16. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio : (a) 16 : 9 (b) 25 : 9 (c) 4 : 1 (d) 5 : 3 [JEE (Main) – 9th Jan 2019 - Shift-1] Q.17. Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index m. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of m is i

n=1.5

screen

D

(a)

2nDλ a ( m - 1)

(b)

nDλ a ( m - 1)

(a)

(c)

Dλ a( m - 1)

(d)

2 Dλ a( m - 1)

(c)

[JEE (Main) – 8th April 2019 - Shift-1]

3 5 4 3



(b) (d)

5 3 5 3

[JEE (Main) – 9th Jan 2019 - Shift-1]

337

OPTICS

Q.18. In a Young’s double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength l = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30° £ q £ 30° is : (a) 640 (b) 320 (c) 321 (d) 641 [JEE (Main) – 9th Jan 2019 - Shift-2] Q.19. In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe 1 at angle rad by using light of wavelength l1. 40 When the light of wavelength l2 is used a bright fringe is seen at the same angle in the same set up. Given that l1 and l2 are in visible range (380 nm to 740 nm), their values are : (a) 625 nm, 500 nm (b) 380 nm, 525 nm (c) 380 nm, 500 nm (d) 400 nm, 500 nm [JEE (Main) – 10th Jan 2019 - Shift-1] Q.20. Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength l such that the first minima occurs directly in front of the slit (S1)? S1 source

1st minima

16. (b) 20. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : In a double slit experiment the distance between two slits is d = 1 mm, wavelength of incident light is l = 632.8 nm, distance between the slits and the screen is D = 100 cm, on screen the distance between a bright fringe and the central bright fringe is y = 1.27 mm. To find : D, the path difference between the waves forming the bright fringe at y = 1.27 mm. ∆=

dy 1 × 10 −3 × 1.27 × 10 −3 = = 1.27 µ m D 100 × 10 −2

Ans. 2. Option (c) is correct. Given : Case 1 : In Young’s double slit experiment, number of fringes observed on a segment of screen is n1 = 16 for incident wavelength l1 = 700 nm,

wavelength is now l2 = 400 nm.

λ 2 ( 5 - 2)

λ (c) 2 (5 - 2 )

To find : n2, number of fringes observed on the screen in case 2.

Screen

(b)

In case 1 :

λ ( 5 - 2)

Path difference between the two interfering waves, for the n1th fringe, on the screen:

λ (d) (5 - 2 )

∆1 = n1λ1

[JEE (Main) – 10th Jan 2019 - Shift-2] Q.21. In a Young’s double slit experiment, the path difference at a certain point on the screen between 1 two interfering waves is th of wavelength. The 8 ratio of the intensity at this point to that at the centre of a bright fringe is close to : (a) 0.94 (b) 0.80 (c) 0.74 (d) 0.85 [JEE (Main) – 11th Jan 2019 - Shift-1] Q.22. In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 mm and a width of 4.05 mm. The number of bright fringes between the first and the second diffraction minima is : (a) 10 (b) 05 (c) 04 (d) 09 [JEE (Main) – 11th Jan 2019 - Shift-2]

ANSWER – KEY 1. (c) 5. (a) 9. (c)

15. (c) 19. (a)

same segment of screen as in case 1, the incident

d 2d



14. (c) 18. (d) 22. (b)

Case 2 : In the same experimental set up and the

P

S2

(a)

13. (b) 17. (a) 21. (d)

2. (c) 6. (d) 10. (b)

In case 2 : Path difference between the two interfering waves, for the n2th fringe, on the screen: ∆ 2 = n2 λ2 As we are observing the same segment of screen in both the cases : n1λ1 = n2 λ2 n2 = n1

Ans. 3. Option (c) is correct. Given : In Young’s double slit experiment, wavelength of incident light is l = 500 nm, the distance between the two slits is d = 0.05 mm. To find : q, the angular width of the fringes formed on the distant screen.

θ = 3. (c) 7. (c) 11. (c)

4. (d) 8. (c) 12. (b)

λ1 700 = 16 × = 28 λ2 400

λ 500 × 10 −9 = = 0.01radians d 0.05 × 10 −3

θ = 0.01 ×

180 = 0.57° p

338 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Ans. 4. Option (d) is correct. Given : Intensity of a plane polarized light incident W normally on a polariser is I = 3.3 2 , cross sectional m area of the polariser is A = 3 × 10–4 m2, angular speed of the polarizer is w = 31.4 rad/s. To find : E, the energy of light passing through the polariser per revolution.

To find : IA : IB : IC, the ratio of intensities measured at points A, B and C.

Time period of revolution of the polariser :

Phase difference at point A :

T =

2p 2p = = 0.2 s ω 31.4

φA = φi +

Energy of light passing through the entire cross sectional area of the polariser per revolution : E = IAT = 3.3 × 3 × 10

−4

−4

× 0.2 ª 2 × 10 J

Ans. 5. Option (a) is correct. Given : Sources S1 and S2 are coherent sources of sound, distance between the two sources is D = 1.5 m wavelength of sound emitted by the two sources is l = 1 m, Case 1 : At location L, a listener observes an intensity minimum Case 2 : At location O, he observes an adjacent intensity maximum.

2p p 2p × (QA − PA ) = + × ( −5) = 0 2 20 λ

Phase difference at point B :

φB = φi +

p 2p p 2p × (QB − PB) = + × (0) = 2 20 2 λ

Phase difference at point C:

φC = φi +

2p p 2p × (QC − PC ) = + × ( 5) = p 2 20 λ

Let intensity emitted by both the sources be Io. Intensity at A : I A = I o cos2

φA = Io 2

Intensity at B : I B = I o cos2

φB I o = 2 2

Intensity at C : IC = I o cos2

φC =0 2

I A : I B : IC = 2 : 1 : 0 Ans. 7. Option (c) is correct. To find : d, the distance S1O. Path difference at location L : S1 L − S 2 L =

2 2 + (1.5)2 − 2 = 0.5 m

Path difference at location O : S1O − S 2O = d − 2 As the listener observes just the adjacent maximum at location O as compared to location L,

λ d − 2 = 0.5 + 2 d = 3m Ans. 6. Option (d) is correct. Given : P and Q are two equally intense coherent sources, distance between the two sources is d = 5m, wavelength of radiation emitted by the two sources is l = 20 m, phase of P is ahead of that p of Q by ∆φi = , points A, B and C are equidistant 2 from the mid point of PQ.

Given : Intensity incident on a polarizer-analyser set is IO, intensity emerging from the polarizeranalyser set is I = 0.1IO To find : q, the angle through which the analyser is to be rotated such that the emerging intensity becomes I¢ = 0. Intensity emerging from the polarizer-analyser set : I = IO cos2 φ = 0.1 IO cosφ = 0.316

φ = 71.58°

...( i )

In equation (i), f is the angle between the optic axis of the polarizer and the analyser. For the emerging intensity from the polarizeranalyser set to be zero : I′ = IO cos2 φ ′ = 0

φ ′ = 90° From equations (i) and (ii) : θ = φ ′ − φ = 90 − 71.58 = 18.4°

...(ii)

339

OPTICS

Ans. 8. Option (c) is correct.

In Young’s double slit interference experiment,

Given : Wavelength of light incident on a single

intensity of light at some point P, when IO is the

slit arrangement is l = 6000 Å, second diffraction

intensity of the central maxima is :

minimum occurs at q2 = 60° from the central

 ∆φ  I P = I0 cos2    2 

maximum. To find : q1, the angle for the first diffraction

IP = IO

minimum. Condition for nth minima in a single slit diffraction

p 0.853 c= os2 8

experiment, when d is the slit width : d sin θ n = nλ

...(i)

For second diffraction minima, equation (i) becomes  :

Ans. 11. Option (c) is correct. Given : The ratio of amplitudes of two interfering waves is a1 1 = a2 3

d sin θ 2 = 2λ d sin 60 = 2 × 6000 × 10 −10 −6

d = 1.38 × 10 m For first diffraction minima, equation (i) becomes : d sin θ1 = λ 1.38 × 10

−6

sin θ1 = 6000 × 10

−10

θ1 = 25° Ans. 9. Option (c) is correct. Given : In Young’s double slit interference

To find :

...(i)

Imax ,the ratio of maximum and minimum Imin

intensities of fringes in the interference pattern. Let intensity corresponding to amplitude a1 be I1 and intensity corresponding to amplitude a2 be I2. Then the ratio of maximum to minimum intensity in the interference pattern will be : ( I1 + I 2 )2 I1 + I 2 + 2 I1I 2 Imax = = Imin ( I1 - I 2 )2 I1 + I 2 - 2 I1I 2

experiment separation between two slits is d = 0.15

As the intensity is directly proportional to square of

mm, wavelength of light incident on the double slit

amplitude, putting I1 = a12 , I 2 = a22 .

arrangement is l = 589 nm, the distance between the screen and the slits is D = 1.5 m. To find : w, the fringe width. The fringe width in the Young’s double slit interference experiment is : w=

λ D 589 × 10 −9 × 1.5 = = 5.9 mm d 0.15 × 10 −3

Ans. 10. Option (b) is correct.

a 2 + a22 + 2 a1a2 Imax =  12 Imin a1 + a22 - 2 a1a2 Now, putting a2 = 3a1, 16 a12 4 Imax = = = 4 :1 Imin 1 4 a12 Ans. 12. Option (b) is correct. Given : In a double slit experiment a thin film is

Given : In a double slit interference experiment the

introduced in front of one of the slits, thickness

path difference between two interfering waves at

of film is t, refractive index of film is m, after

λ some point P on the screen is ∆ = . 8 I To find : P , the ratio of intensity of light at point IO P to the intensity of light at the centre of the bright fringe. The phase difference between two interfering waves at point P : ∆φ =

introduction of the film the maximum at the centre of the fringe pattern shifts by n fringe widths. To find : The value of t in terms of wavelength l. Path difference between two interfering rays, after introduction of thin film : D = ( m - 1)t

...(i)

Fringe width : 2p 2p λ p ×∆ = × = λ λ 8 4

β =

λD a

...(ii)

340 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Width of n fringes : nλ D β¢ = a

...( iii )

Intensity emerging from Polarizer-1 : I1 =

nλ D a ( m - 1)

I 2 = I1 cos2 θ 2 .

Given : In Young’s double slit experiment ratio of widths of slits is w1 : w2 = 4 : 1 I To find : max , ratio of maximum to minimum Imin intensity in the interference pattern.

I2 =

polarizer 3 :

Let, I1 and I2 be the intensities coming from two slits, such that I1 = 4 I2. (as w1 = 4 w2).

I =

I = I 2 cos2 θ1 .

Then the ratio of maximum to minimum intensity in the interference pattern will be : ( I1 + I 2 )2 w1 + w2 + 2 w1w2 Imax = = Imin ( I1 - I 2 )2 w1 + w2 - 2 w1w2

3 3 I0 cos2 60 = I0 8 32

I0 32 = = 10.67. I 3 Ans. 16. Option (b) is correct. Given : In an interference pattern,

Imax 9 = = 9 :1 Imin 1

Imax = 16 Imin

Ans. 14. Option (c) is correct. Given : In a double slit experiment a thin film is introduced in front of one of the slits, thickness of film is t, refractive index of film is m, after introduction of the film the maximum at the centre of the fringe pattern shifts by one fringe width.

To find :

I1 , the ratio of intensities of the interfering I2

waves. ( I1 + I 2 )2 Imax = = 16 Imin ( I1 - I 2 )2

To find : The value of t in terms of wavelength l.

I1 + I 2

Path difference between two interfering rays, after introduction of thin film :

I1 - I 2

...( i )

Path difference associated with shift of one fringe at the centre of fringe pattern : ...(ii)

Equating equations (i) and (ii), t =

I0 3 cos2 30 = I0 2 8

Again, by Malus law, intensity obtained from

In the double slit experiment, the amplitude at a point on screen is : a ∝ w1/2 and intensity is : I ∝ a2 that implies I ∝ w.

λ

I0 2

By Malus law, intensity obtained from Polarizer-2 :

Ans. 13. Option (b) is correct.

D = ( m - 1) t

I0 . I

To find : Ratio

Equating equations (i) and (iii), t =

PHYSICS

λ . m - 1

Ans. 15. Option (c) is correct. Given : Polarizer-1 is at crossed position with polarizer-3, pass axis of Polarizer-2 makes an angle of q1 = 60° with the pass axis of polarizer 3, which means it makes an angle of q2 = 30° with pass axis of Polarizer-1, intensity of unpolarized incident beam falling on Polarizer-1 is I0, intensity of beam transmitted from polarizer-3 = I.

= 4

I1 25 = I2 9 Ans. 17. Option (a) is correct. Given : The refractive index of the glass tank is mglass = 1.5, the refractive index of the liquid filled in the glass tank is m, the light reflected from the liquid-glass interface at the bottom of the tank is never completely polarised. To find : Minimum value of m. As, the light reflected from the liquid-glass interface at the bottom of the tank is never completely polarised : ic > iB

...(i)

341

OPTICS

ic = sin -1

1 , m

...(ii)

is the critical angle at the top surface of the tank and iB = tan -1

mglass

...(iii)

m

Condition for maxima : D = nλ

...(i)

(D is the path difference between the interfering rays) P

is the Brewster’s angle. From equation (i) : sin ic < sin iB

S1

...(iv)



d

Substitute from equations (ii) and (iii) in equation (iv).

S2

1 < m



O

L



1.5 (1.5)2 + m 2

That gives, Also, from diagram :

3

m
n1), as shown in the figure. A ray of light is incident with angle qi from medium 1 and emerges in medium 2 with refraction angle qf with a lateral displacement l.

Sol. Given : Refractive index of air is m1 = 1, refractive index of oil drop is m 2 =

7 , radius of curvature of 4

oil drop is R = 6 cm, refractive index of water is m3 =

4 , depth of water tank is 18 cm, distance of 3

object from water surface is u = 24 cm. To find: location of image, x above the bottom of the tank. Two refractions will take place, one at the spherical surface (air-oil interface) and other at the plane surface (oil water interface). For interface 1 :

n1=constant z

i

7 7 -1 m 2 m1 m 2 - m1 4 1 ; ; = = 4 v1 u R v1 -24 6

1

n(z)

n2=constant

d

v1 = 21cm. The image is formed at v1 = 21 cm below the water

l

2 f



Apply Snell’s law and give relation between n1 and n2. Discuss dependence of the lateral displacement, l of the ray, on various factors. Sol. Given : Refractive index of slab is n(z) thickness of slab is t, angle of incidence at slab is qi, angle of emergence of ray from the slab is qf, n2 > n1. To find: dependence of the lateral displacement, l of the ray, on various factors.

surface, which acts as an object for second interface. For interface 2 : 4 7 4 7 m3 m 2 m3 - m 2 3 4 ; = = 3 4; v2 v1 R v2 21 • v2 = 16 cm. The image is formed at v2 = 16 cm below the water surface or x = 18 – 16 = 2 cm above the bottom of the tank.

347

OPTICS

the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1, S2.

•

•

S1

P1

z O

S2

screen

Q.15. Two coherent monochromatic point sources S1 and S2 of wavelength l = 600 nm are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance d = 1.8 mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Dq. Calculate the total number of fringes between P1 and P2.

y

x

d D 

•

•

S1

S2

P2

d

Sol. Given : Distance between two coherent sources is d = 1.8 mm, wavelength of incident light is l = 600 nm, the angular separation between two consecutive bright spots is Dq. To find : The total number of fringes between P1 and P2. Path difference at P1 : D 1 = S1P1 - S 2 P1 = 0. Path difference at P2 : D 2 = S1P2 - S 2 P2 = d = 1.8 mm.

Discuss the profile of interference pattern generated at the screen. Sol. Given : Wavelength of incident light is l = 600 nm, distance from two coherent sources to screen is D = 3 m, distance between two coherent sources is d = 0.6003 mm. To find : The profile of interference pattern generated at the screen. Path difference at O : D = d= . 0 6003 mm. Using value of l and D, we can also write, D =

As Δ is odd multiple of

λ , point O will appear 2

dark. That is a minima.

Using value of l and d, we can write : λ D 2 = 2 ( 3000 ) . 2 So, order of interference at P2 is 3000 and we observe a maxima at P2. Similarly, order of interference at P1 is 0. So, total number of fringes in first quadrant is 3000, with maximum order at P2. Q.16. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to

( 2000 + 1) λ . 2

P d S1

S2

As we can see from the diagram, locus of points with same path difference from the sources will make a circle. So, fringes will be in form of concentric circles. But as only half of the screen is available, one could see only concentric semi circles.



Hore,

s linear momentum are called ''photons'' Energy of each c

Interference can be explained by wave nature is

.

a

e

v

θ=tan–1

hc e

348 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Dual Nature of Matter and Radiation

Chapter 17 Syllabus

Dual nature of radiation; Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; particle nature of light; Matter waves-wave nature of particle, de Broglie relation; DavissonGermer experiment; de Broglie wavelength of matter waves; Photoelectric effect

Concept Revision (Video Based) Dual Nature of Radiation

Photoelectric Effect     

                  Part -1 De-Broglie Relation      

  

Part - 2

Davisson-Germer Experiment

Part -1     Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.878 × 10–4. The mass of the particle is close to : (a) 1.2 × 10–28 kg (b) 9.7 × 10–28 kg –31 (c) 9.1 × 10 kg (d) 4.8 × 10–27 kg [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to : (a) 1.02 eV (b) 0.61 eV (c) 0.52 eV (d) 0.81 eV [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of the visible light of the given wavelengths is :

(a) 250

(b)

1 500

(c) 500

(d)

1 250

[JEE (Main) – 3rd Sep. 2020 - Shift-2] m Q.4. A particle of mass mA = moving along the 2 x-axis with velocity v0 collides elastically with m another particle B at rest having mass mB = . e If both the particles move along the x-axis after the collision, the change Dl in the wavelength of the particle A, in terms of its de-Broglie wavelength (l0) before the collision is : (a) Dl = 2l0 (b) Dl = 4l0 3 5 (c) Dl = l0 (d) Dl = l 2 2 0 [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.5. Given figure shows few data points in a photoelectric effect experiment for a certain metal. The minimum energy for ejection of electrons from its surface is : (Planck’s constant h = 6.62 × 10–34 J-s)

350 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) VStop (V)

Y

C B

(6, V)

(5.5, 0)

A 5

X

f(1014 Hz)

(b) 2.59 eV (d) 2.27 eV [JEE (Main) – 4th Sep. 2020 - Shift-1] Q.6. In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased : Vs

1/λ

(a) Slope of the straight line get more steep (b) Graph does not change (c) Straight line shifts to right (d) Straight line shifts to left [JEE (Main) – 4th Sep. 2020 - Shift-2] Q.7. An electron, a doubly ionized helium ion (He++) and proton are having the same kinetic energy. The relation between their respective de-Broglie wavelength le¢ lHe++ and lp is : (a) le < lp < lHe++ (b) le > lHe++> lp (c) le < lHe++ = lp (d) le > lp > lHe++ [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.8. Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to : (Given : nitrogen molecule weight : 4.64 × 10–26 kg, Boltzman constant : 1.38 × 10–23 J/K, Planck, constant : 6.63 × 10–34 J-s) (a) 0.34 Å (b) 0.44 Å (c) 0.20 Å (d) 0.24 Å [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.9. An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum) 1/2

1/2



1  2E  (b)  c  m 

1/2

1 E  (d)  c  2m  [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.10. When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie (c) c (2mE)1/2

0

0

is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by : λ0 λ0 (a) (b) 2 2 2 e E t e 2 E 2t 2 1 + 20 2 2+ 2 2 m v0 m v0

λ0

(c) 1+

θ

 E  (a)    2m 

wavelength lA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5)eV. If the de-Broglie wavelength of these photoelectrons lB = 2lA, then the work function of metal B is : (a) 1.5 eV (b) 4 eV (c) 3 eV (d) 2 eV [JEE (Main) – 8th Jan. 2020 - Shift-1] Q.11. An electron (mass m) with initial velocity    If l , v = v i + v j is in an electric field E = − E k. 0

(a) 2.10 eV (c) 1.93 eV

PHYSICS

2 2 2

e E t



(d)

λ0 2 1+

2m 2 v02

0

e 2 E 2t 2 m 2 v02

[JEE (Main) – 8th Jan. 2020 - Shift-2] Q.12. A particle moving with kinetic energy E has de Broglie wavelength l. If energy DE is added to its λ energy, the wavelength become . Value of DE, 2 is  : (a) 2E (c) 3E

(b) 4E (d) E [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.13. Radiation, with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to : (a) 0.8 eV (b) 1.1 eV (c) 1.8 eV (d) 1.6 eV [JEE (Main) – 9th Jan. 2020 - Shift-1] Q.14. An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is : −h | e |E t (a) (b) h | e |E t 2 h (d) − | e |E t

h (c) − | e |E t

[JEE (Main) – 9th Jan. 2020 - Shift-2] Q.15. Two particles move at right angle to each other. Their de-Broglie wavelengths are l1 and l2 respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength l, of the final particle, is given by : 1 1 1 (a) 2 = 2 + 2 (b) λ = λ1λ2 λ λ1 λ2 (c) λ =

λ1 + λ2 2

(d)

2 1 1 = + λ λ1 λ2

[JEE (Main) – 8th April 2019 - Shift-1]

351

DUAL NATURE OF MATTER AND RADIATION

Q.16. The electric field of light wave is given as  Ê 2p x ˆ E = 10 -3 cos Á - 2p ¥ 6 ¥ 10 14 t ˜ x N C Ë 5 ¥ 10 -7 ¯ This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : 12375 Given, E (in eV) = λ (in Å) (a) 2.0 V (b) 0.72 V (c) 0.48 V (d) 2.48 V [JEE (Main) – 9th April 2019 - Shift-1] Q.17. A particle ‘P’ is formed due to a completely inelastic collision of particles ‘x’ and ‘y’ having deBroglie wavelengths ‘lx’ and ‘ly’ respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of ‘P’ is : λx λy λx λy (a) (b) λx + λy λx - λy (c) lx – ly

(d) lx + ly

[JEE (Main) – 9th April 2019 - Shift-2] 2 Q.18. 50 W/m energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to (c = 3 × 108 m/s) : (a) 15 × 10–8 N (b) 20 × 10–8 N –8 (c) 10 × 10 N (d) 35 × 10–8 N [JEE (Main) – 9th April 2019 - Shift-2] Q.19. In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be : 1237 Given E (in eV) = λ (in nm) (a) 1.5 eV (c) 4.5 eV

(b) 3.0 eV (d) 15.1 eV [JEE (Main) – 10th April 2019 - Shift-1] Q.20. Light is incident normally on a completely absorbing surface with an energy flux of 25 Wcm–2. If the surface has an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be : (a) 6.3 × 10–4 Ns (b) 1.4 × 10–6 Ns –3 (c) 5.0 × 10 Ns (d) 3.5 × 10–6 NS [JEE (Main) – 10th April 2019 - Shift-2] Q.21. A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck’s constant h = 6.6 × 10–34 J-s, speed of light c = 3.0 × 108 m/s] (a) 1.5 × 1015 (b) 1.5 × 1016 16 (c) 2 × 10 (d) 1 × 1016 [JEE (Main) – 10th April 2019 - Shift-2] Q.22. The stopping potential Vo (in volt) as a function of frequency (n) for a sodium emitter, is shown in the

figure. The work function of sodium, from the data plotted in the figure, will be : (Given : Planck’s constant (h) = 6.63 × 10–34 J-s, electron charge e = 1.6 × 10–19 C) 3.0 2.0 Vo

1.0 2

(a) 1.82 eV (c) 1.95 eV

4

6 8 n(1014 Hz)

10

(b) 1.66 eV (d) 2.12 eV [JEE (Main) – 12th April 2019 - Shift-1] Q.23. Surface of certain metal is first illuminated with light of wavelength l1 = 350 nm and then, by light of wavelength l2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to : Ê ˆ 1240 ÁË Energy of photon = λ (in nm) eV˜¯ (a) 1.8 (b) 2.5 (c) 5.6 (d) 1.4 [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.24. The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ? (c = 3 × 108 ms–1, h = 6.6 × 10–34 J-s) (a) 6.82 eV (b) 12.5 eV (c) 8.52 eV (d) 7.72 eV [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.25. In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to : (a) 500 keV (b) 100 keV (c) 1 keV (d) 25 keV [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.26. A metal plate of area 1 × 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : [1 eV = 1.6 × 10–19 J] (a) 1014 and 10 eV (b) 1012 and 5 eV 11 (c) 10 and 5 eV (d) 1010 and 5 eV [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.27. If the de-Broglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of

352 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) frequency 6 × 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 × 108 m/s Planck’s constant = 6.63 × 10–34 J-s Mass of electron = 9.1 × 10–31 kg) (a) 1.45 × 106 m/s (b) 1.1 × 106 m/s 6 (c) 1.7 × 10 m/s (d) 1.8 × 106 m/s [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.28. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to :

Ê hc ˆ ÁË e = 1240 nm-V˜¯

(b) 1.5 V (d) 2.0 V [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.29. A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of deλA Broglie wavelengths is close to λB (a) 4.47 (c) 0.07

(b) 10.00 (d) 14.14 [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.30. In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to : (a) 1700 nm (b) 2020 nm (c) 220 nm (d) 250 nm [ JEE (Main) – 12th Jan. 2019 - Shift-2] Q.31. When a certain photosensistive surface is illuminated with monochromatic light of frequency n, the stopping potential for the photo current is – V0/2. When the surface is illuminated by monochromatic light of frequency n/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is : 4 5ν (a) (b) ν 3 3 3ν (d) 2

(c) 2n

[JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY 1. (b) 5. (d) 9. (d) 13. (b) 17. (b) 21. (a) 25. (d) 29. (d)

2. (b) 6. (b) 10. (b) 14. (a) 18. (b) 22. (b) 26. (c) 30. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. Given : Speed of an electron is ve = v, speed of a particle is vp = 5v, the ratio of de-Broglie wavelength of the particle to that of the electron is

λp : λe = 1.878 × 10 −4. To find : mp, the mass of the particle. h λp = mp v p

λe =

(a) 0.5 V (c) 1.0 V

PHYSICS

h me v e

...(i) ...(ii)

From equations (i) and (ii) : λp me v e = mp v p λe me v e = 1.878 × 10 −4 mp v p mp =

9.1 × 10 −31 × v 1.878 × 10 −4 × 5v

= 9.7 × 10 −28 kg

Ans. 2. Option (b) is correct. Given : Case 1 : Wavelength of radiation incident on the metal surface is l1 = 500 nm, maximum kinetic energy of the photoelectrons emitted from the metal surface is E1 = E, Case 2 : Wavelength of radiation incident on the metal surface is l2 = 200 nm, maximum kinetic energy of the photoelectrons emitted from the metal surface is E2 = 3E. To find : f0, the work function of the metal. According to Einstein’s equation for photoelectric effect : hc hc = E1 + φo ; = E + φo ...(i) λ1 λ1 hc hc = E2 + φo ;  = 3E + φo λ2 λ2

...(ii)

From equations (i) and (ii) :

φo =

1  hc  3  −  2  λ1 λ2 

φo =

6.6 × 10 −34 × 3 × 10 8 × 107  3 1  × −  2 5 2

φo = 0.99 × 10 −19 J = 0.62 eV 3. (b) 7. (d) 11. (c) 15. (a) 19. (a) 23. (a) 27. (a) 31. (d)

4. (b) 8. (d) 12. (c) 16. (c) 20. (c) 24. (d) 28. (c)

Ans. 3. Option (b) is correct. Given : Wavelength of light emitted by source-1 is l1 = 1 nm, wavelength of light emitted by source-2 is l2 = 500 nm, power emitted by both the sources is P1 = P2 = P = 200 W. To find : n1 : n2, the ratio of number density of photons emitted by source-1 to that emitted by source-2.

353

DUAL NATURE OF MATTER AND RADIATION

Power emitted by source-1 : hc P1 = P = n1 × λ1 Power emitted by source-2 : hc P2 = P = n2 × λ2

φo = 6.62 × 10 −34 × 5.5 × 1014

...(i)

...(ii)

From equations (i) and (ii) : n1 λ 1 = 1 = n2 λ2 500 Ans. 4. Option (b) is correct. m , initial 2 velocity of particle A is vA = vo along the x axis, mass m of particle B is mB = , initial velocity of particle 3 B is vB = 0, the two particles collide elastically and after the collision they both move along the x axis. To find : Dl, the change in wavelength of particle A, in terms of its de Broglie wavelength (lo) before the collision. Let v1 be the velocity of particle A and v2 be the velocity of particle B after the collision. As the collision is perfectly elastic : v = v2 − v1 ...(i) Given : Mass of particle A is mA =

φo = 36.41 × 10 −20 J = 2.27 eV Ans. 6. Option (b) is correct. Given : A graph of stopping potential (Vs) versus 1 reciprocal of wavelength    for a photo electric λ effect experiment is shown. To find : Change in the graph as the intensity of the incident radiation is increased. The wavelength of the incident radiation as well as the kinetic energy of the photo electrons emitted from the metal surface is both independent of the intensity of the incident radiation. So, the graph does not change upon the increase in intensity of the incident radiation. Ans. 7. Option (d) is correct. Given : le is de Broglie wavelength of an electron, λHe++ is de-Broglie wavelength of a doubly ionized He++ ion, lp is de-Broglie wavelength of a proton, all three particles have same kinetic energy, K. To find : Compare λe ,λHe++ ,λp . h h = ...(i) λe = me v 2 me K

λHe++ =

Also, by law of conservation of momentum : m m m m ...(ii) v + ×0 = v1 + v2 2 3 2 3 From equations (i) and (ii) : v v1 = 5 de-Broglie wavelength of particle A before the collision : h 2h = λo = m mv v 2 de-Broglie wavelength of particle A after the collision : h 10 h λ' = =  m   v  mv  2  5     Change in de-Broglie wavelength of particle A : 8h ∆λ = λ ′ − λo = = 4 λo mv Ans. 5. Option (d) is correct. Given : In a photo electric effect experiment for a certain metal, the minimum frequency of the incident radiation at which the photo electrons are emitted is n = 5.5 × 1014 Hz. To find : fo, the work function of the metal. Let K be the kinetic energy of the photo electrons and its minimum value is zero. φo = hν + K

φo = hν

λp =

h = mHe++ v

h 2mHe++ K

h h = mp v 2 mp K

...(ii) ...(iii)

From equations (i), (iii) and (ii) : 1 1 1 : : λe : λp : λHe++ = me mp mHe++ As,me < mp < mHe++ ;λe > λp > λHe++ Ans. 8. Option (d) is correct. Given : Temperature of nitrogen gas is T = 400 K, mass of nitrogen molecule is m = 4.64 × 10–26 kg, Boltzmann constant is kB = 1.38 × 10–23 J/K, Planck’s constant is h = 6.63 × 10–34 J-s. To find : l, de-Broglie wavelength of a nitrogen molecule. Let the kinetic energy of nitrogen molecule be K. h h h λ = = = 2mK 3 3mk B T 2m × k B T 2

λ =

6.63 × 10 −34 3 × 4.64 × 10 −26 × 1.38 × 10 −23 × 400

= 0.24 Å

Ans. 9. Option (d) is correct. Given : Mass of electron is m, energy of electron = energy of photon = E. λ To find : e , the ratio of de-Broglie wavelength of λp electron to that of photon.

354 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) de-Broglie wavelength of electron, having energy E: h ...(i) λe = 2mE de-Broglie wavelength of photon, having energy E : hc λp = ...(ii) E From equations (i) and (ii) :

λe = λp

h 2mE

×

E 1 E = hc c 2m

λB = 2λA h 2mTB

= 2

h 2mTA

TA = 4 TB

...(i)

Put value of TA from equation (i) in : TB = ( TA − 1.5 ) eV TB = ( 4 TB − 1.5)eV 3TB = 1.5 eV TB = 0.5 eV

Initial momentum of electron : po =

2mvo

Initial de-Broglie wavelength of electron : h h = λo = po 2mvo Velocity of electron at some time t : eE t  v(t ) = vo i + vo j + o k m Momentum of electron at some time t :

Ans. 10. Option (b) is correct. Given : Case a : Energy of photons incident on the surface of metal A is EA = 4.0 eV, maximum kinetic energy of the photoelectrons ejected from the surface of metal A is TA eV, de-Broglie wavelength of the ejected photoelectrons is lA. Case b : Energy of photons incident on the surface of metal B is EB = 4.5 eV, maximum kinetic energy of the photoelectrons ejected from the surface of metal B is TB = (TA – 1.5) eV, de-Broglie wavelength of the ejected photoelectrons is lB = 2lA. To find : fB, the work function of metal B. de-Broglie wavelength of electron, having kinetic energy TA and TB, respectively : h h , λB = λA = 2mTA 2mTB As,

PHYSICS

...(ii)

According to Einstein’s equation for photoelectric effect for metal B : EB = φB + TB 4.5 = φB + 0.5

φB = 4 eV Ans. 11. Option (c) is correct. Given : Mass of electron is m, initial velocity of electron is  v = vo i + vo j , ...(i) initial de-Broglie wavelength of electron is lo, the electron moves under influence of an electric field  E = −E k , o

To find : l, de-Broglie wavelength of electron at time t.

 eE t  p = m 2 vo2 +  o   m 

2

de-Broglie wavelength of electron at time t : h λ = p h

λ =

=

 eE t  m 2 vo2 +  o   m  λo 1+

2

=

1

h 2mvo

1+

e 2 Eo2t 2

2m 2 vo2

e 2 Eo2t 2

2m 2 vo2

Ans. 12. Option (c) is correct. Given : Case-a : de-Broglie wavelength of particle with kinetic energy E is l. Case-b : de-Broglie wavelength of particle with λ kinetic energy E + DE is . 2 To find : Value of DE. de-Broglie wavelength of particle with kinetic energy E : h λ = ...(i) 2mE de-Broglie wavelength of particle with kinetic energy E + DE : λ h = ...(ii) 2 2m( E + ∆E) From equations (i) and (ii) : 2 =

E + ∆E E

4E = E + ∆E ∆E = 3E Ans. 13. Option (b) is correct. Given : Wavelength of radiation incident on a metal surface is l = 6561 Å, radius of the largest circular path followed by the ejected photoelectrons under the influence of magnetic field B = 3 × 10–4 T is r = 10 mm. To find : f, the work function of the metal. Radius of circular orbit of an electron moving in a magnetic field, when m is mass of electron and v is its velocity :

355

DUAL NATURE OF MATTER AND RADIATION

r =

mv eB

v =

eBr 1.6 × 10 −19 × 3 × 10 −4 × 10 × 10 −3 = m 9.1 × 10 −31 = 0.53 × 10 6 m s

Kinetic energy of photoelectrons ejected from the metal surface : 1 1 K = mν 2 = × 9.1 × 10 −31 × ( 0.53 × 10 6 )2 2 2 = 1.28 × 10 −19 J According to Einstein’s equation for photoelectric effect : hc = φ+K λ

φ =

hc 6.6 × 10 −34 × 3 × 10 8 −K = − 1.28 × 10 −19 λ 6561 × 10 −10 = ( 3 − 1.28 ) × 10 −19 J

To find: The de-Broglie wavelength l of the final particle. Momentum of particle-1 : h p1 = λ1 Momentum of particle-2 : h p2 = λ2 Total momentum of the two particle system before collision : p =

Momentum of electron at time t : = p(t ) = mv(t ) e Et de-Broglie wavelength of electron at time t : h h = λ (t ) = p(t ) e Et Rate of change of de-Broglie wavelength of electron at time t : dλ d h  =   dt dt  e Et  dλ h d 1 = dt e E dt  t  dλ h ( −1) = dt e E t2 dλ h = − dt e Et 2 Ans. 15. Option (a) is correct. Given : de-Broglie wavelength of particle 1 is l1, de-Broglie wavelength of particle 2 is l2, the two particles are moving at right angle to each other, the particles suffer a perfectly inelastic collision.

...(i)

As the collision is inelastic the two particles stick together and move as one after collision. By law of conservation of linear momentum, the momentum of final particle after collision will be : p =

p12 + p22 + 2 p1 p2 cos 90 2

2

Ê hˆ Ê hˆ Ê h ˆÊ h ˆ ÁË λ ˜¯ + ÁË λ ˜¯ + 2 ÁË λ ˜¯ ÁË λ ˜¯ cos 90 1 2 1 2

=

φ = 1.72 × 10 −19 J = 1.1eV  Ans.14. Option (a) is correct. Given : Mass of electron is m magnitude of charge on electrons is |e|, initial velocity of electron is zero, the electron is accelerated by a constant electric field E. dλ To find : , the rate of change of de-Broglie dt wavelength of electron at time t. Velocity of electron moving under influence of a constant electric field E at time t : e Et v(t ) = m

p12 + p22 + 2 p1 p2 cos 90

2

p = 1 = λ 1

λ2

=

Ê 1ˆ Ê 1ˆ h = h Á ˜ +Á ˜ λ Ë λ1 ¯ Ë λ2 ¯ 2

Ê 1ˆ Ê 1ˆ ÁË λ ˜¯ + ÁË λ ˜¯ 1 2 1

λ12

+

2

2

1

λ22

Ans. 16. Option (c) is correct. Given : Electric field associated with the light wave falling on a metal plate is Ê 2p x ˆ E = 10 -3 cos Á - 2p ¥ 6 ¥ 1014 t˜ x N C , Ë 5 ¥ 10 -7 ¯ ...(i)

the work function of metal plate is

φ = 2 eV = 2 ¥ 1.6 ¥ 10 -19 . J To find: Vo, the stopping potential of the photoelectrons. From equation (i), the wave vector associated with the given wave is : 2p k = 5 ¥ 10 -7 2p 2p = λ 5 ¥ 10 -7

λ = 5 ¥ 10 -7 m Energy of one photon of light wave of wavelength hc l = 5 × 10–7 m is E = which is partly used to λ overcome the work function of metal f and rest is gained as kinetic energy (eVo) by the photoelectron. So, according to Einstein’s equation for photoelectric effect :

356 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

Radiation pressure on the solar panel due to reflected radiation :

hc = φ + eVo λ eVo = eVo =

p1 =

hc -φ λ 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 5 ¥ 10 -7

= 3.96 ¥ 10

-19

- 2 ¥ 1.6 ¥ 10

- 3.2 ¥ 10

-19

= 0.76 ¥ 10

-19

J

h λx

Momentum of particle y : h λy

Total momentum of the two particle system before collision : p =

px2 + p y2 + 2 px p y cos 180

...( i )

As the collision is inelastic the two particles stick together and move as one after collision. By law of conservation of linear momentum, the momentum of final particle P after collision will be : p =

px2 + p y2 + 2 px p y cos 180 2

=

p =

p2 =

75 I ¥ 100 c

...(ii)

2 Ê hˆ Ê hˆ Ê h ˆÊ h ˆ ÁË λ ˜¯ + Á λ ˜ - 2 ÁË λ ˜¯ Á λ ˜ Ë y¯ x x Ë y¯

h h h = λ λx λy

Ê 25 2 I 75 I ˆ F = ( p1 + p2 ) ¥ A = Á ¥ + ¥ ¥A Ë 100 c 100 c ˜¯ =

1.25I ¥A c

Put given values in the above equation. F =

1.25 ¥ 50 3 ¥ 10 8

λx λy λx - λy

Ans. 18. Option (b) is correct. Given : On surface of solar panel incident solar energy is I = 50 W/m2, portion of incident solar energy reflected by solar panel is 25%. To find : F, force on A = 1 m2, surface area of the panel.

= 20 ¥ 10 -8 N

Ans. 19. Option (a) is correct. Given : In photoelectric effect experiment the threshold wavelength of light is l0 = 380 nm, wavelength of incident light is l = 260 nm. To find : The maximum kinetic energy of photoelectrons. Work function of metal surface :

φ = hν o =

hc λo

...(i)

Energy incident on the metal surface : E = hν =

hc λ

...( ii )

E is partly used to overcome the work function of metal and rest is gained as kinetic energy (KE)max by the photoelectron. So, according to Einstein’s equation for photoelectric effect : E = φ + ( KE)max ( KE)max = E - φ =

Ê1 1ˆ hc hc = hc Á - ˜  λ λo Ë λ λo ¯

( KE)max = ( 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 ) 1 ˆ Ê 1 9 ÁË 260 - 380 ˜¯ ¥ 10 J

1 1 1 = λ λx λy

λ =

...(i)

From equations (i) and (ii),

Ans. 17. Option (b) is correct. Given : de-Broglie wavelength of particle x is lx, de-Broglie wavelength of particle y is ly, both the particles are moving in opposite directions, particle P is formed by inelastic collision between particles x and y. To find : The de-Broglie wavelength l of particle P. Momentum of particle x :

py =

25 2 I ¥ 100 c

Radiation pressure on the solar panel due to absorbed radiation :

-19

0.76 Vo = = 0.48 V 1.6

px =

PHYSICS

( KE)max =

( 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 ) Ê 1 1 ˆ ÁË 260 - 380 ˜¯ ¥ 1.6 ¥ 10 -19 10 9 eV = 1.5 eV

Ans. 20. Option (c) is correct. Given : Energy flux incident on a completely absorbing surface is I = 25 W/cm2, area of surface is A=25 cm2. To find : Dp, the momentum transferred to the surface in Dt = 40 min = (40 × 60) seconds.

357

DUAL NATURE OF MATTER AND RADIATION

Radiation pressure on the surface due to absorbed radiation : I p = c Force on the surface : F = p¥A=

I ¥A c

Momentum imparted in t = 40 min : I Dp = F ¥ Dt = ¥ A ¥ Dt c Put given values in the above equation. 25 Dp = ¥ 25 ¥ 40 ¥ 60 = 5 ¥ 10 -3 N - s 3 ¥ 10 8 Ans. 21. Option (a) is correct. Given : Power of laser is P = 2 mW, wavelength of laser beam is l = 500 nm. n To find : (t = 1sec ond), number of photons that t will be emitted per second. Power of laser : Energy P= time =

number of photons¥ energy of single photon time

nE = t n P P Pλ = = = t E hc hc λ

Ans. 23. Option (a) is correct. Given : Wavelength of light incident on surface of a metal is l1 = 350 nm and l2 = 540 nm, the ratio of maximum speed of photoelectrons for the two v cases is 1 = 2. v2 To find : f, the work function of the metal. According to Einstein’s equation for photoelectric effect : hc 1 = φ + m v12 ...(i) λ1 2 hc 1 = φ + m v22 λ2 2 From equations (i) and (ii) : Ê hc ˆ 1 m v12 ÁË λ - φ ˜¯ 1 2 = 1 hc 2 m v2 -φ 2 λ2 v12 v22

Ê hc ˆ ÁË λ - φ ˜¯ 1 = =4 hc -φ λ2

1240 -φ 350 = 4 1240 -φ 540 3.5 - φ = 4 2.3 - φ

Put given values in the above equation. n 2 ¥ 10 -3 ¥ 500 ¥ 10 -9 = 5 ¥ 1015 = t 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 Ans. 22. Option (b) is correct. Given : A plot of stopping potential Vo as function of incident frequency n for a sodium emitter. 3.0

φ = 1.8 eV Ans. 24. Option (d) is correct. Given : Magnetic field associated with light wave falling on silver plate is B = Bo [sin( 3.14 ¥ 107 ) ct + sin( 6.28 ¥ 107 ) ct], ...(i) work function of silver is f = 4.7 eV. To find : KEmax, maximum kinetic energy of the photoelectrons. Angular frequency associated with the wave :

2.0 Vo

...(ii)

1.0 2

4

6

8

10

n(10 Hz) 14

To find: Work function of sodium. From the graph, the threshold frequency is : 14

ν o = 4 ¥ 10 Hz That implies, work function of sodium will be :

φ = hν o = 6.6 ¥ 10

-34

= 26.4 ¥ 10 -20 J = = 1.66 eV

¥ 4 ¥ 10

26.4 ¥ 10

14

-20

1.6 ¥ 10 -19

eV

ω1 = 6.28 ¥ 107 c and ω2 = 3.14 ¥ 107 c. As we want to calculate the maximum kinetic energy of the photoelectrons we will consider w1, Frequency of incident light :

ν =

ω1 6.28 ¥ 107 c = = 107 ¥ 3 ¥ 10 8 2p 2p

= 3 ¥ 1015  According to Einstein’s equation for photoelectric effect : hν = φ + KEmax KEmax = hν - φ

358 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

So, number of photoelectrons emitted per second from the metal surface : 10 n = N = 1011 100

KEmax = 6.6 ¥ 10 -34 ¥ 3 ¥ 1015 - 4.7 ¥ 1.6 ¥ 10 -19 KEmax = 1.22 ¥ 10 -18 J = 7.7 eV Ans. 25. Option (d) is correct.

Ans. 27. Option (a) is correct.

Given : The resolution of an electron microscope is of the order of the wavelength of electrons used.

Given : Frequency of photon is ν p = 6 ¥ 1014 Hz, de-Broglie wavelength of photon is lp, de-Broglie

To find : E, minimum electron energy required to resolve width of w = 7.5 × 10–12 m.

-3 wavelength of electron is λe = 10 ¥ λp . To find: v, the speed of electron. de-Broglie wavelength of photon :

Momentum associated with electrons of wavelength l will be :

λp =

h p = λ

λe = 10 -3 ¥ λp = 10 -3 ¥ 5 ¥ 10 -7 = 5 ¥ 10 -10 m

2

1 p2 Ê h ˆ E = = ¥  2m ÁË λ ˜¯ 2m

Also,

λe =

Put given values in the above equation. 2

Ê 6.6 ¥ 10 -34 ˆ 1 E = Á ˜ ¥ 2 ¥ 9.1 ¥ 10 -31 Ë 7.5 ¥ 10 -12 ¯ = 4.35 ¥ 10

J =

4.35 ¥ 10

-15

1.6 ¥ 10 -19

v =

eV

Given : Area of metal plate is A = 10–4 m2, intensity of incident radiation on the surface of metal is mW I = 16 2 , work function of metal is f = 5 eV, m energy of incident photons is E¢ = 10 eV, fraction of incident photons which produce photoelectrons is 10%. To find: Maximum energy of a single photoelectron KE and number of photoelectrons emitted per second n.

KE = E '- φ = 10 - 5 = 5 eV Energy incident per unit time on the metal surface : E ˆ Ê ÁË use : I = A.t ˜¯

Number of photons incident per unit time on the metal surface :

that

 produce

= 1.45 ¥ 10 6 m / s



Subtract equation (ii) from equation (i). e( Vo1 - V02 ) = E1 - E2 Vo1 - V02 =

E t N = energy of single incident photon

photons

9.1 ¥ 10 -31 ¥ 5 ¥ 10 -10

Here, Vo1 is the stopping potential for case-a, hc is the energy of the incident light and f is E1 = λ1 the work function of the metal. Similarly, for case-b : eVo2 = E2 - φ ...(ii)

From Einstein’s equation for photoelectric effect, maximum energy of a single photoelectron will be :

IA 16 ¥ 10 -3 ¥ 10 -4 = = = 1012 E¢ 10 ¥ 1.6 ¥ 10 -19

6.6 ¥ 10 -34

Ans. 28. Option (c) is correct. Given : Case-a : Wavelength of incident light in a photoelectric effect experiment is l1 = 300 nm, Case-b : Wavelength of incident light in a photoelectric effect experiment is changed to l2 = 300 nm. To find : The decrease in the stopping potential from case-a to case-b. From Einstein’s equation for photoelectric effect, maximum energy of a single photoelectron in case a, will be : eVo1 = E1 - φ ...(i)

Ans. 26. Option (c) is correct.

Fraction of incident photoelectrons is 10%

h h h = ;v= p e me v me λe

Put given values in the above equation.

ª 25keV

E = IA t

c 3 ¥ 10 8 = = 5 ¥ 10 -7 m ν p 6 ¥ 1014

de-Broglie wavelength of electron :

Energy of electrons with momentum p :

-15

PHYSICS

( E1 - E2 ) e

=

hc Ê 1 1ˆ e ÁË λ1 λ2 ˜¯

=

6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 Ê 1 1 ˆ ÁË 300 - 400 ˜¯ 1.6 ¥ 10 -19 ¥ 10 9

Vo1 - V02 = 1.0 V

359

DUAL NATURE OF MATTER AND RADIATION

Ans. 29. Option (d) is correct. Given : Mass of particle A is m, charge on particle A is q, particle A is accelerated by a potential difference of VA = 50 V, mass of particle B is 4m, charge on particle B is q, particle B is accelerated by a potential difference of VB = 2500 V. λ To find : A , the ratio of de-Broglie wavelengths of λB the two particles. de-Broglie wavelength of particle A : h h h = = ...(i) λA = pA 2mE A 2mqVA In equation (i), pA is momentum of particle A, EA = qVA is the energy of particle A accelerated by a potential difference of VA. de-Broglie wavelength of particle B : h h = λB = pB 2( 4 m ) EB =

h

...( ii )

2( 4 m )qVB

In equation (ii), pB is momentum of particle B, EB = qVB is the energy of particle B accelerated by a potential difference of VB. From equations (i) and (ii),

λA = λB

4 VB VA

=2¥

2500 = 14.14  50

Ans. 30. Option (d) is correct. Given : Energy of electron entering the mercury vapour is E1 = 5.6 eV, energy of the electron emerging from mercury vapour is E2 = 0.7 eV. To find : The minimum wavelength of photons emitted by mercury atoms. Maximum possible energy absorbed by the mercury vapour : E = E1 - E2 = 5.6 - 0.7 = 4.9 eV Maximum wavelength of photon emitted by the mercury vapour in de-excitation :

λmax =

hc 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8 = = 250 nm E 4.9 ¥ 1.6 ¥ 10 -19

Ans. 31. Option (d) is correct. Given : Case-a : When frequency of incident monochromatic light at a photosensitive surface is V, the stopping V potential for the photocurrent is - o , 2 Case-b : When frequency of incident monochromatic ν light at a photosensitive surface is , the stopping 2 potential for the photocurrent is –V0. To find : v0, the threshold frequency for photoelectric emission.

From Einstein’s equation for photoelectric effect, the energy of incident light En is partly used to overcome the work function f0 = hn0 of metal and rest is gained as kinetic energy (eV0) by the photoelectron. Eν = hν o + eVo ...(i) Equation (i) for case a : hν

Ê V ˆ = hν o + e Á - o ˜ Ë 2¯

Equation (ii) for case b : Êν ˆ h Á ˜ = hν o + e ( - Vo ) Ë 2¯

...(ii)

...(iii)

Subtract equation (iii) from equation (ii). eVo Êν ˆ hÁ ˜ = Ë 2¯ 2 hν

= eVo

...(iv)

Substitute (iv) in (ii). hν hν = hν o 2

νo =

3 ν 2

Subjective Questions (Chapter Based) Q.1. When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3A, the stopping potential V is . If the threshold wavelength for the metallic 4 surface is nl then value of n will be : [JEE (Main) – 2nd Sep. 2020 - Shift-1] Sol. Given : Case-1 : For wavelength of incident radiation is l, stopping potential is V, Case-2 : For wavelength of incident radiation is 3l, V stopping potential is , 4 Threshold wavelength for the metallic surface is lth = nl. To find : The value of n. hc Let φo = be the work function for the metallic λth surface. From Einstein’s equation for photoelectric effect : hc = φo + eV ...(i) λ hc V = φo + e 3λ 4

...(ii)

From equations (i) and (ii) : 2 hc 3 = eV 3 λ 4 eV =

8 hc 9 λ

...(iii)

360 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Put value of eV from equation (iii) in equation (i). hc hc 8 hc = + λ λth 9 λ

λth = 9λ = nλ Q.2. A beam of electrons of energy E scatters from a target having atomic spacing of 1A. The first maximum intensity occurs at q = 60°. Then E (in eV) is........ (Plank's constant h = 6.64 × 10–34 J-s, 1 eV = 1.6 × 10–19 J, electron mass m = 9.1 × 10–31 kg). [JEE (Main) – 5th Sep. 2020 - Shift-1] Sol. Given : Energy of a beam of electrons is E, the beam is scattered from a target having atomic spacing of d = 1Å, the (n = 1) peak in the diffraction pattern occurs at q = 60°. To find : The value of E. Condition for maximum in the diffraction pattern : 2d sin θ = nλ de-Broglie wavelength associated with an electron in the beam : 2d sin θ λ = = 3Å n Energy of the electron : h λ = 2Em E = E =

h2 2mλ 2

4 − φo 2.5 − φo

φo = 2eV Q.4. A beam of electromagnetic radiation of intensity 6.4 × 10–5 W/cm2 is comprised of wavelength, l = 310 nm. It falls normally on a metal (work function j = 2eV) of surface area of 1 cm­­2. If one in 103 photons ejects an electron, total number of electrons ejected in 1 s is 10x. (hc = 1240 eVnm, leV = 1.6 × 10–19 J), then x is _______. [JEE (Main) – 7th Jan. 2020 - Shift-1] Sol. Given : Intensity of radiation incident on a metal −5 2 surface is I = 6.4 × 10 W / cm , wavelength of radiation is l = 310 nm, one in 103 incident photons eject an electron from the metal surface, total number of photoelectrons ejected from the metal surface in 1 second is

n = 10 x To find : The value of x. Energy of single incident photon : hc E = λ

...(i)

...(ii)

Total number of photoelectrons ejected from the metal surface in 1 second : n′ n = ...(iv) 10 3

( 6.64 × 10 −34 )2 2 × 9.1 × 10 −31 × ( 3 × 10 −10 )2

Q.3. The surface of a metal is illuminated alternately with photons of energies E1 = 4eV and E2 = 2.5 eV respectively, the ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is.......... [JEE (Main) – 5th Sep. 2020 - Shift-2] Sol. Given : Case 1 : Photons of energy incident on a metal surface is E1 = 4 eV, Case 2 : Photons of energy incident on a metal surface is E2 = 2.5 eV, The ratio of maximum speed of photo electrons emitted in the two cases is v1 : v2 = 2. To find : f0, the work function of the metal. Let m be the mass of a photo electron. From Einstein’s equation of photoelectric effect : 1 1 E1 = mv12 + φo ; mv12 = E1 − φo ...(i) 2 2 1 2 1 mv2 + φo ;  mv22 = E2 − φo 2 2

2

 v1  E1 − φo   = E2 − φo  v2 

Total number of photons incident on the metal surface in 1 second : I ...(iii) n′ = E

E = 0.81 × 10 −17 J = 50 eV

E2 =

From equations (i) and (ii) :

4 =

n = 9

PHYSICS

...(ii)

From equations (ii), (iii), (iv) : I Iλ n = = 10 3 E 10 3 hc 6.4 × 10 −5 × 310 × 10 −9 = 1011 = 10 3 × 6.6 × 10 −34 × 3 × 10 8

....(v)

On comparing equations (i) and (v), we get : x = 11 Q.5. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is Sol. Given : Work function of silver is f1 = 4.6 eV, work function of sodium is f2 = 2.3 eV. To find : The ratio of the slope of stopping potential versus frequency plot for silver (Vo vs n) to that of sodium. Einstein’s equation for photoelectric effect : Eν = φo + eVo ...( i ) Equation (i) for silver : hν = φ1 + e( Vo )silver ...( ii )

361

DUAL NATURE OF MATTER AND RADIATION







( Vo )silver =

hν φ1 e e

...(iii)

Comparing equation (iii) with equation of straight line, we see that the slope of (Vo)silver vs n plot for h silver will be . e Similarly, the slope of (Vo)sodium vs n plot for sodium h will also be . e The ratio of the two slopes will be : h h : = 1:1 = 1 e e

Q.6. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A × 10Z (where l < A < 10). The value of ‘Z’ is Sol. Given : Radius of silver sphere is r = 1 cm, work function of silver is f = 4.7 eV, the sphere is suspended from an insulating thread in free space, wavelength of radiation incident on silver sphere is l = 200 nm, maximum photoelectrons emitted from sphere is



Z

n = A ¥ 10 .

Put given values in the above equation. 200 ¥ 10 -9

= 4.7 ¥ 1.6 ¥ 10 -19 + 1.6 ¥ 10 -19 Vo





Vo = 1.5 V So, potential acquired by the sphere after losing n electrons is Vo = 1.5 V. Also, potential acquired by the sphere after attaining charge q : 1 q Vo =  4p ε o r Put given values in the above equation. q ˆ Ê 1.5 = 9 ¥ 10 9 Á Ë 1 ¥ 10 -2 ˜¯ q = 1.6 ¥ 10 -12 C q = ne ; n =

q = 107 e

On comparing with equation I, we get Z = 7.

l (mm)

V0 (Volt)

0.3

2.0

0.4

1.0

0.5

0.4

Given that c = 3 × 108 m s–1 and e = 1.6 × 10–19 C, Planck’s constant (in units of J-s) found from such an experiment is Sol. Given : 3 readings of a photoelectric effect experiment, l1 = 0.3 mm, Vo1 = 2.0 V, l2 = 0.4 mm, Vo2 = 1.0 V, l3 = 0.5 mm, Vo3 = 0.4 V. To find : The value of Planck’s constant, h. Einstein’s equation for photoelectric effect : 1 1 = (φ + eVo ) λ hc 1 e Vo + constant = ...(i) λ hc



|f is the work function| of metal. Equation I represents e Ê1 ˆ . a straight line Á vs Vo ˜ with slope Ëλ ¯ hc



Also, slope of the straight line will be :

...(i)

To find : Value of Z. Due to loss of photoelectrons a positive charge is attained by the sphere. The sphere losses electrons until the potential acquired by it is equal to the stopping potential Vo. From Einstein’s equation for photoelectric effect : hc = φ + eVo λ 6.6 ¥ 10 -34 ¥ 3 ¥ 10 8

Q.7. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (l) of incident light and the corresponding stopping potential (V0) are given below:

Ê 1 1ˆ ÁË λ - λ ˜¯ 1 2



Vo1 - Vo 2

=

e hc

Ê y 2 - y1 ˆ ÁË use slope = x - x ˜¯ 2 1

Put given values in the above equation : 1 ˆ Ê 1 10 6 Á Ë 0.3 0.4 ˜¯ 1.6 ¥ 10 -19 1 = ¥ ( 2 - 1) h 3 ¥ 10 8 On solving :

h = 6.4 ¥ 10 -34 J - s Q.8. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, then what is the work function of the metal? Sol. Given : Two wavelengths are incident on a metal surface l1 = 248 nm, l2 = 310 nm, the maximum speed of photoelectrons corresponding to wavelength 1 is u1, the maximum speed of photoelectrons corresponding to wavelength 2 is u2, u1 : u2 = 2 : 1, hc = 1240 eV nm. To find : f, the work function of the metal. Energy corresponding to wavelength-1 : hc 1240 = E1 = eV = 5 eV λ1 248

362 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)



Energy corresponding to wavelength-2 : hc 1240 = E2 = eV = 4 eV λ2 310 Einstein’s equation for photoelectric effect for wavelength-1 : E1 = φ + KEmax 1 KEmax 1 = E1 - φ = 5 - φ

...( i )

Einstein’s equation for photoelectric effect for wavelength-2 : E2 = φ + KEmax 2



KEmax 2 = E2 - φ = 4 - φ



KEmax1 ∝ u12

As, and (i) and (ii), we have :

KEmax 2 ∝ u22 ,

...( ii )



Êu ˆ KEmax 1 5 -φ = Á 1˜ = KEmax 2 4 -φ Ë u2 ¯ 4 5 -φ = 1 4 -φ

φ =

11 = 3.7 eV 3



from equations

2





Q.9. A particle A of mass m and initial velocity v m collides with a particle B of mass which is at 2 rest. The collision is head on, and elastic. What will be the ratio of the de-Broglie wavelengths lA to lB after the collision? Sol. Given : Mass of particle A is m, velocity of particle A m is v, mass of particle B is , velocity of particle B is 2 0, the collision between two particles is head on and elastic. To find : The ratio of de-Broglie wavelengths lA to lB after the collision.

Momentum of particles before collision : p before = m ¥ v +

pafter = m ¥ vA + vA +



vB - v A vB - v A = =1 uA - uB v

vB - v A = v

...(iii)

From equations (ii) and (iii) : vA =

v 4 ,vB = v 3 3

...(iv)

From equation (iv) : h = mvA

h 3h = Ê v ˆ mv mÁ ˜ Ë 3¯

...(v)

From equation (iv) :

λB =

...(ii)

For elastic collision, coefficient of restitution :

λA =

...( i )

m ¥ vB = mv 2

vB = v 2





m ¥ 0 = mv 2

Let vA be velocity of particle A and vB be velocity of particle B after collision. As momentum in elastic collision is conserved, momentum of particles after collision will be :

e =

PHYSICS

h h 3h = = Ê m ˆ Ê 4 ˆ 2mv Ê mˆ ÁË 2 ˜¯ vB ÁË 2 ˜¯ ÁË 3 v˜¯

...(vi)

From equations (v) and (vi) :

λA = 2 λB 

.

s.

s

Doesn't explain Zeeman's and Stark's effect

spectra

Atoms & Nuclei Part-1

components of different wavelength appear

ATOMS AND NUCLEI

363

rs

e lum o V ar cle u N

Nuclear Rad ius

e ton Iso

Isoba

Iso to pe s

Atoms & Nuclei Part-2

e.g.,

n

e.g.,

200 me

moderate penetrating power (100 times )

gave showed

showed

364 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Atoms and Nuclei

Chapter 18 Syllabus

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum; Composition and size of nucleus, atomic masses; isotopes, isobars, isotones; Radioactivity-alpha, beta and gamma particles/rays and their properties; radioactive decay law; Decay constant; Half-life and mean life; Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number, nuclear fission and fusion. Characteristic and continuous X-rays, Moseley’s law;

Topic-1

Bohr’s Atomic Model

LIST OF TOPICS : Topic-1 : Bohr’s Atomic Model

.... P. 365

Topic-2 : Radioactivity

.... P. 370

Concept Revision (Video Based) Alpha Particle Scattering

Bohr’s Model

      

           

Part -1         

Part - 2

Hydrogen Spectrum

Energy Levels

  Part -1         

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to : 1 1 (a) 4 (b) n n (c)

1 n

3



(d)

1 n2

[JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3 × 10–15) A1/3 m. It follows that the mass density of n nucleus is of the order of : (Mprot. ≅ Mneut: ≈ 1.67 × 10–27 kg) (a) 1010 kg m–3 (b) 1024 kg m–3 3 –3 (c) 10 kg m (d) 1017 kg m–3 [JEE (Main) – 3rd Sep. 2020 - Shift-2]

Q.3. If the potential energy between two molecules A B is given by U = 6 + 12 , then at equilibrium, r r separation between molecules and the potential energy are : 1/6

1/6

B (a)   A

 2B  (b)    A

,0 1/6

 2B  (c)    A

, −

A2 2B

,

A2 4B

1/6

 B  (d)  2A   

, −

A2 2B

[JEE (Main) – 6th Sep. 2020 - Shift-1] Q.4. The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10–16 s. The frequency of revolution of the electron in its first excited state (in s–1) is : (a) 5.6 × 1012 (b) 1.6 × 1014 (c) 7.8 × 1014 (d) 6.2 × 1015 [JEE (Main) – 7th Jan. 2020 - Shift-1]

366 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.5. The graph which depicts the results of Rutherford gold foil experiment with a-particles is : q : Scattering angle Y : Number of scattered a-particles detected (Plots are schematic and not to scale) (a)

(b) Y Y Y

Y Y Y

1. 1. 1. 00

2. 2. 2. 0θ θ

θ π π

00

π

(c)

0θ θ

θ π π

π

(d) Y Y Y

Y Y Y



0 θ θ

θ π π

π

00

(b) n = 5 (d) n = 6 [JEE (Main) – 12th April 2019 - Shift-1] Q.12. The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, l1/l2, of the photons emitted in this process is : 27 20 (a) (b) 5 7 (c)

0 θ θ

θ π π

π

[JEE (Main) – 8th Jan. 2020 - Shift-1] Q.6. The energy required to ionise a hydrogen like ion in its ground state is 9 Rydberg. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state? (a) 8.6 nm (b) 24.2 nm (c) 11.4 nm (d) 35.8 nm [JEE (Main) – 9th Jan. 2020 - Shift-2] Q.7. Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is : (a) n = 2 ® n = 3 (b) n = 1 ® n = 4 (c) n = 2 ® n = 5 (d) n = 2 ® n = 4 [JEE (Main) – 8th April 2019 - Shift-1] Q.8. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be : (a) 889.2 nm (b) 488.9 nm (c) 642.7 nm (d) 388.9 nm [JEE (Main) – 9th April 2019 - Shift-1] Q.9. A He+ ion is in its first excited state. Its ionization energy is : (a) 48.36 eV (b) 54.40 eV (c) 13.60 eV (d) 6.04 eV [JEE (Main) – 9th April 2019 - Shift-2] Q.10. In Li++, electron in first Bohr orbit is excited to a level by a radiation of wavelength l. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of l? (Given : h = 6.63 × 10–34 Js; c =3 × 108 ms-1) (a) 11.4 nm (b) 9.4 nm (c) 12.3 nm (d) 10.8 nm [JEE (Main) – 10th April 2019 - Shift-2] Q.11. An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength l, energy

1240 eV ): λ (in nm)

(a) n = 4 (c) n = 7

4. 4. 4.

3. 3. 3. 00

E=

PHYSICS

7 5

(d)

9 7

[JEE (Main) – 12th April 2019 - Shift-2] Q.13. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is : (a) 3.5 Å (b) 6.6 Å (c) 12.9 Å (d) 9.7 Å [JEE (Main) – 12th April 2019 - Shift-2] Q.14. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be : (hc = 12500 eV-Å) (a) 25a0 (b) 9a0 (c) 4a0 (d) 16a0 [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.15. In a hydrogen like atom, when an electron jumps from the M - shell to the L - shell, the wavelength of emitted radiation is l. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be : 16 27 λ λ (a) (b) 25 20 (c)

25 λ 16

(d)

20 λ 27

[JEE (Main) – 11th Jan. 2019 - Shift-2] Q.16. A particle of mass m moves in a circular orbit in 1 2 a central potential field U( r ) = kr . If Bohr’s 2 quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as : 1 (a) rn ∝ n , En ∝ n (b) rn ∝ n , En ∝ n (c) rn ∝ n, En ∝ n

2 (d) rn ∝ n , En ∝

1 n2

th



[JEE (Main) – 12 Jan. 2019 - Shift-1]

ANSWER – KEY 1. (c) 5. (a) 9. (c) 13. (d)

2. (d) 6. (c) 10. (d) 14. (d)

3. (b) 7. (d) 11. (b) 15. (d)

4. (c) 8. (b) 12. (a) 16. (a)

367

ATOMS AND NUCLEI

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. Given : The initial state of an electron in a hydrogen atom (Z = 1) is ni = n + 1, the final state of the electron in the hydrogen atom is nf = n, n >> 1. To find : The dependence of frequency of radiation emitted in the process on n. From Rydberg’s equation the energy of the radiation emitted when the electron makes transition from ni = n + 1 to nf = n :  1 1  ∆E = hν = R y Z 2  2 −  n ( n + 1)2  

ν = As n  1,ν ∝

R y  2n + 1    h  n 2 ( n + 1)2 

−27 is M ≈ 1.67 × 10 kg. To find : r, the density of nucleus. 3 AM AM ρ = = 3 4 p R 3 4p R 3

3 × A × 1.67 × 10 −27  4 × 3.14 × (1.3 × 10 −15 )3 × A

= 1.8 × 1017 kg m −3 = 107 kg m −3 Ans. 3. Option (b) is correct. Given : Potential energy between two molecules is A B U = 6 − 12 . r r To find : r, the separation between two molecules and Uo, their potential energy at equilibrium. Force acting between two molecules : dU 6 A 12 B F = − = 7 − 13 dr r r At equilibrium : F = 0 6A r7

−

12 B

A−

r 13 2B r6

= 0

Z2

13 23 = , T2 . Here T2 is the time period of 2 Z Z2 revolution of electron in 1st excited state (n = 2). T 1 Therefore, 1 = T2 8 So, = T1

Eion = 9 R y , R y = 13.6 eV = 13.6 ×1.6 × 10 −19 . J To find : l3Æ1, the wavelength corresponding to the electron transition from the second excited state (n = 3) to ground state (n = 1) for the hydrogen like ion. Ionization energy for hydrogen like ion : = R y Z2 9R y ; Z = 3

Using Rydberg’s equation for (n = 3) to (n = 1) transition : hc 1   1 = R y Z2  2 − 2  λ3 →1 3  1

1/6

 2B  r =    A 

A  2B 1 / 6      A  

1 1 = = 7.8 × 1014 s−1 T2 1.28 × 10 −15

Ans. 5. Option (a) is correct. Given : Four plots depicting results from the Rutherford’s a scattering experiment, on x axis is the scattering angle q, on y axis is the number of a particles detected (Y). To find : the correct graph. • As most of the space in an atom is empty, the a particles go through an atom without much of deflection. So, Y has large value at small angles (q). • Most of the mass of an atom is concentrated inside the nucleus, which is positively charged and has much smaller volume as compared to the volume of an atom. So, very few a particles suffer a head on collision and are scattered at an angle q = (p) radian. Both of the above points are well depicted in graph a. Ans. 6. Option (c) is correct. Given : Energy required to ionise a hydrogen like ion in its ground state is

= Eion

= 0

Uo =

n3

ν2 =

3

Ans. 2. Option (d) is correct. Given : Mass number of nucleus is A, radius of nucleus is R = (1.3 × 10–15) A1/3 m, mass of a nucleon

ρ =

to :

T2 = T1 × 8 = 1.6 × 10 −16 × 8 = 1.28 × 10 −15 s

1 n

Ans. 4. Option (c) is correct. Given : Time period of revolution of electron in its ground state (n = 1) is T1 = 1.6 × 10–16 s. To find : n2, the frequency of revolution of electron in its 1st excited state (n = 2). Time period of revolution of electron in nth orbit of an atom with atomic number Z is proportional

6

−

B  2B 1 / 6      A  

12

A2 = 4B

6.6 × 10 −34 × 3 × 10 8 λ3 →1 1  = 13.6 × 1.6 × 10 −19 × 3 2  1 −  9  

368 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) λ3 →1 =

6.6 × 10 −34 × 3 × 10 8 13.6 × 1.6 × 10 −19 × 8

= 11.4 nm

Ans. 7. Option (d) is correct. Given : Incident radiation is from transition ni = 2 to nf = 1 of hydrogen atoms, the radiation is incident on He+ ions, He+ ions are in states n = 1 and n = 2. To find : the possible transition of He+ ions. Energy of radiation coming from hydrogen atoms (Z = 1) : Ê 1 1ˆ E = 13.6 ( Z )2 Á 2 - 2 ˜ eV ÁË n f ni ˜¯

Energy levels of electron in He+ (Z = 2) ion : 13.6 ( Z 2 ) n2

E1 = -54.4 eV ,E2 = -13.6 eV , E3 = -6.04 eV ,E4 = -3.4 eV

Ê 1 1ˆ E1 ∝ Z 2 Á 2 - 2 ˜ ÁË n f ni ˜¯

...(i)

Energy of second Balmer line in hydrogen spectrum : Ê 1 1ˆ E2 ∝ Z2 Á 2 - 2 ˜ ÁË n f ni ˜¯ 1ˆ hc Ê 1 ∝ 12 Á 2 - 2 ˜ Ë λ2 2 4 ¯

Put value of l1 in equation (iii). λ2 = 488.9 nm

n2

Put (n = 2) in above equation : E2 =

13.6 ( 2 )2 22

= 13.6 eV

E4 = -13.6

Z2 nf2

= -7.65 eV

Energy of level ni = 1 of Li++ : E1 = -13.6

Z2 ni2

= -122.4 eV

Energy required for transition from ni = 1 to nf = 4 : E = E4 - E1 = 114.75 eV Wavelength of incident radiation : hc 1240 eV.nm λ = = = 10.8 nm  E 114.75 eV

1ˆ hc Ê 1 ∝ 12 Á 2 - 2 ˜ Ë2 λ1 3 ¯

From equations (i) and (ii), λ2 20 = λ1 27

13.6 ( Z 2 )

Energy of level nf = 4 of Li++ :

So, a photon of energy E = 10.2 eV will cause a transition n = 2 ® n = 4 in a He+ ion. Ans. 8. Option (b) is correct. Given : Wavelength of first Balmer line in hydrogen spectrum (ni = 3 to nf = 2) is l1 = 660 nm. To find : l2, wavelength of second Balmer line in hydrogen spectrum (ni = 4 to nf = 2). Energy of first Balmer line in hydrogen spectrum:

hc 3 ∝ λ2 16

En =

nf = 4

We can see : E2 - E4 ª E.

hc 5 ∝ λ1 36

Ans. 9. Option (c) is correct. Given : He+ ion is in its first excited state (n = 2). To find : E2, its ionisation energy. Energy levels of electron in He+ (Z = 2) ion :

Ans. 10. Option (d) is correct. Given : Initial state of electron in Li++ is (ni = 1), final state of electron is nf, on de-excitation the atom comes back to (ni = 1) state, total number of spectral lines observed are N = 6. To find : l, wavelength of incident radiation. Number of spectral lines produced as an excited electron falls back to ground state (ni = 1) from state nf : n f ( n f - 1) N = =6 2

Ê 1 1ˆ E = 13.6 Á - ˜ = 10.2 eV Ë 1 4¯

En =

PHYSICS

...(ii)

...(iii)

Ans. 11. Option (b) is correct. Given : An excited He+ ion makes an overall transition nf ® (ni = 1), wavelengths of two photons emitted in succession are l1 = 108.5 nm, l2 = 30.4 1240eV nm, E = λ ( nm ) To find : Value of nf. As there were two photons emitted in succession, the He+ ion makes an intermediate transition from state nf ® m emitting photon of wavelength l1 followed by transition m ® (ni = 1) emitting another photon of wavelength l2. Energy for transition nf ® m : Ê 1 1ˆ E1 = 13.6 Z 2 Á 2 - 2 ˜ ÁË m n f ˜¯ Ê 1 1ˆ hc = 13.6 ¥ 2 2 ¥ Á 2 - 2 ˜ ÁË m λ1 n f ˜¯

...(i)

369

ATOMS AND NUCLEI

De-Broglie wavelength of electron in orbit (n = 3)  : h h λB = = ...(iii) 3h mv

Energy for transition m ® (ni = 1) : Ê 1 1 ˆ E2 = 13.6 Z 2 Á 2 - 2 ˜ Ë ni m ¯ Ê 1 1 ˆ hc = 13.6 ¥ 2 2 ¥ Á 2 - 2 ˜ λ2 Ë ni m ¯

2p ( 4.65 ¥ 10 -10 )

...( ii )

From equations (i) and (ii), Ê 1 1ˆ hc hc + = 13.6 ¥ 2 2 ¥ Á 2 - 2 ˜ λ1 λ2 ËÁ ni n f ˜¯

...(iii)

Put given values in equation (iii) : Ê 1 1240 1240 1ˆ + = 54.4 Á 2 - 2 ˜ ÁË 1 108.5 30.4 n f ˜¯

2p ( 4.65 ¥ 10 -10 ) = 9.7 Å 3

λB =

Ans. 14. Option (d) is correct. Given : Initial state of hydrogen atom is ni = 1, wavelength of photon absorbed by the hydrogen atom is l = 980 Å, hc = 12500 eV-Å. To find : rnf, the radius of the excited state nf of hydrogen atom. Energy of photon absorbed when electron makes a transition from (ni = 1) to nf :

nf = 5

E =

Ans. 12. Option (a) is correct. Given : 1st transition of an electron in H atom is from third excited state (n = 4) to second excited state (n = 3), 2nd transition of the electron is from n = 3 to n = 2 state. λ To find : 1 , the ratio of the respective wavelengths λ2 of the photons emitted in this process. Energy of photon emitted when the electron makes transition from n = 4 to n = 3 state : 1ˆ Ê 1 E1 ∝ Á 2 - 2 ˜ Ë3 4 ¯ hc 1ˆ Ê 1 ∝ Á 2 - 2˜ Ë3 λ1 4 ¯

...(i)

Energy of photon emitted when the electron makes transition from n = 3 to n = 2 state : 1ˆ Ê 1 E2 ∝ Á 2 - 2 ˜ Ë2 3 ¯ hc 1ˆ Ê 1 ∝ Á 2 - 2˜ Ë2 λ2 3 ¯

...(ii)

From equations (i) and (ii), 1ˆ Ê 1 ÁË 2 - 2 ˜¯ λ1 3 = 5 ¥ 144 = 20 = 2 1 ˆ 36 λ2 7 7 Ê 1 ÁË 2 - 2 ˜¯ 3 4

Ê 12500 1ˆ = 13.6 ¥ Á 1 - 2 ˜ ÁË 980 n f ˜¯ nf = 4 Radius of orbit (nf = 4) : rn f = n 2f ao

( ao is the Bohr ' s radius)

rn f = 4 2 ao = 16 ao Ans. 15. Option (d) is correct. Given : Wavelength of radiation emitted when an electron jumps from M-shell (n = 3) to L shell (n = 2) is l. To find : l¢, wavelength of radiation emitted when an electron jumps from N-shell (n = 4) to L shell (n = 2). Energy of photon emitted when electron makes a transition from (n = 3) to (n = 2) state : 1ˆ Ê 1 E ∝ Á 2 - 2˜ Ë2 3 ¯ 1ˆ hc Ê 1 ∝ Á 2 - 2˜ Ë λ 2 3 ¯

...(i)

Energy of photon emitted when electron makes a transition from (n = 4) to (n = 2) state :

Ans. 13. Option (d) is correct. Given : Electron in a hydrogen atom is in second excited state (n = 3), radius of the orbit of electron is r = 4.65 Å To find : lB, the de-Broglie wavelength of the electron. By Bohr’s model : nh mvr = ...(i) 2p Put given values in equation (i). nh 3h mv = = 2p r 2p( 4.65 ¥ 10 -10 )

Ê 1 hc 1ˆ = 13.6 ¥ Á 2 - 2 ˜ ÁË ni n f ˜¯ λ

...(ii)

1ˆ Ê 1 E¢ ∝ Á 2 - 2 ˜ Ë2 4 ¯ hc 1ˆ Ê 1 ∝ Á 2 - 2˜ Ë λ¢ 2 4 ¯ From equations (i) and (ii), 1ˆ Ê 1 ÁË 2 - 2 ˜¯ 3 = 5 ¥ 16 ¥ λ λ¢ = λ ¥ 2 1 ˆ 36 3 Ê 1 ÁË 2 - 2 ˜¯ 2 4

λ¢ =

20 λ 27

...(ii)

370 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 16. Option (a) is correct. Given : Mass of particle is m, the particle moves in a circular orbit in a central potential field 1 U( r ) = kr 2 . (i) 2

and substitute for vn in equation (iii). Ê nh ˆ mÁ Ë 2≠ mrn ˜¯

mv r

2

= krn2

rn4 ∝ n 2 rn ∝

n

1 2 1 2 krn + mvn 2 2

En =

= kr

From equation (iii),

For nth orbit :

1 2 1 2 krn + krn 2 2

En = mvn2 = krn rn

...(iv)

Energy of particle in orbital n : En = potential energy + kinetic energy

...( ii )

As the particle gets trapped in this potential, the force is acting like a centripetal force. 2

nh 2p

mvnrn =

To find : Variation of radii of possible orbitals rn and energy levels En with quantum number n. Magnitude of force for a conservative field : dU d Ê 1 2 ˆ F = = kr ˜ ¯ dr dr ÁË 2 F = kr

PHYSICS

En = krn2

...(iii)

En ∝ n

Use Bohr’s quantisation rule :

[ use equation (iv)]

Topic-2

Radioactivity Concept Revision (Video Based) Alpha, Beta, Gamma Particles

Radioactivity       

 

Part -1      Part - 2

    

              Part -1    Part - 2

Radioactive Decay Law     

 

     Part -1           Part - 2

  Part -1    Part - 2

    

 

Mass Defect and Binding Energy

Nuclear Fission and Fusion              

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV given that the Avogadro number, N = 6.023 × 1026 per kilo mole and 1 eV = 1.6 × 10–19 J. The power output of the reactor is close to : (a) 54 MW (b) 60 MW (c) 125 MW (d) 35 MW [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. In a radioactive material, fraction of active material 9 remaining after time t is . The fraction that was 16 t remaining after is : 2

3 4 4 (c) 5 (a)

7 8 3 (d) 5 (b)

[JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. Find the Binding energy per nucleon for 120 20 Sn. Mass of proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and mass of tin nucleus msn = 119.902199 U. (take 1U = 931 MeV) (a) 9.0 MeV (b) 8.5 MeV (c) 8.0 MeV (d) 7.5 MeV [JEE (Main) – 4th Sep. 2020 - Shift-2]

371

ATOMS AND NUCLEI

Q.4. Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives T1/2(A) : T1/2(B) : T1/2(C) are in the ratio : ln R 64-

A

2- C 0-

B 5

10

t (yrs)

(a) 2 : 1 : 3 (c) 4 : 3 : 1

(b) 3 : 2 : 1 (d) 2 : 1 : 1 [JEE (Main) – 5th Sep. 2020 - Shift-1] Q.5. A radioactive nucleus decays by two different processes. The half-life for the first process is 10s and that for the second is 100s. The effective half life of the nucleus is close to : (a) 12 sec. (b) 55 sec. (c) 6 sec. (d) 9 sec. [JEE (Main) – 5th Sep. 2020 - Shift-2] Q.6. You are given that Mass of 73 Li = 7.0160 u, Mass of 24 He = 4.0026 u and Mass of 11 He = 1.0079 u When 20 g of 73 Li is converted into 24 He by proton capture, the energy liberated, (in kWh), is : [Mass of nucleon = 1 GeV/c2] (a) 1.33 × 106 (b) 6.82 × 105 6 (c) 8 × 10 (d) 5.5 × 105 [JEE (Main) – 6th Sep. 2020 - Shift-1] Q.7. Given the masses of various atomic particles mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u, mv = 0, md = 2.0141 u, where p = proton, n ∫ neutron, e ∫ electron, v ∫ antineutrino and d ∫ deuteron. Which of the following process is allowed by momentum and energy conservation : (a) n + p Æ d + y (b) p Æ n + e+ + υ (c) n + n Æ deuterium atom (electron bound to the nucleus) (d) e+ + e– Æ g [JEE (Main) – 6th Sep. 2020 - Shift-2] Q.8. The activity of a radioactive sample falls from 700 s–1 to 500 s–1 in 30 minutes. Its half life is close to : (a) 72 min (b) 62 min (c) 66 min (d) 52 min [JEE (Main) – 7th Jan. 2020 - Shift-2] Q.9. A nucleus A, with a finite de-broglie wavelength lA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The deBroglie wavelengths lB and lC of B and C are respectively : (a) lA, 2lA (b) 2lA, lA λA λ (c) λA , (d) A , λA 2 2

[JEE (Main) – 8th April 2019 - Shift-2]

Q.10. The ratio of mass densities of nuclei of 40Ca and 16 O is close to : (a) 1 (b) 0.1 (c) 5 (d) 2 [JEE (Main) – 8th April 2019 - Shift-2] Q.11. Two radioactive materials A and B have decay constants 10l and l, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time : 1 1 (a) (b) 11 λ 9λ (c)

11 10 λ

(d)

1 10 λ

[JEE (Main) – 10th April 2019 - Shift-1] Q.12. Two radioactive substances A and B have decay constants 5l and l respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to 2

Ê 1ˆ become Á ˜ will be : Ë e¯ (a)

1 2λ

(b)

1 4λ

(c)

1 λ

(d)

2 λ

[JEE (Main) – 10th April 2019 - Shift-2] Q.13. Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be : (a) 3 : 8 (b) 1 : 8 (c) 8 : 1 (d) 9 : 8 [JEE (Main) – 12th April 2019 - Shift-2] Q.14. A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 × 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for halflives of A and B would then be respectively : (a) 5 days and 10 days (b) 10 days and 40 days (c) 20 days and 5 days (d) 20 days and 10 days [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.15. At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio RB of their activities after time t itself decays RA with time t as e–3t. If the half-life of A is ln 2, the half-life of B is : ln 2 (a) 4ln 2 (b) 2 (c)

ln 2 4

(d) 2ln 2

[JEE (Main) – 9th Jan. 2019 - Shift-2] Q.16. Using a nuclear counter the count rate of emitted particles from a radioactive source is measured.

372 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to : (a) 200 (b) 150 (c) 400 (d) 360 [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.17. Consider the nuclear fission Ne20 ® 2He4 + C12 Given that the binding energy/nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement : (a) energy of 9.72 MeV will be supplied (b) 8.3 MeV energy will be released (c) energy of 3.6 MeV will be released (d) energy of 11.9 MeV has to be supplied [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.18. In a radioactive decay chain, the initial nucleus 232 is 90 Th. At the end there are 6 a-particles and 4 b-particles which are emitted. If the end nucleus is A Z X, A



and Z are given by : (a) A = 208; Z = 80 (b) A = 202; Z = 80 (c) A = 208; Z = 82 (d) A = 200; Z = 81 [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (b) 5. (d) 9. (d) 13. (d) 17. (a)

2. (a) 6. (d) 10. (a) 14. (c) 18. (c)

3. (b) 7. (a) 11. (a) 15. (c)

4. (a) 8. (b) 12. (a) 16. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. 235 , Given : Fuel used in a nuclear reactor is U 92 mass of the fuel used in t = 30 days is m = 2 kg, energy released per fission reaction is E = 200 MeV, Avogadro’s number is N = 6.023 × 1026 per kilo mole, 1eV = 1.6 × 10–19 J. To find : P, power output of the reactor. 235 Number of U 92 atoms in m = 2 kg of fuel :

2 × 6.023 × 10 26 n = 235 Energy released on fission of m = 2 kg of fuel : E′ = nE =

2 × 6.023 × 10 26 × 200 × 10 6 × 1.6 × 235 10 −19 J

Power output of the reactor in t = 30 days : P =

E′ 2 × 6.023 × 10 26 × 200 × 10 6 × 1.6 × 10 −19 = t 235 × 30 × 24 × 60 × 60

P ª 60MW Ans. 2. Option (a) is correct. Given : In a radioactive sample, fraction of material N1 9 remaining after time t is = , where No is N o 16

PHYSICS

the number of nuclei originally present in the radioactive sample and N1 is the number of nuclei present in the radioactive sample at time t. N2 , the fraction of material remaining To find : No t after time . 2 Let l be the decay constant for the given radioactive material. At time t : N1 9 = e −λt = No 16 At time

t : 2

N2 = e −λt / 2 No 1/2

N2  9  = ( e − λt )1 / 2 =   No  16  N2 3 = No 4

Ans. 3. Option (b) is correct. Given : Mass of proton is mp = 1.00783 U, mass of neutron is mn = 1.00867 U, mass of tin nucleus is mSn = 119.902199 U, 1 U=931 MeV. To find : BE, the binding energy per nucleon for Sn120 50 . Mass defect : ∆m = 50mp + (120 − 50 )m n − mSn ∆m = 50 × 1.00783 + 70 × 1.00867 − 119.902199 ∆m = 1.096201U Energy corresponding to the mass defect : ( ∆m)c 2 = 1.096201U × 931 = 1020.563 MeV 120 Binding energy per nucleon for Sn50 :

BE =

( ∆m)c 2 (1020.563) = = 8.5 MeV 120 120

Ans. 4. Option (a) is correct. Given : lnR versus time t(years) graph for three radioactive substances A, B and C. ln R 642- C 0-

A B 5

10

t (yrs)

To find : T1/2 (A) : T1/2 (B) : T1/2 (C), the ratio of their half lives. Let lA, lB, lC, be the decay constants for the radioactive substances A, B and C respectively. For substance A : R = R o e − λA t

373

ATOMS AND NUCLEI

ln R = ln R o − λA t

...(i)

From equation (i), lA will be the slope of the graph lnR versus time t for substance A. 3 λA = ...(ii) 5 Similarly :

λB =

6 2 ,λC = 5 5

T1 / 2 ( A ) : T1 / 2 ( B) : T1 / 2 (C ) =

...(iii)

E′ 0.2 × 1010 = kWh time 3600

P = 5.5 × 10 5 kWh

Ans. 5. Option (d) is correct. Given : A radioactive nucleus decays by two different processes; half life for process 1 is t1 = 10 s, half life for process 2 is t2 = 100 s. To find : teff, the effective half life of the radioactive nucleus. Let l1 be the decay constant for process 1 and l2 be the decay constant for process 2. If No is the number of radioactive nuclei in the sample at time t = 0, then number of radioactive nuclei remaining in the sample after time t will be : N = N o e −( λ1 + λ2 )t dN = −( λ1 + λ2 )N dt From above equation, the effective decay constant for the two processes will be : λeff = ( λ1 + λ2 ) ln 2 ln 2 ln 2 = + teff t1 t2 1 1 1 = + teff 10 100 100 ª 9 s 11

Ans. 6. Option (d) is correct. 4 Given : Mass of Li 73 is mLi = 7.0160 u, mass of He 2 1 is mHe = 4.0026 u, mass of H1 is mH = 1.0079 u, initial mass of Li used is M = 20 g, mass of nucleon GeV is m= u 1 2 . n 1= c To find : Energy liberated in kWh for the following reaction,

∆m = 0.0187 u

6.023 × 10 23 × 20 = 17.21 × 10 23 7

E′ = 17.21 × 10 23 × 0.0187 = 0.32 × 10 23 GeV

P =

5 5 5 : : 3 6 2

Mass defect for the above reaction : ∆m = 1.0079 + 7.0160 − 2 × 4.0026 

n =

Power liberated in n = 17.21 × 1023 reactions :

ln 2 ln 2 ln 2 : : λA λB λC

Li 73 + H11 → 2 He42

Number of Li atoms in M = 20 g sample :

E′ = 0.2 × 1010 kJ

T1 / 2 ( A ) : T1 / 2 ( B) : T1 / 2 (C ) = 2 : 1 : 3

teff =

E = ( ∆m)c 2 = 0.0187 GeV

Energy released in n = 17.21 × 1023 reactions :

From equations (ii) and (iii) : T1 / 2 ( A ) : T1 / 2 ( B) : T1 / 2 (C ) =

Energy released in the above reaction :

...(i)

Ans. 7. Option (a) is correct. Given : Mass of proton is mp = 1.0072 u, mass of neutron is mn = 1.0087 u, mass of electron is me = 0.000548 u, mass of anti neutrino is mν = 0u, mass of deuteron is md = 2.0141 u. To find : A process allowed by energy and momentum conservation. a. n01 + p11 → d12 + γ , the reaction is balanced and Dm = +0.0018 u, the excess energy corresponding to Dm is taken away by the g particle. So, the reaction is allowed. 1 1 0 b. p1 → n0 + e +1 + ν , the reaction is balanced and Dm = –0.002048 u, as Dm is negative the reaction does not seem to follow the law of conservation of energy. 1 1 2 0 c. n0 + n0 → d1 + e −1 , the reaction is balanced and Dm =+0.001252 u, but there is no particle on the right hand side of the reaction to take away the excess energy corresponding to Dm. So, the reaction is not allowed.

d. e −01 + e +01 → γ , the reaction is balanced and Dm = +0.001096 u, but the reaction does not seem to follow the law of conservation of linear momentum. To conserve momentum we need two g particles. Ans. 8. Option (b) is correct. Given : Activity of a radioactive sample at t1 is A1 = 700 s–1 and at t2 is A2 = 500 s–1, where t2 – t1 = 30 min. To find : t1/2, half-life of the sample. Activity of a sample at time t1, when N1 is the number of nuclei present in the sample at t1 and No is number of nuclei present in the sample at t = 0 and l is the decay constant: A1 = λ N1 = λ N o e − λt1

...(i)

A 2 = λ N 2 = λ N o e − λ t2

...(ii)

Similarly, From equations (i) and (ii) : A1 = e( λ ( t2 −t1 )) A2

...(iii)

374 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) de-Broglie wavelength of C : h h h = λC = = ( m)vC Ê 4 vA ˆ Ê vˆ mÁ ˜ mÁ Ë 2¯ Ë 2 ˜¯

700 = e 30 λ 500 7 ln   = 30λ 5

Ê h ˆ = Á = λA Ë ( 2m)vA ˜¯

λ = 0.0112  t1 / 2 =

0.693 ª 62 min λ

Ans. 9. Option (d) is correct. Given : de-Broglie wavelength of nucleus A is lA, A undergoes a spontaneous fission into two nuclei B and C, mass of B = mass of C = m, velocity of B is in same direction as A, velocity of C is in opposite direction as A, 1 vC = - vB . 2 To find : de-Broglie wavelengths lB and lC of B and C. Before fission A

C

B

Ans. 10. Option (a) is correct. Given : Nucleus-1 is Ca40 and nucleus-2 is O16. To find : R, the ratio of mass densities of the two nuclei. Mass density of nucleus is a constant and does not depend on specific element. So, R = 1 : 1 = 1 Ans. 11. Option (a) is correct. Given : Decay constant of material A is 10l, decay constant of material B is l, at time t = 0 number of nuclei in material A = number of nuclei in material B = No . To find : Time T, at which the ratio of number of nuclei of A to that of B is NA 1 ...(i) = NB e N A = N o e -10 λ T

v

vB = v Velocity of C :

As seen from the diagram, momentum of the system before fission : 2mvA Momentum of the system after fission : mv mv mvB + mvC = mv =  2 2 As the momentum needs to be conserved for the entire process : mv 2mvA = 2 de-Broglie wavelength of A : h λA = ( 2m) vA

...(i)

NB = Noe -λT From equation (ii) and (iii) : NA = e -9 λ T NB

...(iii)

...(iv)

T =

1 9λ

Ans. 12. Option (a) is correct. Given : Decay constant of material A is 5l, decay constant of material B is l, at time t = 0 number of nuclei in material A = number of nuclei in material B = N o. To find : Time T, at which the ratio of number of nuclei of A to that of B is NA Ê 1ˆ = Á ˜ Ë e¯ NB

2

...(i)

Number of nuclei present in material A at t = T : ...(ii)

N A = N o e -5λ T

...(ii)

–lt

(use : N = Noe for a radioactive material) Number of nuclei present in material B at t = T :

de-Broglie wavelength of B : h h h λB = = = ( m) vB mv m ( 4 vA ) 1Ê h ˆ λA = = Á 2 Ë ( 2m) vA ˜¯ 2

(use : N = Noe for a radioactive material) Number of nuclei present in material B at t = T :

Compare equations (iii) and (i) : 9λ T = 1

v 2

v = 4 vA

...(ii)

–lt

Mass of A = mass of B + mass of C = m + m = 2m Velocity of A : vA Velocity of B :

vC = -

...(iv)

Number of nuclei present in material A at t = T :

After fission v/2

PHYSICS

NB = Noe -λT ...(iii)

From equation (ii) and (iii), NA = e -4 λ T NB

...(iii)

...(iv)

375

ATOMS AND NUCLEI

Compare equations (i) and (iv) : 4λ T = 2 1 T = 2λ Ans. 13. Option (d) is correct. Given : Half-life of nuclei A is t1A/ 2 = 10 min, half-

life of nuclei B is t1B/ 2 = 20 min, in a sample at time t = 0 number of nuclei of A = number of nuclei of B = N o. To find : At time T = 60 min, the ratio of decayed numbers of nuclei A and B. Number of nuclei present in material A at time t=T: No N N N NA = = 60 /o10 = 6o = o T / t1A/ 2 64 2 2 2 Number of nuclei decayed in material A by time t=T: N 63 1 - NA = 1 - o = No 64 64 Number of nuclei present in material B at time t=T: No N N N NB = = 60 /o20 = 3o = o T / t1B/ 2 8 2 2 2 Number of nuclei decayed in material B by time t=T: N 7 1 - N B = 1 - o = No 8 8 Ratio of decayed numbers of nuclei A and B : 1 - NA 63 8 9 = ¥ = = 9:8 1 - NB 64 7 8

Half-life of material B will be given as : 0.693 t1B/ 2 = λB

...(v)

From equations (iv) and (v) : t1A/ 2 λ 4 = B = B λA 1 t1 / 2 The only option which shows this ratio is option c. t1A/ 2 = 20 days and t1B/ 2 = 5 days. Ans. 15. Option (c) is correct. Given : At time t = 0 activity of radioactive material A R oA = activity of radioactive material B R oB , after time t, RB = e -3t , ...(i) RA half-life of A is t1A/ 2 = ln 2.

To find : tB1/ 2 , half-life of B. o Let at t = 0 number of nuclei in A be N A and o number of nuclei in B be N B . Let lA and lB be decay constants for materials A and B respectively. So, activity of A at t = 0 :

R oA = λA N oA

...(i)

N So, number of nuclei in sample B will be o and let 2 decay constant be lB. Then, activity of radioactive material B will be : λ N R B = B o = 20mCi ...(ii) 2 ...(iii)

...(ii)

Activity of B at t = 0 : R oB = λB N oB R oA

=

R oB

Half-life of A : t1A/ 2 =

...(iii)

;

λA N oA = λB N oB

To find : t1A/ 2 and t1B/ 2 , half-lives of A and B. Let initial number of nuclei in sample A be No and the decay constant be lA. Then, activity of radioactive material A will be :

From equations (i) and (ii) : λA 1 = λB 4

...(iv)

As,

Ans. 14. Option (c) is correct. Given : A radioactive material A has activity of RA = 10 mCi, another radioactive material B has activity of RB = 20 mCi, at time t = 0 number of nuclei in sample A is twice the number of nuclei in sample B, decays 1Ci = 3.7 ¥ 1010 . sec

R A = λA N o = 10mCi

Half-life of material A will be given as : 0.693 t1A/ 2 = λA

...(iv)

0.693 = ln 2 λA

0.693 =1 ...(v) ln 2 Number of nuclei present in material A after time t :

λA =

N oA e - λA t So, activity of material A after time t : R A = λA N oA e - λA t

...(vi)

Similarly, activity of material B after time t : R B = λB N oB e - λB t

...(vii)

From equations (vi) and (vii) : RB λ N o e - λB t = B oB - λ t = e -3t RA λA N A e A

...(viii)

Substitute from equation (iv) in equation (viii) : λB - λA = 3 ...(ix) Substitute from (v) in (ix). λB = 4 tB1 / 2 =

0.693 0.693 ln 2 = = λB 4 4

376 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 16. Option (a) is correct. Given : Count rate of emitted particles from a radioactive source at t = 0 is R1 = 1600 count s-1 , at t = 8 s the count rate is R 2 = 100 count s-1 . To find : R, count rate observed at t = 6s. Activity of a radioactive material at t = 0 is R1 = 1600 count s-1 .

Ans. 18. Option (c) is correct. Given : Initial nucleus in a radioactive decay chain 232 is Th 90 , the end product includes 4b particles, 6a

particles and a nucleus X ZA . To find : Value of A and Z. The overall nuclear fission reaction can be written as : 232 Th 90 Æ X ZA + 6α 24 + 4 β -01

Its activity at one half-life (t = 1t1 / 2 ) :

A = 208

Its activity at two half-lives (t = 2t1 / 2 ) :

From reaction (i) : 90 = Z + ( 6 ¥ 2 ) - ( 4 ¥ 1)

R 2(1 / 2 ) = 400 count s-1 Its activity at three half-lives (t = 3t1 / 2 ) : -1

Z = 82 ...(i)

Its activity at four half-lives (t = 4t1 / 2 ) : R 4(1 / 2 ) = 100 count s-1 = R 2

...(ii)

From equation (ii), t = 8 s is same as four half-lives. That implies, t1 / 2 =

8 = 2 s 4

So, three half-lives will be 3t1/2 = 3 × 2 = 6 s and from equation (i), we can see that the count rate observed at t = 6 s is R = 200 count s

-1

Ans. 17. Option (a) is correct.

Subjective Questions (Chapter Based) Q.1. In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 305 Å. The corresponding difference for the Paschan series in Å is : [JEE (Main) – 4th Sep. 2020 - Shift-1] Sol. Given : the difference between the largest and shortest wavelength of the Lyman series of L L − λmin = 305Å. hydrogen atom is, λmax

P P To find : λmax − λmin , the corresponding difference in Paschen series. For Lyman series :

Given : Binding energy/nucleon of Ne20 is 8.03 MeV, binding energy/nucleon of He4 is 7.07 MeV, binding energy/nucleon of C12 is 7.86 MeV.

L λmin

To find : Amount of energy to be supplied to the nuclear fission reaction, Ne20 ® 2He4 + C12.

L λmax

Energy absorbed or released in a nuclear reaction = binding energy of products – binding energy of reactants. Binding energy of products : BE p = 2 ¥ ( B.E. of He4 ) + ( B.E. of C12 )

BEp = 2 ¥ 4 ¥ 7.07 + 12 ¥ 7.86 = 56.56 + 94.32 = 150.88 eV Binding energy of reactants : BEr = B.E.of Ne 20 BEr = 20 ¥ 8.03 = 160.6 MeV So,

...(i)

From reaction (i) : 232 = A + ( 6 ¥ 4 ) + 0

R1 / 2 = 800 count s-1

R 3(1 / 2 ) = 200 count s

PHYSICS

DE = BE p - BEr = 150.88 - 160.6 = -9.72 MeV

Thus, a total energy of ΔE = 9.72 MeV needs to be supplied to the nuclear fission reaction.

1 1

1  1  1 L = R  2 − 2  = R ;λmin = R ∞  1 1  3 4  1 L = R  2 − 2  = R ;λmax = 3R 2  4 1

L L λmax − λmin =

1 1 = 305 Å ; = 915 Å 3R R

For Paschen series : 1 P λmin

1 P λmax

1  R P 9  1 = R  2 − 2  = ; λmin = R 9 ∞  3 1  7 144  1 P = R 2 − 2  = R ;λmax = 7R 4  144 3

P P λmax − λmin =

144 9 − = 10587.86 Å 7R R

Q.2. A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest and the atom recoils and goes to its first excited state. The initial kinetic N . The value of energy of the particle (in eV) is 4 N is : (Given the mass of the hydrogen atom to be 1 GeV/c2)........

[JEE (Main) – 5th Sep. 2020 - Shift-1]

377

ATOMS AND NUCLEI

Sol. Given : Mass of a particle is m = 200 MeV/c2 , N initial kinetic energy of the particle is K = eV , 4 GeV 5m , initial mass of hydrogen atom = is M 1= c2 kinetic energy of hydrogen atom is 0, after collision, velocity of particle is 0 and the hydrogen atom recoils and goes to its first excited state. To find : The value of N. Let the initial velocity of the particle be v and the final velocity of hydrogen atom be v ¢. By law of conservation of linear momentum : mv + 0 = 0 + 5 mv′ v v′ = 5

Sol. Given : Number of electrons in a hydrogen type ionized atom with atomic number Z is 1, in that atom the photon emitted during transition from ni = 2 to nf = 1 state has 74.8 eV more energy than the photon emitted during transition from ni = 3 to nf = 2, ionization energy of hydrogen atom is E = 13.6 eV. To find : Value of Z. Energy of photon emitted during transition from ni = 2 to nf = 1 : Ê 1 1ˆ DE2 Æ1 = EZ 2 Á 2 - 2 ˜ ÁË n f ni ˜¯



1ˆ Ê 1 = 13.6 ¥ Z 2 ¥ Á 2 - 2 ˜ Ë1 2 ¯



Loss of kinetic energy during the collision: 2

1 2 1 41 v  4 N N mv − ( 5m)   =  mv 2  = × =  2 2 5 5 2 5     5 4 The lost kinetic energy is used to place hydrogen atom in the first excited state. N = 10.2 eV 5



DE2 Æ1 = 13.6 ¥

Ê 1 1ˆ DE3Æ 2 = EZ 2 Á 2 - 2 ˜ ÁË n f ni ˜¯



To find : l4Æ2, wavelength corresponding to the transition (n = 4) to (n = 2). Using Rydberg’s equations for hydrogen atom (Z = 1) for transition (n = 3) Æ (n = 2) : 1 1   1 ∝  2 − 2 ...(i) λ3 → 2 3  2 Similarly, for transition (n = 4) Æ (n = 2) : 1 1   1 ∝  2 − 2 λ4 → 2 4  2

...(ii)

From equations (i) and (ii) :

λ4 → 2

1   1  2 − 2 5 16 2 3  λ = = × × 6561 × 1  3 → 2 36 3  1 −  2  42  2 10 −10 = 486 nm

Q.4. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is .................. .

1ˆ Ê 1 = 13.6 ¥ Z 2 ¥ Á 2 - 2 ˜ Ë2 3 ¯



...(i)

Energy of photon emitted during transition from ni = 3 to nf = 2 :

N = 51eV Q.3. The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is _______. [JEE (Main) – 8th Jan. 2020 - Shift-2] Sol. Given : In an hydrogen atom wavelength corresponding to the transition (n = 3) to (n = 2) is l3Æ2 = 6561 Å.

3 ¥ Z2 4

DE2 Æ1 = 13.6 ¥

5 ¥ Z2 36

...(ii)

From equations (i) and (ii), 3 5 13.6 ¥ ¥ Z 2 = 74.8 + 13.6 ¥ ¥ Z2 4 36 13.6 ¥

22 ¥ Z 2 = 74.8 36 Z2 = 9

Z = 3 Q.5. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf  . Vi and Vf are respectively the initial and final potential energies V of the electron. If i = 6.25, then the smallest Vf possible nf is Sol. Given : An electron in hydrogen atom makes a transition from an energy state ni to another energy state nf, the ratio of potential energies of the electron in the two states is Vi = 6.25. Vf To find : The smallest possible value of nf. Potential energy of electron in hydrogen atom revolving in nth state is : 2.72 Vn = - 2 ...(i) n

378 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

From equation (i) : Vi Vf

nf

Ê 2.72 ˆ Á - 2 ˜ n2 Ë ni ¯ f = = 2 = 6.25  2.72 n i - 2 nf 5 = 2.5 = 2

ni So, smallest possible value of nf is nf = 5. Q.6. A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/e = 1.237 × 10–6 eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is Sol. Given : Wavelength of light incident on hydrogen atom is l = 970 Å, ground state energy of hydrogen atom is = –13.6 eV, hc = 1.237 ¥ 10 -6 eV - m. e

To find : The number of lines present in the emission spectrum. Energy of incident light : hc 12500 Eλ = = = 12.9 eV ...(i) λ 970 (use hc = 12500 eV. Å) After absorbing the incident radiation let the electron jump to energy state n. Energy of state n : 13.6 En = - 2 ...(ii) n So, E + Eλ = En From equations (i) and (ii), 13.6 -13.6 + 12.9 = - 2 n n = 4 Total number of spectral lines in the emission spectra of hydrogen atom when the electron will de-excite from n = 4 state : n ( n - 1) N = =6 2

Q.7. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is Sol. Given : Half-life of a radioisotope is t1/2 = 1386 s, activity of isotope sample is 103 dps. To find : Percentage of the initial number of nuclei that will decay in the first t = 80 s. Let initial number of nuclei in the radioisotope sample be No.



PHYSICS

So, number of nuclei present in the sample after time t will be : N = No e - λt Number of nuclei decayed in the sample after time t will be : N d = N o - N = N o (1 - e - λ t ) l is the decay constant : 0.693 0.693 λ = = = 5 ¥ 10 -4 s t1 / 2 1386

...(i)

...(ii)

Put t = 80 s and value of l from equation (ii) in equation (i). N d = N o (1 - e - λ t ) = N o (1 - e -5 ¥10



-4

¥ 80

)

N o1 = [1 - ( -5 ¥ 10 -4 ¥ 80 )] No (expand e - x )

So, percentage of the initial number of nuclei that will decay in the first t = 80 s : Nd ¥ 100 = 5 ¥ 10 -4 ¥ 80 ¥ 100 = 4 No

Q8. For a radioactive material, its activity A and rate of dN change of its activity R are defined as A = – dt dA and R = – , where N(t) is the number of nuclei dt at time t. Two radioactive sources P (mean life t) and Q (mean life 2t) have the same activity at t = 0. Their rates of change of activities at t = 2t R n are Rp and RQ, respectively. If P = , then the RQ e value of n is Sol. Given : Mean-life of radioactive source P is t, meanlife of radioactive source Q is 2t, at t = 0 activity of P = activity of Q Ap = AQ, at t = 2t rate of change of R n their activities are P = . RQ e

To find : Value of n. Let number of nuclei in radioactive source P at t = 0 0 be N P , number of nuclei in radioactive source Q





0 at t = 0 be NQ . Decay constant of radioactive source P be : lP Decay constant of radioactive source Q be : lQ 1 λP τ So, = =2 ...( i ) 1 λQ 2τ

Activity of P : AP = -

dN P = λP N P dt

...(ii)

Activity of Q : AQ = -

dN Q dt

= λQNQ

...(iii)

379

ATOMS AND NUCLEI



At At





t = t : A P = λP N P = RP = -



0 t = 0 ( A P = A Q ) :λP N 0P = λQNQ

At

λP N 0P e - λPt

dA P = λP2 N 0P e - λP t dt

0 t = t : A Q = λQNQ = λQNQ e

RQ = -

dA Q dt

0 = λQ2 NQ e

RP = RQ

=



- λQ t

- λQ t

n e



...(v)

From equations (v) and (vi), at t = 2t :

λP2 N 0P e - λP ( 2τ ) 0 - λQ ( 2τ ) λQ2 NQ e

...(iv)

Use equations (iv) and (i), RP e - λP ( 2τ ) n = 2 - λ ( 2τ ) = RQ e e Q Put λP = 2 e -2

...(vi)



e

-1

1 1 ,λ = in the above equation : τ Q 2τ =

n e



2 n = e e



n = 2





of band theory

Zero kelvin (0K)

Electronic Devices

s

B

A

B

A

A

OR Gate

B

A

NOT Gate

AND Gate

A+B

N F1 NP

trivalent (B, Al) elements

used as a voltage regulator

A B

E

A

C

NOR Gate

AB

B

A+B

b

Bias

Signal in

P F2 PN

NAND Gate

AB

B

e

E

c

C

0V

Signal out

Load resistor

+V

380 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) PHYSICS

Electronic Devices

Chapter 19 Syllabus

Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias, diode as a rectifier; I-V characteristics of LED, photodiode, solar cell and Zener diode; Zener diode as a voltage regulator. Junction transistor, transistor action, characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator Thi. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch.

Concept Revision (Video Based)    

Semiconductor Diodes

I-V Characteristics in Forward and Reverse Bias     

              

 

Part -1    Part - 2

Light Emitting Diode

Zener Diodes       

Part -1    Part - 2 Calculating Current through Zener Diode Characteristics of Zener Diode

Transistors    Logic Gates

PN Junction Diode   

    

Part -1   

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. In the following digital circuit, what will be the output a ‘Z’, when the input (A, B) are (1, 0), (0, 0), (1, 1), (0, 1) A

Z

B

(a) 1, 1, 0, 1 (c) 1, 0, 1, 1

(b) 0, 1, 0, 0 (d) 0, 0, 1, 0 [JEE (Main) – 2nd Sep. 2020 - Shift-2] Q.2. When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to

be connected in series with the diode so that the current does not exceed the safe limit is : (a) 300 W (b) 200 W (c) 50 W (d) 100 W [JEE (Main) – 3rd Sep. 2020 - Shift-1] Q.3. If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is : Planck’s constant h = 6.63 × 10–34 J-s. Speed of light c = 3 × 108 m/s (a) 1.1 eV (b) 1.5 eV (c) 2.0 eV (d) 3.1 eV [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.4. Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in the figure below, the time variation of the output voltage is : (Graphs drawn are schematic and not to the scale)

(a) V= 0

(a) V= 0 (c) V= 0

(c) V= 0

382 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

V0 variation with input voltage linearly increasing with time, is given by (Vinput = 0V at t = 0)

R1 10 V

Vo

Vin

V= 0

B A

–10 V

Vin

6V

(a) (a) V= 0

t

(b) t

(a) V= V= 0 (b) (c) V=00

tt t

t t

VO

t

(a)

(b) V= 0 (d) V= 0

Vo

t t

Vo 4V

(a)

6V

(b)

time t tt

(d) V= V 0o

t

(a)

(d) t

RL=100 

4V 100 

(b) V= 0

(c)

(b) V= (c) V= V=000 (d)

(d) V= 0

Vo

[JEE (Main) – 4th Sep. 2020 - Shift-1] (c) Q.5. Which of the following will NOT be observed time when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed? (a) Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED (b) Multimeter shows NO deflection in both cases i.e., before and after reversing the probes is the chosen component is metal wire. (c) Multimeter shows NO deflection in both cases i.e., before and after reversing the probes if the chosen component is capacitor. (d) Multimeter shows an equal deflection in both cases i.e., before and after reversing the probes if the chosen component is resistor. [JEE (Main) – 3rd Sep. 2020 - Shift-2] Q.6. Identify the operation performed by the circuit given below:

B (a) C (c)

time

Vo Vo

6V 6V 4V 4V

(b) (c) (a)

V Voo

6V 4V6V

(d) (b)

time time 6V (c) 4V

Vo Vo

time time 6V 6V 4V 4V

(d) (c)

Vo

6V 4V

(d)

time time

(d)

time

Vo

Vo 4V

(a)

6V

(b)

time

time th

[JEE (Main)6V – 5 Sep. Vo Vo 2020 - Shift-2] 6V Q.9. Identify the correct output 4Vsignal Y in the given 4V (c) combination of gates (as shown(d)n) for the given inputs A and B. time time B Y

A

A

B 5

(a)

A

(a) NOT (c) NAND

(b) 4V

time t

PHYSICS

10

(a)

15

t

(b) t (c)

(c) (b) (a) (d) (c)

5

10

15

tt

25

t

(b) t (d) (b)

(d)

tt

(b) AND (a) t 5 10 15 t (b) 5 10 15 tt t t (d) OR (d) [JEE (Main) – 4th Sep. 2020 - Shift-2] (c) (d) Q.7. With increasing biasing voltage of a photodiode, t 5 10 15 t the photocurrent magnitude : [JEE (Main) – 6th Sep. 2020 - Shift-1] (a) remains constant Q.10. Which of the following gives a reversible (b) increases initially and saturates finally operation? (c) increases linearly (a) (b) (a) (b) (d) increases initially and after attaining certain (a)(a) (a) (b)(b) (b) value, it decreases [JEE (Main) – 5th Sep. 2020 - Shift-1] (c) (d) Q.8. Two Zener diodes (A and B) having breakdown (c) (d) (c)(c) (c) (d)(d) (d) voltages of 6V and 4V respectively, are connected as shown in the circuit below. The output voltage [JEE (Main) – 7th Jan. 2020 - Shift-1]

383

ELECTRONIC DEVICES

Q.11. In the figure, potential difference between A and B is : 10 k W

A

10 k W

30 V 10 k W

Q.15. Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm) + + + +

B

R

(a) 15 V

(b) zero

(c) 10 V

(d) 5 V [JEE (Main) – 7th Jan. 2020 - Shift-2]

R

B

– – – –

CV VC = , QB 2 e

VC CV (d) = QA VC = , QB (c) QA CV = = , QB e e 2

+5V

[JEE (Main) – 9th Jan. 2020 - Shift-2] Output-Y

B

5V

+ + + +

(a) QA = VC, QB = CV = (b) QA

Q.12. Boolean relation at the output stage-Y for the following circuit is : A

A

– – – –

Q.16. The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. 200  IZ

(a) A + B

(b) A + B

(c) A ◊ B

(d) A ⋅ B

9 V— –

[JEE (Main) – 8th Jan. 2020 - Shift-1] Q.13. In the given circuit, value of Y is : 1

Y 0

800 



The current IZ through the Zener is : (a) 10 mA (b) 17 mA (c) 15 mA (d) 7 ma [JEE (Main) – 8th April 2019 - Shift-11] Q.17. A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its DC current gain is Amp, RC = 1 kW and VCC = 10V. What is the minimum base current for VCE to reach saturation  ?

(a) toggles between 0 and 1 (c) 0

— – VB

(d) will not execute JEE (Main) – 8th Jan. 2020 - Shift-2] Q.14. The current i in the network is : 5Ω D

10 Ω

C 110 Ω

D 5Ω

— –V CC

—

—

(a) 40 mA

10 Ω 5Ω

i

RC

RB

(b) 1

9V

(c) 7 mA

—

(b) 100 mA (d) 10 mA [JEE (Main) – 8th April 2019 - Shift-2]

Q.18. An NPN transistor is used in common emitter configuration as an amplifier with 1 kW load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 mA change in the base current of the amplifier. The input resistance and voltage gain are :

(a) 0.3 A

(b) 0 A

(a) 0.33 kW, 1.5

(c) 0.2 A

(d) 0.6 A

(c) 0.67 kW, 200 (d) 0.33 kW, 300 [JEE (Main) – 9th April 2019 - Shift-1]

[JEE (Main) – 9th Jan. 2020 - Shift-2]

(b) 0.67 kW, 300

384 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.19. The logic gate equivalent to the given logic circuit is :

•

• (400, 20)

A •

Y B

• (300, 15)

Ic • (mA)

(a) NAND (c) NOR

(b) OR (d) AND [JEE – Mains – 9th April 2019 - Shift-2] Q.20. An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100 W and the output load resistance is 10 kW. The common emitter current gain b is : (a) 102 (b) 60 (c) 6 ×102 (d) 104 [JEE (Main) – 10th April 2019 - Shift-1] Q.21. The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is, RL = 4 kW. The series resistance of the circuit is Ri = 1 kW. If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?

•

• (200, 10) • (100, 5) • • • Ib (A)

•

(a) 2.5 × 104, 2.5 × 106 (b) 5 × 104, 5 × 106 (c) 5 × 104, 5 × 105 (d) 5 × 104, 2.5 × 106 [JEE (Main) – 12th April 2019 - Shift-1] Q.24. Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V. If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current ? Is Rs=2 k IL IZ

Ri VB — –

PHYSICS

RL=4 k RL

(a) 2.5 mA (c) 7.5 mA

(a) 0.5 mA; 6 mA (b) 1 mA; 8.5 mA (c) 0.5 mA; 8.5 mA (d) 1.5 mA; 8.5 mA [JEE (Main) – 10th April 2019 - Shift-2] Q.22. The truth table for the circuit given in the fig. is : A B

Y

A B Y 0 0 1 (a) 0 1 1 1 0 1 1 1 1

A B Y 0 0 1 (b) 0 1 0 1 0 0 1 1 0

A B Y 0 0 1 (c) 0 1 1 1 0 0 1 1 0

A B Y 0 0 0 (d) 0 1 0 1 0 1 1 1 1

[JEE (Main) – 12th April 2019 - Shift-1] Q.23. The transfer characteristic curve of a transistor, having input and output resistance 100 W and 100 kW respectively, is shown in the figure. The Voltage and Power gain, are respectively :

(b) 1.5 mA (d) 3.5 mA [JEE (Main) – 12th April 2019 - Shift-2] Q.25. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019m–3 and their mobility is 1.6 m2/(V◊s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to : (a) 2 Wm (b) 4 Wm (c) 0.4 Wm (d) 0.2 Wm [JEE (Main) – 9th Jan. 2019 - Shift-1] Q.26. Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of Vo changes by   : (assume that the Ge diode has large breakdown voltage) Ge Vo

12 V

(a) 0.8 V (c) 0.2 V

Si

5K

(b) 0.6 V (d) 0.4 V [JEE (Main) – 9th Jan. 2019 - Shift-2]

385

ELECTRONIC DEVICES

Q.27. To get output ‘l’ at R, for the given logic gate circuit the input values must be :

(b) AB + AB

(a) AB + AB (c) AB

(d) AB [JEE (Main) – 12th Jan. 2019 - Shift-1]

X

Q.32.

P

R

IC

Y Q ||

(a) X = 0, Y = 1 (b) X = 1, Y = 1 (c) X = 1, Y = 0 (d) X = 0, Y = 0 [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.28. For the circuit shown below, the current through the Zener diode is :

vi



5 k

120 V — –

10 k

50 V

(a) 9 mA (c) Zero

(b) 5 mA (d) 14 mA [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.29. In the given circuit the current through Zener Diode is close to :

12 V

R1

500 

R2 1500 

V2 = 10 V

R2

(a) 0.0 mA (c) 6.7 mA

(b) 6.0 mA (d) 4.0 mA [JEE (Main) – 11th Jan. 2019 - Shift-1] Q.30. The circuit shown below contains two ideal diodes, each with a forward resistance of 50 W. If the battery voltage is 6 V, the current through the 100 W resistance(in Amperes) is : D1

RB

||

C | B

RC v0

E VCC

IE

VBB

In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5 V, bdc = 200, RB = 100 kW, RC = 1 kW and VBE = 1.0 V, The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively : (a) 25 mA and 3.5 V (b) 20 mA and 3.5 V (c) 25 mA and 2.8 V (d) 20 mA and 2.8 V [JEE (Main) – 12th Jan. 2019 - Shift-2]

ANSWER – KEY

1. (d) 5. (c) 9. (b) 13. (c) 17. (a) 21. (c) 25. (c) 29. (a)

2. (d) 6. (b) 10. (b) 14. (a) 18. (b) 22. (c) 26. (d) 30. (b)

3. (d) 4. (c) 7. (b) 8. (d) 11. (c) 12. (d) 15. (d) 16. (a) 19. (b) 20. (a) 23. (d) 24. (d) 27. (c) 28. (a) 31. (c) 32. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. Given : A logic gate circuit To find : The value of Z, for (A, B) = (1, 0), (0, 0), (1, 1), (0, 1). A

150 

Z

75 k B

— –

D2 100 k

6V

(a) 0.036 (c) 0.027

(b) 0.020 (d) 0.030 [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.31. The output of the given logic circuit is : A

B

W

X

Y

Z

1

0

1

1

1

0

0

0

1

0

0

0

1

1

0

1

0

1

0

1

1

1

1

0

Ans. 2. Option (d) is correct. Y

B

A

Given : Voltage drop across a diode, when forward biased is V = 0.5 V, the safe limit for current that can flow through the diode is I = 10 mA, emf of the battery used in the circuit shown is E = 1.5 V.

386 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) 0.5 V 1V

1.5 V

To find : The value of R, such that the current through the diode does not exceed I. Voltage drop across R : VR = E − V = 1.5 − 0.5 = 1V Value of R : R =

VR 1 = = 100 Ω I 10 × 10 −3

Ans. 3. Option (d) is correct. Given : A semiconductor photodiode can detect a photon with a wavelength of l = 400 nm. To find : Eg, the band gap energy of the photodiode. Eg =

hc 6.63 × 10 −34 × 3 × 10 8 = λ 400 × 10 −9

= 4.9725 × 10 −19 J = 3.10 eV Ans. 4. Option (c) is correct. Given : The breakdown voltage for the Zener diodes used in the circuit is VZ = 6 V. To find : Vo(t) the time variation of the output voltage for the given input waveform Vin(t). The peak of the input waveform is at Vinp = 10V , which exceeds the zener breakdown voltage. The circuit has two Zener diodes in reverse polarity so if one is forward biased the other will be reverse biased. a. For Vin < VZ, the output voltage simply follows the input voltage. b. For Vin > VZ : The reverse biased diode will be in conduction mode and the output will be constant. Ans. 5. Option (c) is correct. Given : A multimeter is operating in a resistance mode. To find : Which of given options will not happen when the two probes of multimeter are connected across a component. When the chosen component is a capacitor, the multimeter will show deflection and read the resistance of the capacitor. Ans. 6. Option (b) is correct. Given : A logic gate circuit To find : The operation performed by the circuit. A B C

W X Y

PHYSICS

A

B

C

0

0

0

1

0

0

0

1

1

0

0

1

0

1

0

1

0

0

0

1

1

1

0

0

1

1

0

0

0

1

0

1

0

1

0

1

0

0

0

1

1

1

0

0

0

1

1

1

0

0

0

1

W=A X=B Y =C 1 1 1

Z 0

Z = 1, only when all inputs are 1. So, the given circuit behaves as an AND gate. Ans. 7. Option (b) is correct. Given : A biasing voltage is applied across a photodiode. To find : The variation of magnitude of photo current as we increase the biasing voltage. With increasing biasing voltage the photo current through the photodiode will initially increase and then finally saturate. Ans. 8. Option (d) is correct. Given : Breakdown voltage for Zener diode-1 is V1 = 6V, breakdown voltage for Zener diode-2 is V2 = 4V, the input voltage Vin(t) varies linearly with time, at t = 0, Vin(t) = 0. To find : Vo(t) the output voltage as function of time. a. For Vin < V2, the output voltage simply follows the input voltage, that is Vo(t) varies linearly with time. b. For V1 > Vin > V2 : Zener diode-2 is in breakdown region, so we see a change in slope of the output voltage. c. For Vin > V1 : Zener diode-1 is also in breakdown region, so we see a constant output voltage independent of time. Ans. 9. Option (b) is correct. Given : A logic gate circuit To find : The correct output signal Y for given input signals A and B. A Y B A

B 5

10

15

25

t

The truth table for the above circuit: A

B

Y

0

0

0

1

0

1

0

1

1

1 1 1 So, the output Y is always 1 except when both inputs are zero.

387

ELECTRONIC DEVICES Y 1

Y 5

10

15

20

0

t

Ans. 10. Option (b) is correct. Given : Four logic gates. To find : The one which gives reversible operation. Since the logic gate b has only one input its operation is reversible. Ans. 11. Option (c) is correct. Given : A circuit diagram as shown below, 10 k W

30 V

To find : The value of Y. 1

0

1 0 0

A

1 0

1 0

10 kΩ

10 kΩ

0

Y

1

1 B

To find : VAB, the potential difference between A and B. As the diode in the circuit is forward biased, it will behave as a simple wire. So, the given circuit effectively becomes : 10 kΩ

A

Ans. 14. Option (a) is correct. Given : A circuit as shown below, 5

10 

D 5 i

10 

110  D

30 V 5 kΩ

5

To find : The current i in the network. Since, the diodes in the circuit are reverse biased they will not conduct and the circuit effectively becomes :

B

VAB =

30 × 5 = 10V 5 + 10

10 

Ans. 12. Option (d) is correct. Given : A logic gate circuit,

5 +5V

A

9V

i

10 

Output-Y 5V

B

5

To find : The Boolean relation at the output stage Y. First part of the circuit shown above is an OR gate and the second part of the circuit is a NOT gate. So, the output at Y will be : = Y

= A + B A.B

Ans. 13. Option (c) is correct. Given : A logic gate circuit,

Effective resistance of the circuit : R eff = 5 + 10 + 5 + 10 = 30Ω Current through the circuit : 9 = i = 0.3 A 30 Ans. 15. Option (d) is correct. Given : Capacitors A and B are identical with capacitance C, both are charged to the same

388 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) potential V = 5 V, at t = 0 they are connected to the circuits shown below, at t = CR charge on capacitor A is QA and charge on capacitor B is QB. + + + +

R

A

– – – –

+ + + +

R

B

— – VB —

a To find : Value of QA and QB. Charge on capacitor A at t = 0 :

b

—

—

VCE = VCC - IC RC = 0

Charge on capacitor B at t = 0 :

IC =

QBo = VC In circuit a, the diode is reverse biased. So, charge on capacitor A at t = CR : QA = VC In circuit b, the diode is forward biased. So, charge on capacitor B at t = CR : VC QB = VCe −t / CR = VCe −CR / CR = e Ans. 16. Option (a) is correct. Given : reverse breakdown voltage of a Zener diode connected as shown in the circuit below is VZ = 5.6 V. 200  9 V— –

— –V CC

To find : The minimum base current for VCE to reach saturation. At saturation, VCE = 0

QAo = VC

I1

RC

RB

– – – –

PHYSICS

IZ

800 

As for a common emitter NPN transistor β DC = saturation base current will be : IB =

VZ = 5.6 V Voltage drop across the R2 = 800 W resistor : VZ = 5.6 V Current through the R2 = 800 W resistor : VZ 5.6 = = 7 mA I2 = R 2 800

I Z = I1 - I 2 = 17 - 7 = 10mA Ans. 17. Option (a) is correct. Given : A common emitter amplifier circuit built using an NPN transistor is as shown below, DC current gain bDC = Ampere, RC = 1 kW, VCC = 10 V.

Vin (10 ¥ 10 -3 ) = = 0.67 k W DI B 15 ¥ 10 -6

Voltage gain : R DIC R L β¥ L = ¥ R in DI B R in Voltage gain =

3 ¥ 10 -3 15 ¥ 10 -6

¥

1 ¥ 10 3 0.67 ¥ 10 3

ª 300

Ans. 19. Option (b) is correct. Given : A logic circuit as shown below, A

Y1 Y

V1 = V - VZ = 9 - 5.6 = 3.4 V

Current IZ through the Zener diode :

IC 10 -2 = = 40 m A β DC 250

R in =

Voltage drop across the R1 = 200 W resistor : Current through the R1 = 200 W resistor : V1 3.4 I1 = = = 17 mA R1 200

IC , IB

Ans. 18. Option (b) is correct. Given : An NPN transistor is used in a common emitter configuration as an amplifier, load resistance is RL = 1 kW, signal voltage applied across the base-emitter is Vin = 10 mV, change in collector current due to the signal voltage is DIC = 3 mA, change in base current due to the signal voltage is DIB = 15 mA. To find : The input resistance Rin and the voltage gain. Input resistance :

I2

To find : The current IZ through the zener diode. Voltage drop cross the Zener diode :

VCC 10 = = 10 -2 A RC 1000

B

Y2

To find : The equivalent logic gate. Truth table for the give logic circuit : A

B

Y1 = A

Y2 = B

Y = Y1 .Y2

0

0

1

1

0

0

1

1

0

1

1

0

0

1

1

1

1

0

0

1

389

ELECTRONIC DEVICES

The output Y of the give logic circuit resembles to the output of OR gate. Ans. 20. Option (a) is correct. Given : An NPN transistor is used in a common emitter configuration as an amplifier; power gain is Ap = 60 dB, input circuit resistance is Rin = 100 W, output load resistance is RL = 10 kW. To find : b, the common emitter current gain. Power gain : Ê output power ˆ A p = 10 log 10 Á Ë input power ˜¯

Maximum voltage drop across Ri = 1 kW resistor : V max = VBmax - VZ = 16 - 6 = 10 V Maximum current through the Ri = 1 kW resistor :

Maximum current through the Zener diode : Imax = Iimax - I L = 10 - 1.5 = 8.5 mA Z Ans. 22. Option (c) is correct. Given : A logic circuit as shown below,

ÊP ˆ 10 log 10 Á o ˜ =  60 Ë Pin ¯

A B

Po = 10 6  Pin Also,

Y X

To find : The correct truth table for the circuit.

ÊR ˆ Po = β2Á L ˜ Pin Ë R in ¯

A

B

X = A+B

Y = AX

0

0

0

1

P R = o ¥ in Pin R L

0

1

1

1

1

0

1

0

1

1

1

0

β2

= 10 6 ¥

β = 10

100 10 ¥ 10

= 10 4

3

2

Ans. 21. Option (c) is correct. Given : A circuit of a voltage regulator constructed by using a Zener diode, breakdown voltage of Zener diode is VZ = 6 V , R L = 4 kW ,Ri = 1kW , VBmin = 8 V ,VBmax = 16 V.

Ans. 23. Option (d) is correct. Given : Transfer characteristic curve of a transistor, R in = 100 W ,R out = 100 kW.

• • Ic • (mA)

Ri

•

VB — –

RL

To find : Imin and Imax Z , minimum and maximum Z values of the current through the Zener diode. Voltage drop across the load resistance : VZ = 6V Current through the load resistance : V 6 IL = Z = = 1.5 mA R L 4 ¥ 10 3 Minimum voltage drop across Ri = 1 kW resistor : V min = VBmin - VZ = 8 - 6 = 2 V Minimum current through the Ri = 1 kW resistor : Imin i

min

V = Ri

=

2 1 ¥ 10 3

= 2 mA

Minimum current through the Zener diode : Imin = Iimin - I L = 2 - 1.5 = 0.5 mA Z Similarly,

V max 10 = = 10 mA Ri 1 ¥ 10 3

Iimax =

• (400, 20) • (300, 15) • (200, 10) • (100, 5) • • • Ib (A)

•

To find : The voltage gain AV and the power gain AP. As the characteristic curve is plotted between IC and IB, we can say the that the transistor is in common emitter configuration. I So, gain β = C IB Voltage gain : AV =

Vout IC ¥ R out = Vin I B ¥ R in

...(i)

2 Pout IC ¥ R out = 2 Pin I B ¥ R in

...(ii)

Power gain : AP =

Choose any value of IC and IB from the given graph and put in equations (i) and (ii). AV =

5 ¥ 10 -3 ¥ 100 ¥ 10 3 100 ¥ 10 -6 ¥ 100

= 5 ¥ 10 4

390 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) AP =

( 5 ¥ 10 -3 )2 ¥ 100 ¥ 10 3 (100 ¥ 10 -6 )2 ¥ 100

= 2.5 ¥ 10 6 Ans. 24. Option (d) is correct. Given : A circuit of a voltage regulator constructed by using a Zener diode, breakdown voltage of Zener diode is

PHYSICS

Ans. 26. Option (d) is correct. Given : Ge diode starts conducting at VGe = 0.3 V, Si diode starts conducting at VSi = 0.7 V. To find : The change in the potential Vo when the ends of the Ge diode are overturned in the figure shown below. Ge Vo

VZ = 6 V ,R L = 4 kW , RS = 2 kW ,

Si

12 V

5K

min max Vin = 10 V , Vin = 16 V.

Is

In the above figure both the diodes are forward biased. So, voltage drop across the parallel combination of the two diodes is VGe = 0.3 V. That implies, Vo = 12 –VGe = 12 – 0.3 = 11.7 V When the ends of the Ge diode are overturned, only Si diode is forward biased and will conduct. So, voltage drop across the parallel combination of the two diodes is VSi = 0.7 V. That implies, Vo = 12 – VSi = 12 – 0.7 = 11.3 V The change in the potential Vo : 11.7 – 11.3 = 0.4 V

Rs=2 k IL IZ RL=4 k

To find : Imax Z , maximum value of the current through the Zener diode. Voltage drop across the load resistance :

Ans. 27. Option (c) is correct. Given : A logic circuit as shown below, X1

X

VZ = 6V P

Current through the load resistance : VZ 6 IL = = = 1.5 mA R L 4 ¥ 10 3

Q

Maximum voltage drop across RS = 2 kW resistor : V

max

=

max Vin

- VZ = 16 - 6 = 10 V

Maximum current through the RS = 2 kW resistor : ISmax =

R

Y

V max 10 = = 5 mA RS 2 ¥ 10 3

Y1

To find : What combination of values of X and Y will give output 1 at R. Truth table for the logic circuit shown above : X Y X1 = X Y1 = Y

Maximum current through the Zener diode :

P = X1 Q = XY1 R = P+Q +Y

Imax = ISmax - I L = 5 - 1.5 = 3.5 mA Z

0 0

1

1

1

1

0

Ans. 25. Option (c) is correct. Given : Density of electrons in an n-type semiconductor is ne = 1019 m–3, mobility of electrons is me = 1.6 m2/V ◊ s. To find : r, resistivity of the semiconductor. Conductivity of an n-type semiconductor :

1 0

0

1

0

0

1

0 1

1

0

1

1

0

1 1

0

0

1

1

0

σ e = ne e m e

...(i)

From equation (i), resistivity of the semiconductor : 1 1 ...(ii) ρ = = σ e ne em e Put given values in equation (ii). 1 ρ = ª 0.4 Wm 19 10 ¥ 1.6 ¥ 10 -19 ¥ 1.6

From the above truth table : X = 1, Y = 0 will give R = 1. Ans. 28. Option (a) is correct. Given : A circuit diagram as shown below, 5 k 120 V — –

IL

I IZ

50 V

10 k

To find : Iz, the current through the Zener diode.

391

ELECTRONIC DEVICES

Voltage drop across the RL = 10 k W resistor : VZ = 50V

middle arm of the circuit is open. Equivalent resistance of the circuit will be : R eq = R + 150 + R ¢ = 50 + 150 + 100 = 300W

Current through the load resistance : VZ 50 IL = = = 5 mA R L 10 ¥ 10 3

So, current through the R¢ = 100 W resistor will be : V 6 I = = = 0.020 A R eq 300

Voltage drop across RS = 5 kW resistor : V = 120 - VZ = 120 - 50 = 70V

Ans. 31. Option (c) is correct. Given : A logic circuit as shown below, A D A

Current through the RS = 5 kW resistor : V 70 I = = = 14 mA RS 5 ¥ 10 3

C

Current through the Zener diode : I Z = I - I L = 14 - 5 = 9mA Ans. 29. Option (a) is correct. Given : A circuit diagram as shown below, 12 V

R1

B

E B To find : The output of the given circuit. Output at C :

500 

C = ( A.B)

R2 1500 

V2 = 10 V

R2

To find : IZ, the current through the Zener diode. Large amount of current flows through a zener diode if the applied reverse voltage reaches its breakdown voltage. Otherwise the current through the diode is zero. Voltage drop across R2 : V2 = 10 V Sum of currents through both R2 : V2 2 V2 2 ¥ 10 = = I2 = -1 R2 1500 Ê 1 1 ˆ ÁË R + R ˜¯ 2

Y

2

Output at D : D = ( A.C ) = A.( A.B) D = A + ( A.B) = A + A.B (Use de-Morgan’s principle : x.y = x + y ) ( x = x) Output at E : E = B + C = B + ( A.B) E = B + A + B = 1 + A = 1

( x.y = x + y )

( use x + x = 1,1 + x = 1) Output at Y :

= 13.3 mA

Y = D.E = ( A + A.B).(1) = A + A.B + 1

Voltage drop across R1 : V1 = 12 - V2 = 12 - 10 = 2 V

( x.y = x + y ) Y = A.A.B + 0

Current through R1 : V1 2 I1 = = = 4 mA R1 500 We are getting I2 > I1, which is not possible. So, Zener diode will never reach its breakdown voltage and current through it will be zero. Ans. 30. Option (b) is correct. Given : A circuit consisting of two ideal diodes, resistance of each diode when in forward bias is R = 50 W. D1 150 

Y = A.( A + B) = A.A + A.B = 0 + A.B = A.B ( x = x , x.y = x + y ,. x x = 0 ) Ans. 32. Option (a) is correct. Given : For the circuit diagram shown below, VBB can vary from 0 to 5 V, VCC = 5 V, bDC = 200, RB = 100 kW, RC = 1 kW, VBE = 1.0 V. IC ||

RB

75 k vi

— –

D2

( x + y = x .y ,1 = 0 )

VBB

C | B

|| RC

E IE

VCC

v0

100 k

6V To find : The current through the R¢ = 100 W resistor. In the circuit shown above, diode D1 is forward biased and diode D2 is reverse biased. So, the

To find : The minimum base current IB and the input voltage VBB at which the transistor will go to saturation.

392 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) At saturation, VCE = 0 VCE = VCC - IC RC = 0

biased and have built in potential of Vd = 0.7 V, N

VCC 5 = = 5 mA RC 1 ¥ 10 3

IC =

4V O

M

To find : VA, the voltage at point A. For the given circuit : Voltage drop between N and M is

So, saturation base current will be : IC 5 ¥ 10 -3 = = 25 m A β DC 200

IB =

A

Vin= 12.7 V

IC , IB

β DC =

Vin = 12.7 V

From Kirchhoff ’s voltage rule : VBB = I B R B + VBE

Voltage drop between N and A is

At saturation :

Voltage drop between A and O is

Vd = 0.7 V VA = ?

VBB = 25 ¥ 10 -6 ¥ 100 ¥ 10 3 + 1

By applying Kirchhoff ’s voltage law to loop MNAO :

= 2.5 + 1 = 3.5 V

Vin + Vd + VA = 0

Subjective Questions (Chapter Based) Q.1. The output characteristics of a transistor is shown in the figure. When VCE is 10V and IC = 4.0 mA, then value of bac is ________ (IB) 60 µA

(IC) in mA

8

50 µA

6

40 µA 30 µA

4

12.7 − 0.7 − VA = 0 VA = 12V Q.3. The circuit shown below is working as a 8 V de regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both zener diodes are identical) _______. 200 Ω

20 µA 10 µA

2 2

4

6

PHYSICS

8 10 12 (VCE) in volts

200 Ω Vo

Vin= 12.7 V

14

8V

[JEE (Main) – 6th Sep. 2020 - Shift-2] Sol. Given : Output characteristics of a transistor To find : bac, when VCE = 10 V and IC = 4.0 mA ∆IC = 4.5 − 3 = 1.5 mA ∆I B = 30 − 20 = 10m A

β ac =

∆IC = 150 ∆I B

[JEE (Main) – 9th Jan. 2020 - Shift-2] Sol. Given : A circuit shown below works as a Vo = 8V regulated dc voltage source, both the diodes are identical. 200 Ω

Q.2. Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. Built in potential in each diode is 0.7 V. For the input voltages shown in the figure, the voltage (in volt) at point A is _________.

i 200 Ω

Vin= 12 V

Vo

8V

A Vin= 12.7 V

4V

[JEE (Main) – 9th Jan. 2020 - Shift-1] Sol. Given : Both the diodes shown in the circuit below are ideal, have negligible resistance when forward

To find : P, the power dissipated in each diode when the input voltage is Vin = 12 V. Current through the circuit : Vin − Vo 12 − 8 4 i = = = = 0.01A 200 + 200 200 + 200 400

393

ELECTRONIC DEVICES

Voltage drop across each diode : Vo 8 V = = = 4V d 2 2 The power dissipated in each diode : P = iV = 0.01 × 4 = 40 mW Q.4. What will be the reading of the ammeter for the silicon diode in the given circuit? 200 

Sol. Given : A combination of two ideal diodes D1 and D2 connected to a source E = 25 V. To find : The current through the circuit. Diode D1 is forward biased but diode D2 is reverse biased. So, the outer arm of the circuit can be thought off as open. Current through the circuit : E 25 25 I = = = = 1.67 A R eq 5 + 10 15 Q.7. Give truth table for combination of logic gates shown below. E NAND A

•

3V

C

Sol. Given : a circuit comprising of a silicon diode, R = 200 W resistor, a DC source with E = 3 V and an ammeter. To find : The reading of the ammeter. Potential drop across the silicon diode : V = 0.7 V

NAND

•

A B C = (A.B) D = ( B.C )

Q.5. Temperature of a piece of copper and a piece of undoped silicon is raised from 300 K to 400 K. Describe the behaviour of resistance of both the materials with changing temperature. Sol. Given : Initial temperature of copper and silicon piece is Ti = 300 K, final temperature of copper and silicon piece is Tf = 300 K. To find : Variation of resistance of copper and silicon within the given temperature range. As copper is a metal its resistance will increase linearly as temperature rises from Ti = 300 K to Tf = 400 K For undoped silicon the resistance will decrease exponentially with increasing temperature due to increase in number of electron-hole pairs in the system. Q.6. Determine the current flowing in the circuit, where two ideal diodes are connected in parallel as shown below.

F = D+E

0 0

1

1

1

1

1 0

1

1

0

1

0 1

1

0

1

1

1 1

0

1

1

1

X

A

Z B

Y

Sol. Given : A logic circuit as shown above. To find : What combination of values of A and B will give output 0 at Z. Truth table for the logic circuit shown above :

D2

25 V 7

E = A.C

Q.8. To get output of 0 at Z, for the given logic gate circuit, what must be the input values at A and B?

5

10 

F

D NAND B Sol. Given : A logic circuit as shown above. To find : The truth table for the given circuit.

Current through the circuit : E - V 3 - 0.7 I = = = 11.5 mA R 200

D1

OR



A

B

X = A.B

Y=B

Z=X+Y

0

0

1

1

1

1

0

1

1

1

0

1

1

0

1

1

1

0

0

0

From the above truth table : A = 1, B = 1 will give Z = 0.



k

394 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

Communication Systems

Chapter 20 Syllabus

Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and Frequency Modulation, Bandwidth of signals, Bandwidth of Transmission medium, Basic Elements of a Communication System (Block Diagram only).

Concept Revision (Video Based) Propagation of Electromagnetic Waves in Atmosphere

 

Modulation

    

              Part -1     Part - 2 Amplitude Modulation      

 

     

           Part -1   

Part - 2

Bandwidth of Signal  

 

Part -1    Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. An amplitude modulated waves is represented by expression Vm = 5 (1 + 0.6 cos 6280t) sin (211 × 104t) volts. The minimum and maximum amplitudes of amplitude modulated wave are, respectively : (a) 5 V, 8 V (b) 5 V, 8 V 2 3 (c) 3 V, 5 V (d) V, 5 V 2 [JEE (Main) – 2nd Sep. 2020 - Shift-1] Q.2. The wavelength of the carrier waves in a modern optical fiber communication network is close to : (a) 2400 nm (b) 1500 nm (c) 600 nm (d) 900 nm [JEE (Main) – 8th April 2019 - Shift-1] Q.3. In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be : (Radius of the Earth = 6.4 × 106 m). (a) 20 m (b) 51 m (c) 32 m (d) 40 m [JEE (Main) – 8th April 2019 - Shift-2]

Q.4. A signal Acoswt is transmitted using V0sinw0t as carrier wave. The correct amplitude modulated (AM) signal is : A A (a) V0sinw0 t, sin(w0 – w)t + sin(w0 + w)t 2 2 (b) V0 sin[w0(1 + 0.01 Asinwt)t] (c) (V0 sinw0t + Acoswt) (d) (V0 + A) coswt sinw0t [JEE (Main) – 9th April 2019 - Shift-1] Q.5. The physical sizes of the transmitter and receiver antenna in a communication system are : (a) independent of both carrier and modulation frequency (b) inversely proportional to carrier frequency (c) inversely proportional to modulation frequency (d) proportional to carrier frequency [JEE (Main) – 9th April 2019 - Shift-2] Q.6. A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are : (a) 4; 1 × 108 Hz (b) 4; 2 × 108 Hz (c) 0.25; 2 × 108 Hz (d) 0.25; 1 × 108 Hz [JEE (Main) – 10th April 2019 - Shift-1]

396 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.7. Given below in the left column are different modes of communication using the kinds of waves given in the right column. A. Optical Fibre P. Ultrasound Communication B. Radar Q. Infrared Light C. Sonar R. Microwaves D. Mobile S. Radio Waves Phones From here, the options given below, find the most appropriate match between entries in the left and the right column. (a) A-Q, B-S, C-R, D-P (b) A-S, B-Q, C-R, D-P (c) A-Q, B-S, C-P, D-R (d) A-R, B-P, C-S,D-Q [JEE (Main) – 10th April 2019 - Shift-1] Q.8. In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin (20000 pt) while modulating signal is given by, m(t) = 2 sin (2000 pt). The values of modulation index and lower side band frequency are : (a) 0.5 and 10 kHz (b) 0.4 and 10 kHz (c) 0.3 and 9 kHz (d) 0.5 and 9 kHz [JEE (Main) – 12th April 2019 - Shift-2] Q.9. In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108 m/s, h = 6.6 × 10–34 J-s) (a) 3.75 × 106 (b) 3.86 × 106 (c) 6.25 × 105 (d) 4.87 × 105 [JEE (Main) – 9th Jan. 2019 - Shift-2] Q.10. A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given : radius of earth = 6.4 × 106 m). (a) 65 km (b) 48 km (c) 80 km (d) 40 km [JEE (Main) – 10th Jan. 2019 - Shift-1] Q.11. The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot ? (a) 2750 kHz (b) 2900 kHz (c) 2250 kHz (d) 2000 kHz [JEE (Main) – 10th Jan. 2019 - Shift-2] Q.12. An amplitude modulated signal is given by V(t) = 10[1+0.3 cos (2.2 × 104t)] sin (5.5 × 105t). Here t is in seconds. The side band frequencies (in kHz) are, (Given p = 22/7) (a) 1785 and 1715 (b) 892.5 and 857.5 (c) 178.5 and 171.5 (d) 89.25 and 85.75 [JEE (Main) – 11th Jan. 2019 - Shift-1]

PHYSICS

Q.13. An amplitude modulated signal is plotted below : V (t) 10 V 8V t

8 s 100 s



Which one of the following best describes the above signal ? (a) (9 + sin (2.5p × 105 t)) sin (2p × 104t) V (b) (1 + 9 sin(2p × 104 t)) sin (2.5p × 105t) V (c) (9 + sin (2p × 104 t)) sin (2.5p × 105t) V (d) (9 + sin (4p × 104 t)) sin (5p × 105t) V [JEE (Main) – 11th Jan. 2019 - Shift-2] Q.14. A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index ? (a) 0.3 (b) 0.5 (c) 0.6 (d) 0.4 [JEE (Main) – 12th Jan. 2019 - Shift-1] Q.15. To double the covering range of a TV transmission tower, its height should be multiplied by : 1 (a) (b) 2 2 (c) 4

(d) 2 [JEE (Main) – 12th Jan. 2019 - Shift-2]



ANSWER – KEY

1. (b) 5. (b) 9. (c) 13. (c)

2. (b) 6. (c) 10. (a) 14. (c)

3. (c) 7. (c) 11. (d) 15. (c)

4. (a) 8. (d) 12. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. Given : An amplitude modulated wave, Vm = 5(1 + 0.6 cos 6280t ) sin( 211 × 10 4 t )V ...(i) To find : Amin and Amax, the minimum and maximum amplitudes of the amplitude modulated wave. From equation (i) : A max + A min = 5 ...(ii) 2 A max − A min = 5 × 0.6 = 3 2

...(iii)

From equations (ii) and (iii) : = A max = 8 V , A min 2 V Ans. 2. Option (b) is correct. Given : Modern optical fibre communication network.

397

COMMUNICATION SYSTEMS

To find : Wavelength of carrier waves in the network. In modern optical fibre communication network, the signals are transmitted by waves of wavelength l = [1310 – 1550] nm. Ans. 3. Option (c) is correct. Given : In a line of sight radio communication distance between the transmitting and receiving antenna is d = 50 km, height of receiving antenna is h = 70 m, radius of earth is R = 6.4 × 106 m. To find : h¢, minimum height of the transmitting antenna. In line-of-sight radio communication, distance between the transmitting and receiving antenna is given as : d =

Ans. 6. Option (c) is correct. Given : Frequency of signal is fm = 100 MHz, associated peak voltage is Am = 100 V, frequency of carrier wave is fc = 300 GHz, associated peak voltage is Ac = 400 V. To find : The modulation index MI and difference between the two side band frequencies. Modulation index : A m 100 = = 0.25  MI = A c 400 Range of frequencies in case of amplitude modulation is (fc – fm) to (fc + fm). Bandwidth : ( fc + fm ) - ( fc - fm ) = 2 fm = 2 ¥ 100 ¥ 10 6 = 2 ¥ 10 8 Hz

2 Rh + 2 Rh¢

=

2R ( h + h¢ )

...(i)

Put given values in equation (i). 50 ¥ 10 3 =

2 ¥ 6.4 ¥ 10 6 ( 70 + h¢ )

h¢ ª 32m Ans. 4. Option (a) is correct. Given : Transmitted signal is A m = A cos ωt,

...(i)

carrier wave used in transmission is A o = vo sin ωot

...(ii)

To find : the correct amplitude modulated (AM) signal. The modulated wave will be given as : Y = ( A m + A o )sin ωot From equations (i) and (ii), Y = ( A cos ωt + vo )sin ωot Y = vo sin ωot + A sin ωot cos ωt Y = vo sin ωot +

A [sin(ωo + ω )t 2 + sin(ωo - ω )t]

A Y = vo sin ωot + sin(ωo - ω )t 2 A + sin(ωo + ω )t 2 Ans. 5. Option (b) is correct. Given : A communication system. To find : The factors on which the physical size of transmitter and receiver antenna depends. Size of antenna is directly proportional to the wavelength of the signal : h ∝ λ h ∝

c ν

h ∝

1 ν

(ν is frequency of the signal )

Ans. 7. Option (c) is correct. Given : Different modes of communication in left column and kinds of waves used in right column. To find : The most appropriate match between the entries in the left and in the right column. We know : • Optical fibre communication uses infrared light. • Radar uses radio waves. • Sonar uses ultrasound waves and • Mobile phones use microwaves. Ans. 8. Option (d) is correct. Given : In an amplitude modulator circuit the carrier wave is C(t ) = 4 sin( 20000p t ), ...( i ) Modulating wave is m(t ) = 2 sin( 2000p t ).

...(ii)

To find : Value of modulation index MI and lower side band frequency. From equation (i) : C(t ) = A c sin ωc t = 4 sin( 20000p t ) A c = 4 ,fc =

ωc 20000p = = 10000 ...(iii) 2p 2p

From equation (ii) : m(t ) = A m sin ωmt = 2 sin ( 2000p t ) A m = 2 ,fm =

ωm 2000p = = 1000 2p 2p

...(iv)

From equations (iii) and (iv) : Am 2 = = 0.5  MI = Ac 4 Range of frequencies in case of amplitude modulation is (fc – fm) to (fc + fm). Again, from equations (iii) and (iv), the lower side of band frequency is : fc - fm = 10000 - 1000 = 9kHz Ans. 9. Option (c) is correct. Given : In a communication system wavelength of signal is l = 800 nm, only 1% of signal frequency can be used as bandwidth.

398 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) To find : Number of channels that can be accommodated for transferring TV signal of bandwidth 6 MHz. Frequency of signal : f =

c 3 ¥ 10 8 = = 3.75 ¥ 1014 Hz λ 800 ¥ 10 -9

6 ¥ 10

d =

2 Rh¢ + 2 Rh

d =

2R ( h¢ + h )

...( i )

2 ¥ 6.4 ¥ 10 6 ( 140 + 40 )

d = 65km Ans. 11. Option (d) is correct. Given : Modulation frequency of an AM radio station is fm = 250 kHz, fm is 10% of carrier frequency fc. To find : f2 the second accepted frequency. Carrier frequency : fc = 10 ¥ fm = 2500kHz Lower limit of bandwidth : fc - fm = 2500 - 250 = 2250kHz Upper limit of bandwidth : fc + fm = 2500 + 250 = 2750kHz For accepted frequency, two bandwidths should not overlap. So, either f2 = fc - 2 fm = 2000 kHz  f2 = fc + 2 fm = 3000 kHz

Ans. 12. Option (d) is correct. Given : Amplitude modulated signal is V(t ) = 10[1 + 0.3 cos( 2.2 ¥ 10 4 t )]sin( 5.5 ¥ 10 5 t ). ...(i) To find : The side band frequencies in kHz. From equation (i), the lower band frequency is : fL =

10 V 8V

V (t)

t

5.5 ¥ 10 5 - 2.2 ¥ 10 4 = 84 kHz 2p

8 s 100 s

To find : The equation describing the above signal. Equation of an amplitude modulated wave : Y = ( A c + A m sin ωmt )sin ωc t

...(i)

Maximum amplitude : A c + A m = 10V

Put given values in equation (i).

Or

5.5 ¥ 10 5 + 2.2 ¥ 10 4 = 91kHz 2p

= 6.25 ¥ 10 5

Ans. 10. Option (a) is correct. Given : Height of TV transmission tower is h¢ = 140 m, height of receiving antenna is h = 40 m, radius of earth is R = 6.4 × 106 m. To find : d, the maximum distance up to which signals can be broadcasted in line of sight mode. The maximum distance up to which signals can be broadcasted in line of sight mode :

d =

fU =

So, option (d) is nearest to the answer.

So, number of channels that can be accommodated for transferring TV signal of bandwidth 6 MHz : 6

Again, from equation (i), the upper band frequency is :

Ans. 13. Option (c) is correct. Given : An amplitude modulated signal,

Total bandwidth used for communication : 1 ¥ 3.75 ¥ 1014 Hz = 3.75 ¥ 1012 Hz 100

3.75 ¥ 1012

PHYSICS

...(ii)

Minimum amplitude : A c - A m = 8V

...(iii)

From equations (ii) and (iii) : A c = 9 V ,A m = 1V

...(iv)

Angular frequency of carrier wave : 2p 2p ωc = = = 2.5p ¥ 10 5 s-1 Tc 8 ¥ 10 -6

...(v)

Angular frequency of signal wave : 2p 2p ωm = = = ( 2p ¥ 10 4 )s-1 ...(vi) Tm 100 ¥ 10 -6 Put values from equations (iv), (v) and (vi) in equation (i). Y = [9 + 1 ¥ sin( 2p ¥ 10 4 t )] sin( 2.5p ¥ 10 5 t )V Y = [9 + sin( 2p ¥ 10 4 t )]sin( 2.5p ¥ 10 5 t )V Ans. 14. Option (c) is correct. Given : A 100 V carrier wave whose amplitude varies between Vmax = 160 V and Vmin = 40 V. To find : MI, the modulation index. V - Vmin 160 - 40 120 = = = 0.6 MI = max Vmax + Vmin 160 + 40 200 Ans. 15. Option (c) is correct. Given : Height of TV transmission tower is h, nh will be the tower’s new height if its covering range is doubled. To find : The value of n. Range of TV transmitting tower : d =

2Rh

399

COMMUNICATION SYSTEMS

When the range is doubled : 2d = 2 2 Rh = 2 R( nh ) nh = 4 h



n = 4

Subjective Questions (Chapter Based) Q.1. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ? Sol. Given : Carrier frequency for a telephonic communication service is fc = 10 GHz, only 10% of it is utilized for transmission. To find : Number of channels that can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz. 10% of carrier frequency : 10 10%of fc = ¥ 10 ¥ 10 9 = 1GHz 100



To find : Maximum modulated frequency which could be detected by the circuit. Maximum modulated frequency which could be detected by the circuit : 1 f = 2p mRC =

1 2p ¥ 0.6 ¥ 100 ¥ 10 3 ¥ 250 ¥ 10 -12

= 10.62 kHz ( m = 60% = 0.6 ) Q.4. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. What is the maximum distance up to which it can detect object located on the surface of the earth? (Radius of earth = 6.4 × 106 m) Sol. Given : Power of radar is P =1 kW, operating frequency of radar is f = 10 GHz, height at which radar is located h = 500 m, radius of earth is R = 6.4 × 106 m . To find : r, the range of radar.

If each channel requires a bandwidth of 5 kHz, then number of channels that can be broadcasted simultaneously : 1 ¥ 10

9 3

h R

= 2 ¥ 10 5

5 ¥ 10 Q.2. A signal of 2 kHz frequency is amplitude modulated on a carrier wave of frequency 20 MHz. What frequencies will be included in the resultant signal? Sol. Given : Frequency of the signal is fm = 2 kHz, frequency of carrier wave is fc = 20 MHz. To find : The frequencies included in the resultant signal. Frequencies of the resultant signal will be : fc – fm, fc + fm. fc - fm = ( 20000 - 2 )kHz = 19998 kHz fc = 20MHz fc + fm = ( 20000 + 2 )kHz = 20002 kHz Q.3. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 picofarad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. Sol. Given : A diode detector uses a capacitor C = 250 pF connected in parallel with a load resistance R = 100 kW, the detector is used to detect an amplitude modulated wave of 60% modulation.



r

R

Range of radar : r =

( R + h )2 - R 2 = h 2 + 2 Rh

As h  R,

r = 2 Rh = 2 ¥ 6.4 ¥ 10 6 ¥ 500 = 8 ¥ 10 4 m Q.5. By modulating signal a 350 V carrier wave is varied between 300 V and 400 V. What is the modulation index? Sol. Given : A 350 V carrier wave whose amplitude varies between Vmax = 400 V and Vmin = 300 V. To find : MI, the modulation index. V - Vmin 400 - 300 100 1 = = = MI = max Vmax + Vmin 400 + 300 700 7



APPENDIX-A

A.1  SI Units SI Base Unit Base Quantity







Name

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

Electric current

ampere

A

Temperature

kelvin

K

Amount of substance

mole

mol

Luminous intensity

candela

s

cd

A.2  Some Derived SI Units Quantity

Name

Plane angle Frequency Force Pressure Energy Power Electric charge Electric potential Capacitance Electric resistance Magnetic flux Magnetic field Inductance

radian hertz newton pascal joule watt coulomb volt farad ohm weber tesla henry

Symbol ω Hz N Pa J W C V F Ω φ B L

Expression in Terms of Base Units m/m s–1 kg · m/s2 kg/m · s2 kg · m2/s2 kg · m2/s3 A·s kg · m2/A · s3 A2 · s4/kg · m2 kg · m2/A2 · s3 kg · m2/A · s2 kg /A · s2 kg · m2/A2 · s2

Expression in Terms of SI Units

J/m N/m2 N·m J/s W/A C/V V/A T· m Nb/m2 or tesla T · m2/A

APPENDIX-B

B.1  Conversion Factors Length m



1 meter 1 centimeter 1 kilometer 1 inch 1 foot 1 mile

1 10–2 103 2.540 × 10–2 0.3048 1609

cm

km

2

–3

10 1 105 2.540 30.48 1.609 × 105

10 10–5 1 2.540 × 10–5 3.048 × 10–4 1.609

in.

ft

mi

39.37 0.3937 3.937 × 104 1 12 6.336 × 104

3.281 3.281 × 10–2 3281 × 103 8.333 × 10–2 1 5280

6.214 × 10–4 6.214 × 10–6 0.6214 1.578 × 10–5 1.894 × 10–4 1

Mass



1 kilogram 1 gram 1 slug 1 atomic mass unit Note : 1 metric ton = 1000 kg.

kg

g

slug

u

1 10–3 14.59 1.660 × 10–27

103 1 1.459 × 104 1.660 × 10–24

6.854 × 10–2 6.852 × 10–5 1 1.137 × 10–28

6.024 × 1026 6.024 × 1023 8.789 × 1027 1

401

APPENDIX

Time s



1 second 1 minute 1 hour 1 day 1 year

min

h –2

1.667 × 10 1 60 1440 5.259 × 105

1 60 3600 8.640 × 104 3.156 × 107

day –4

2.778 × 10 1.667 × 10–2 1 24 8.766 × 103

yr –5

1.157 × 10 6.944 × 10–4 4.167×10–2 1 365.2

3.169×10–8 1.901×10–6 1.141×10–4 2.738×10–5 1

Speed m/s



1 meter per second 1 centimeter per second 1 foot per second 1 mile per hour Note : 1 mi/min = 60 mi/h = 88 ft/s.

cm/s 2

10 1 30.48 44.70

1 10–2 0.3048 0.4470

ft/s

mi/h

3.281 3.281 ×10–2 1 1.467

2.237 2.237 ×10–2 0.6818 1

Force



1 newton 1 pound

N

lb

1 4.448

0.2248 1

Energy, Energy Transfer 1 joule 1 foot-pound 1 electron volt 1 calorie 1 British thermal unit 1 kilowatt-hour



1 joule 1 foot-pound 1 electron volt 1 calorie 1 British thermal unit 1 kilowatt-hour

J

ft · lb

eV

1 1.356 1.602 × 10–19 4.186 1.055 × 103 3.600 × 106

0.7376 1 1.182 × 10–19 3.087 7.779 × 102 2.655 × 106

6.242 × 1018 8.464 × 1018 1 2.613 × 1019 6.585 × 1021 2.247 × 1025

cal

Btu

kWh

0.2389 0.3239 3.827 × 10–20 1 2.520 × 102 8.601 × 105

9.481 × 10–4 1.285 × 10–3 1.519 × 10–22 3.968 × 10–3 1 3.413 × 102

2.778 × 10–7 3.766 × 10–7 4.450 × 10–26 1.163 × 10–6 2.930 × 10–4 1

Pressure



1 pascal 1 atmosphere 1 centimeter mercurya 1 pound per square inch 1 pound per square foot

Pa

atm

1 1.013 × 105 1.333 × 103 6.895 × 103 47.88

9.869 × 10–6 1 1.316 × 10–2 6.805 × 10–2 4.725 × 10–4 lb/in.2

cm Hg 1 pascal 1 atmosphere 1 centimeter mercurya 1 pound per square inch 1 pound per square foot

–4

7.501 × 10 76 1 5.171 3.591 × 10–2

lb/ft2 –4

1.450 × 10 14.70 0.1943 1 6.944 × 10–3

At 0°C and at a location where the free-fall acceleration has its “standard” value, 9.80665 m/s2.

2.089 × 10–2 2.116 × 103 27.85 144 1

402 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

B.2 Conversions of useful physical quantities Length

Acceleration

1 in. = 2.54 cm (exact) 1 m = 39.37 in. = 3.281 ft 1 ft = 0.3048 m 12 in. = 1 ft 3 ft = 1 yd 1 yd = 0.9144 m 1 km = 0.621 mi 1 mi = 1.609 km 1 mi = 5280 ft 1 μm = 10–6 m = 103 nm 1 light-year = 9.461 × 1015 m



4

2

1 m3 = 106 cm3 = 6.102 × 104 in.3 1 ft3 = 1 728 in.3 = 2.83 × 10–2 m3 1 L = 1000 cm3 = 1.0576 qt = 0.0353 ft3 1 ft3 = 7.481 gal = 28.32 L = 2.832 × 10–2 m3 1 gal = 3.786 L = 231 in.3

1 bar = 105 N/m2 = 14.50 lb/in.2



1 atm = 760 mm Hg = 76.0 cm Hg



1 atm = 14.7 lb/in.2 = 1.013 × 105 N/m2



1 Pa = 1 N/m2 = 1.45 × 10–4 lb/in.2



1 yr = 365 days = 3.16 × 107 s



1 day = 24 h = 1.44 × 103 min = 8.64 × 104 s



1 J = 0.738 ft · lb



1 cal = 4.186 J



1 Btu = 252 cal = 1.054 × 103 J



1 eV = 1.602 × 10–19 J



1 kWh = 3.60 × 106 J

Power

1000 kg = 1 t (metric ton) 1 slug = 14.59 kg 1 u = 1.66 × 10–27 kg = 931.5 MeV/c2



1 hp = 550 ft · lb/s = 0.746 kW



1 W = 1 J/s = 0.738 ft · lb/s



1 Btu/h = 0.293 W

Some Approx imations Useful for Estimation Problems

1 N = 0.2248 lb 1 lb = 4.448 N

Velocity



Energy

1 m = 10 cm = 10.76 ft 1 ft2 = 0.0929 m2 = 144 in.2 1 in.2 = 6.452 cm2

Force

1 ft/s2 = 0.3048 m/s2 = 30.48 cm/s2

2

Mass



Time

Volume

1 m/s2 = 3.28 ft/s2 = 100 cm/s2

Pressure

Area 2



1 mi/h = 1.47 ft/s = 0.447 m/s = 1.61 km/h 1 m/s = 100 cm/s = 3.281 ft/s 1 mi/min = 60 mi/h = 88 ft/s



1 m ≈ 1 yd





1 m/s ≈ 2 mi/h



1 kg ≈ 2 lb





1 yr ≈ π × 107 s



1N≈





60 mi/h ≈ 100 ft/s





1 km ≈



1 lb 4 1 1 L ≈ gal 4

1 mi 2

APPENDIX-C

C.1  Important Constants Symbol



Meaning

Best Value

Approximate Value

8

8

c

Speed of light in vacuum

2.99792458 × 10 m/s

3.00 × 10 m/s

G

Gravitational constant

6.67408(31) × 10–11 N . m2 /kg2

6.67 × 10–11 N . m2 /kg2

NA

Avogadro’s number

6.02214129(27) × 1023

6.02 × 1023

k

Boltzmann’s constant

1.3806488(13) × 10–23 J/K

1.38 × 10–23 J/K

R

Gas constant

8.3144621(75) J/ mol . K

8.31 J/ mol · K = 1.99 cal / mol . K = 0.0821 atm . L/mol . K

s

Stefan-Boltzmann constant

5.670373(21) × 10–8 W/m2 . K 9

2

5.67 × 10–8 W/m2 · K 2

8.99 × 109 N . m2/ C2

k

Coulomb force constant

8.987551788… × 10 N . m / C

qe

Charge on electron

–1.602176565(35) × 10–19 C

–1.60 × 10–19 C

ε0

Permittivity of free space

8.854187817… × 10–12 C2/N . m2

8.85 × 10–12 C2/N . m2

–7

µ0

Permeability of free space

4π × 10 T . m/A

1.26 × 10–6 T . m/A

h

Planck’s constant

6.62606957(29) × 10–34 J . s

6.63 × 10–34 J . s

403

APPENDIX



C.2  Submicroscopic Masses Symbol





Approximate Value –31

9.11 × 10–31 kg

Electron mass

9.10938291(40) × 10

mp

Proton mass

1.672621777(74) × 10–27 kg

1.6726 × 10–27 kg

mn

Neutron mass

1.674927351(74) × 10–27 kg

1.6749 × 10–27 kg

u

Atomic mass unit

1.660538921(73) × 10–27 kg

1.6605 × 10–27 kg

kg

C.3  Solar System Data

Earth

Moon



mass

1.99 × 1030 kg

average radius

6.96 × 108 m

Earth-sun distance (average)

1.496 × 1011 m

mass

5.9736 × 1024 kg

average radius

6.376 × 106 m

orbital period

3.16 × 107 s

mass

7.35 × 1022  kg

average radius

1.74 × 106 m

orbital period (average)

2.36 × 106 s

Earth-moon distance (average)

3.84 × 108 m

C.4  Metric Prefixes for Powers of Ten and Their Symbols Prefix tera

Symbol

Value

T

10

Prefix

Symbol

Value

12

deci

d

10–1

9

centi

c

10–2

giga

G

10

mega

M

106

milli

m

10–3

kilo

k

10

3

micro

µ

10–6

hecto

h

102

nano

n

10–9

1

pico

p

10–12

femto

f

10–15

deka

da

10

–

–

100 ( = 1 )





Best Value

me

Sun



Meaning

C.5  Selected British Units

Volume

1 liter (L) = 10–3 m3 1 U.S. gallon (gal) = 3.785 ×10–3 m3

Length

Force





1 inch (in.) = 2.54 cm (exactly)

Mass

1 metric ton = 103 kg

1 mile (mi) = 1.609 km

1 atomic mass unit (u) = 1.6605 × 10–27 kg

1 pound (ld) = 4.448 N

Time 3

Energy

1 British thermal unit (Btu) = 1.055 × 10 J

Power

1 horsepower (hp) = 746 W

Pressure

1 lb / in2 = 6.895 × 103 Pa

1 angstrom (Å) = 10 Area

m

1 acre (ac) = 4.05 × 103 m2 1 square foot (ft2) = 9.29 ×10–2 m2 1 barn (b) = 10–28 m2

1 degree (°) = 1.745 × 10–2 rad 1 minute of arc (‘) = 1 / 60 degree 1 second of arc (“) = 1 / 60 minute of arc

1 astronomical unit (au) = 1.50 ×1011m –10

1 mile per hour (mph) = 1.609 km/h 1 nautical mile per hour (naut) = 1.852 km/h

1 light year (ly) = 9.46 × 1015 m 1 nautical mile = 1.852 km

1 year (y) = 3.16 × 107 s 1 day (d) = 86,400 s

Speed Angle

C.6  Other Units Length

1 solar mass = 1.99 × 1030 kg

1 foot (ft) = 0.3048 m

1 grad = 1.571 × 10–2 rad Energy

1 kiloton TNT (kT) = 4.2 × 1012 J 1 kilowatt hour (kW . h) = 3.60 × 106 J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J

404 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

1 electron volt (eV) = 1.60 × 10-19J Pressure

1 atmosphere (atm) = 1.013 × 105 Pa

Nuclear decay rate



C = 2πr = πd

1 torricelli (torr) = 1 mm Hg = 133.3 Pa

Area of a circle with radius r or diameter d

A = πr2= πd2/ 4

1 curie (Ci) = 3.70 × 1010 Bq

Area of a sphere with radius r

A = 4πr2

Volume of a sphere with radius r

V = (4 / 3) (πr3)





C.7  Useful Formulae Circumference of a circle with radius r or diameter d

1 millimeter of mercury (mm Hg) = 133.3 Pa

PHYSICS

C.8  The Greek Alphabet Alpha

A

a

Eta

H

η

Nu

Beta

N

ν

Tau

T

τ

B

b

Theta

Θ

θ

Xi

Ξ

ξ

Upsilon

ϒ

υ

Gamma

Γ

g

Iota

I

ι

Omicron

O

o

Phi

Φ

φ

Delta

∆

d

Kappa

K

κ

Pi

Π

p

Chi

X

χ

Epsilon

E

e

Lambda

Λ

l

Rho

P

ρ

Psi

ψ

ψ

Zeta

Z

z

Mu

M

m

Sigma

Σ

s

Omega

Ω

ω

APPENDIX-D

Symbols, Dimensions, and Units of Physical Quantities Quantity

Common Symbol

Unit

Dimensions

Unit in Terms of Base SI Units

L/T2

m/s2

Acceleration

→

a

m/s2

Amount of substance

n

MOLE

Angle

θ, φ  α

mol

radian (rad)

1

rad/s2

T–2

s–2

ω  L

rad/s2

T–1

s–1

kg.m2/s

ML2/T

kg . m2/s

Angular velocity

 ω

rad/s

T–1

s–1

Area

A

m2

L2

m2

farad (F)

Q2T2/ML2

A2.s4/kg.m2

coulomb (C)

Q

A.s

C/m

Q/L

Angular acceleration Angular frequency Angular momentum

Atomic number

Z

Capacitance

C

Charge Line Charge density

q, Q, e l

2

A⋅s/m

2

Surface Charge density

s

C/m

Q/L

Volume Charge density

ρ

C/m3

Q/L3

A⋅s/m3

Conductivity

s

1/Ω⋅ m

Q T/ML

A2⋅s3/kg⋅m3

Current

I

AMPERE

Q/T

A

Current density

J

A/m2

Q/TL2

A/m2

Density

ρ

kg/m3

M/L3

kg/m3

Dielectric constant

κ  P

C⋅m

QL

A⋅s⋅m

V/m

ML/QT2

kg⋅m/A⋅s3

Electric dipole moment Electric field

 E

2

A⋅s/m2 3

405

APPENDIX Electric flux Electromotive force

ΦE e

Energy

E, U, K

Entropy

S  F

Force Frequency Heat

f/ν Q

V⋅m

ML3/QT2

kg⋅m3/A⋅s3

volt (V)

ML2/QT2

kg⋅m2/A⋅s3

2

2

joule (J)

ML /T

kg⋅m2/s2

J/K

ML2/T2K

kg⋅m2/s2⋅K

2

newton (N)

ML/T

kg⋅m/s2

hertz (Hz)

T-1

s-1

joule (J)

ML2/T2

kg⋅m2/s2

henry (H)

ML /Q

kg⋅m2/A2⋅s2

Meter

L

m

µ

N ⋅ m/T

QL2/T

A⋅m2

Magnetic field

 Β

tesla (T) (=Wb/m2)

M/QT

kg/A⋅s2

Magnetic flux

ΦB

weber (Wb)

ML2/QT

kg⋅m2/A⋅s2

m, M

Kilogram

M

kg

C

J/mol ⋅ K

Inductance Length Displacement

L ,L

d, h

Position

 x, y, z, r 

Mass Molar specific heat

2

 ∆x , ∆r

Distance Magnetic dipole moment

2

kg⋅m2/s2⋅mol⋅K

2

2

kg⋅m2

I  P

kg ⋅ m

ML

kg ⋅ m/s

ML/T

kg⋅m/s

Time Period

T

S

T

s

Permeability of free space

m0

N/A2 (=H/m)

ML/Q2

kg⋅m/A2⋅s2

Permittivity of free space

ε0

C2/N ⋅ m2 (=F/m)

Q2T2/ML3

Moment of inertia Momentum

Potential

V

volt (V) (=J/C)

Power

P

watt (W)(=J/s)

2

ML /QT

2

ML2/T3 2

2

kg⋅m2/A⋅s3 kg⋅m2/s3 kg/m⋅s2

Pressure

P

pascal (Pa)(=N/m )

M/LT

Resistance

R

ohm (Ω)(=V/A)

ML2/Q2T 2

kg⋅m2/A2⋅s3

c

J/kg ⋅ K

L /T K

m2/s2⋅K

Speed

v

m/s

L/T

m/s

Temperature

T

Kelvin

K

K

Time

Specific heat

2

A2⋅s4/kg⋅m3

t 

Second

T

s

Torque

τ

N⋅m

ML2/T2

kg⋅m2/s2

Velocity

v

m/s

L/T

m/s

Volume

V

m3

L3

m3

Wavelength

l

m

L

Work

W

joule (J) (=N⋅ m)

m 2

2

ML /T

kg⋅m2/s2

406 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

PHYSICS

APPENDIX-E

Indian Space Research Organisation (ISRO) - - - - [1975 to 2020]

India has been successfully launching satellites of many types since 1975. Satellites have been launched from various vehicles, including those launched by American, Russian and European rockets, as well as those launched indigenously by India. The organization responsible for India’s satellite program is the Indian Space Research Organisation (ISRO). Satellites

Launch Date

Launch Vehicle

Aryabhata

19-Apr-75

u-11 Interkosmos

Bhaskara-I

07-Jun-79

C-1 Interkosmos

Rohini Technology Payload

10-Aug-79

SLV-3

Rohini RS-1

18-Jul-80

SLV-3

Rohini RS-D1

31-May-81

SLV-3

Ariane Passenger Payload Experiment

19-Jun-81

Ariane-1 (V-3)

Bhaskara-II

20-Nov-81

C-1 Intercosmos

INSAT-1A

10-Apr-82

Delta 3910 PAM-D

Rohini RS-D2

17-Apr-83

SLV-3

INSAT-1B

30-Aug-83

Shuttle [PAM-D]

Stretched Rohini Satellite Series (SROSS-1)

24-Mar-87

ASLV

IRS-1A

17-Mar-88

Vostok

Stretched Rohini Satellite Series (SROSS-2)

13-Jul-88

ASLV

INSAT-1C

21-Jul-88

Ariane-3

INSAT-1D

12-Jun-90

Delta 4925

IRS-1B

29-Aug-91

Vostok

INSAT-2DT

26-Feb-92

Ariane-44L H10

Stretched Rohini Satellite Series (SROSS-C)

20-May-92

ASLV

INSAT-2A

10-Jul-92

Ariane-44L H10

INSAT-2B

23-Jul-93

Ariane-44L H10+

IRS-1E

20-Sep-93

PSLV-D1

Stretched Rohini Satellite Series (SROSS-C2)

04-May-94

ASLV

IRS-P2

15-Oct-94

PSLV-D2

INSAT-2C

07-Dec-95

Ariane-44L H10-3

IRS-1C

29-Dec-95

Molniya

IRS-P3

21-Mar-96

PSLV-D3

INSAT-2D

04-Jun-97

Ariane-44L H10-3

IRS-1D

29-Sep-97

PSLV-C1

INSAT-2E

03-Apr-99

Ariane-42P H10-3

Oceansat-(IRS-P4)

26-May-99

PSLV-C2

INSAT-3B

22-Mar-2000

Ariane-5G

GSAT-1

18-Apr-01

GSLV-D1

Technology Experiment Satellite (TES)

22-Oct-01

PSLV-C3

INSAT-3C

24-Jan-02

Ariane-42L H10-3

Kalpana-1(METSAT)

12-Sep-02

PSLV-C4

INSAT-3A

10-Apr-03

Ariane-5G

GSAT-2

08-May-03

GSLV-D2

INSAT-3E

28-Sep-03

Ariane-5G

407

APPENDIX RESOURCESAT-1(IRS-P6)

17-Oct-03

PSLV-C5

EDUSAT

20-Oct-04

GSLV-F01

HAMSAT

05-May-05

PSLV-C6

CARTOSAT-1

05-May-05

PSLV-C6

INSAT-4A

22-Dec-05

Ariane-5GS

INSAT-4C

10-Jul-06

GSLV-F02

CARTOSAT-2

10-Jan-07

PSLV-C7

Space Capsule Recovery Experiment(SRE-1)

10-Jan-07

PSLV-C7

INSAT-4B

12-Mar-07

Ariane-5ECA

INSAT-4CR

02-Sep-07

GSLV-F04

CARTOSAT-2A

28-Apr-08

PSLV-C9

IMS-1 (Third World Satellite − TWsat)

28-Apr-08

PSLV-C9

Chandrayaan-1

22-Oct-08

PSLV-C11

RISAT-2

20-Apr-09

PSLV-C12

ANUSAT

20-Apr-09

PSLV-C12

Oceansat-2(IRS-P4)

23-Sep-09

PSLV-C14

GSAT-4

15-Apr-10

GSLV-D3

CARTOSAT-2B

12-Jul-10

PSLV-C15

StudSat

12-Jul-10

PSLV-C15

GSAT-5P /INSAT-4D

25-Dec-10

GSLV-F06

RESOURCESAT-2

20-Apr-11

PSLV-C16

Youthsat

20-Apr-11

PSLV-C16

GSAT-8 / INSAT-4G

21-May-11

Ariane-5VA-202

GSAT-12

15-Jul-11

PSLV-C17

Megha-Tropiques

12-Oct-11

PSLV-C18

Jugnu

12-Oct-11

PSLV-C18

RISAT-1

26-Apr-12

PSLV-C19

SRMSAT

26-Apr-12

PSLV-C18

GSAT-10

29-Sep-12

Ariane-5VA-209

SARAL

25-Feb-13

PSLV-C20

IRNSS-1A

01-Jul-13

PSLV-C22

INSAT-3D

26-Jul-13

Ariane-5

GSAT-7

30-Aug-13

Ariane-5

Mars Orbiter Mission (MOM)

05-Nov-13

PSLV-C25

GSAT-14

05-Jan-14

GSLV-D5

IRNSS-1B

04-Apr-14

PSLV-C24

IRNSS-1C

16-Oct-14

PSLV-C26

GSAT-16

07-Dec-14

Ariane-5

IRNSS-1D

28-Mar-15

PSLV-C27

GSAT-6

27-Aug-15

GSLV-D6

Astrosat

28-Sep-15

PSLV-C30

GSAT-15

11-Nov-15

Ariane 5 VA-227

IRNSS -1E

20-Jan-16

PSLV-C31

IRNSS -1F

10-Mar-16

PSLV-C32

IRNSS-1G

28-Apr-16

PSLV-C33

Cartosat-2C

22-Jun-16

PSLV-C34

CartoSat-2E

8 September 2016,

INSAT-3DR

Pratham

26 September 2016,

PSLV-C35

408 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) GSAT-18

6 October 2016,

Ariane-5 ECA

ResourceSat-2A

7 December 2016,

PSLV-C36

CartoSat-2D

15 February 2017,

PSLV-C37

South Asia Satellite (GSAT-9)

5 May 2017,

GSLV Mk.II[3

GSAT-19

05-Jun-17

GSLV Mk.III-D1

NIUSat[

23-June 2017,

PSLV-C38

GSAT-17

29 June 2017,

Ariane-5 ECA

IRNSS-1H

02-Sep-17

PSLV–C39

CartoSat-2F

10-January 2018,

PSLV-C40

GSAT-6A

29-Mar-18

GSLV-F08

IRNSS-11

12-Apr-18

GSLV-F08, PSLV-C41

GSAT-29

01-Nov-18

GSLV Mk III D2

HySIS

29-Nov-18

PSLV-C43

GSAT-7A

19-Dec-18

GSLV Mk.II-F11

Microsat-R

23-Jan-19

PSLV-C44

EMISAT

01-Apr-19

PSLV-C45

PS4 Stage attached with ExseedSat-2, AMSAT, ARIS and AIS payloads

01-Apr-19

PSLV-C45

Risat-2B

21-May-19

PSLV-C46

Chandrayaan-2

22-Jul-19

Chandrayaan-2

PHYSICS

ll