Oswaal NTA CUET (UG) Chapterwise Question Bank Physics (For 2024 Exam) 9789359582351, 9359582352

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For 2024 Exam

Highly Recommended

CHAPTER-WISE

QUESTION BANK Includes SOLVED PAPERS (2021 - 2023)

PHYSICS Section-II (Domain Specific Subject) Strictly as per the Latest Exam Pattern issued by NTA

The ONLY book you need to #AceCUET(UG)

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100% Updated

Previous Years’ Questions

Revision Notes

Concept Videos

800+ Questions

With 2023 CUET Exam Paper

(2021-2023) for Better Exam Insights

for Crisp Revision with Smart Mind Maps

for Complex Concepts Clarity

for Extensive Practice

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1st EDITION

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YEAR 2024 "9789359582351"

CUET (UG)

SYLLABUS COVERED

PUBLISHED BY OSWAAL BOOKS & LEARNING PVT. LTD.

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This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

Kindle Edition (2)

Preface Welcome to the ultimate resource for your Common University Entrance Test (CUET) preparation! The Common University Entrance Test (CUET) marks a significant shift in the admission process for UG programs in Central Universities across India. The introduction of CUET aims to create a level playing field for students nationwide, regardless of their geographical location, and revolutionize the way students connect with these prestigious institutions. CUET (UG), administered by the esteemed National Testing Agency (NTA), is a prestigious all-India test that serves as a single-window opportunity for admissions. The NTA consistently provides timely notifications regarding the exam schedule and any subsequent updates. The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET (UG) scores are mandatory required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG) provided by NTA. Oswaal CUET (UG) Question Bank is your strategic companion designed to elevate your performance and simplify your CUET journey for success in this computer-based test.

Here’s how this book benefits you: • 100% Updated with 2023 CUET Exam Paper • Previous years Questions (2021-2023)for Better Exam insights • Revision Notes for Crisp Revision with Smart Mind Maps • Concept Videos for complex concepts clarity • 800+ questions for Extensive Practice Almost 1.92 million candidates registered for CUET (UG) in 2023. Candidates have been quite anxious about appearing for CUET (UG), however, with the right preparation strategy and resources, you can secure a good rank in CUET (UG). We believe that with dedication, hard work, and the right resources, you can conquer CUET and secure your place in the Central Universities of your choice. Good luck with your preparations, with this trusted companion on your journey to academic success! All the best! Team Oswaal Books

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Contents l Know your CUET(UG) Exam

6 - 6

l Latest CUET (UG) Syllabus

7 - 8

l Examination Paper CUET 2023



11 - 16

1 - 14

Chapter-1 – Electrostatics Chapter-2 – Current Electricity

15 - 26

Chapter-3 – Magnetic Effects of Current and Magnetism

27 - 38

Chapter-4 – Electromagnetic Induction and Alternating Currents

39 - 52

Chapter-5 – Electromagnetic Waves

53 - 61

Chapter-6 – Optics

62 - 80

Chapter-7 – Dual Nature of Matter and Radiation

81 - 91 92 - 106

Chapter-8 – Atom and Nuclei Chapter-9 – Electronic Devices

107 - 124

Chapter-10 – Communication Systems

125 - 136

qqq

LOOK OUT FOR 'HIGHLY LIKELY QUESTIONS'

These questions are selected by Oswaal Editorial Board. These are highly likely to be asked in the upcoming examinations.

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2 Languages + 6 Domain Specific Subject + General Test

OR

3 Languages + 5 Domain Specific Subjects + General Test

Subject/Language Choice

Objective Type with MCQs

CBT

Mode of Test

Test Pattern

Know Your CUET (UG) Exam SECTIONS

SECTION I (A) 13 Languages

Tested through reading Comprehension (i) Factual (ii) Literary (iii) Narrative

SECTION III SECTION I (B)

SECTION II

20 Languages

Domain Specific Subjects ( 27 Subjects)

General Test (Compulsory)

INCLUDES : NCERT Model syllabus (only of 12th Standard) is available on all the Subjects

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• • • • • •

General Knowledge Current Affairs General Mental Ability Numerical Ability Quantitative Reasoning Logical & Analytical Reasoning

Latest CUET (UG) Syllabus PHYSICS -322 Note: There will be one Question Paper which will have 50 questions out of which 40 questions need to be attempted.

Unit I : Electrostatics Electric charges and their conservation. Coulomb’s law – force 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, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside). Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, the electrical potential energy of a system of two point charges, and electric dipoles in an electrostatic field. Conductors and insulators, free charges, and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, the combination of capacitors in series and in parallel, the capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor, Van de Graff generator.

Unit II : Current Electricity Electric current, the flow of electric charges in a metallic conductor, drift velocity and mobility, and their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity. Carbon resistors, colour code for carbon resistors; series and parallel combinations of resistors; temperature dependence of resistance. The internal resistance of a cell, potential difference, and emf of a cell, combination of cells in series and in parallel. Kirchhoff ’s laws and simple applications. Wheatstone bridge, Metre Bridge. Potentiometer – principle, and applications to measure potential difference, and for comparing emf of two cells; measurement of internal resistance of a cell.

Unit III : Magnetic Effects of Current and Magnetism Concept of the magnetic field, Oersted’s experiment. Biot - Savart law and its application to current

carrying circular loop. Ampere’s law and its applications to infinitely long straight wire, straight and toroidal solenoids. Force on a moving charge in uniform magnetic and electric fields. Cyclotron. Force on a current-carrying conductor in a uniform magnetic field. The force between two parallel current carrying conductors – definition of ampere. Torque experienced by a current loop in a 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. The magnetic dipole moment of a revolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniform magnetic field; bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferromagnetic substances, with examples. Electromagnets and factors affecting their strengths. Permanent magnets.

Unit IV : 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; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, wattless current. AC generator and transformer.

Unit V : Electromagnetic Waves Need for displacement current. Electromagnetic waves and their characteristics (qualitative ideas only). Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays) including elementary facts about their uses. Unit VI : Optics Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal reflection, and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula.

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Contd... Magnification, power of a lens, combination of thin lenses in contact combination of a lens and a mirror. Refraction and dispersion of light through a prism.

details should be omitted; only the conclusion should be explained).

Scattering of light–blue colour of the sky and reddish appearance of the sun at sunrise and sunset.

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.

Optical instruments: Human eye, image formation, and accommodation, correction of eye defects (myopia and hypermetropia) using lenses. Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers. Wave optics: Wave front and Huygens’ Principle, reflection, and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygens’ Principle. Interference, Young’s double hole experiment and expression for fringe width, coherent sources, and sustained interference of light. Diffraction due to a single slit, width of central maximum. Resolving the power of microscopes and astronomical telescopes. Polarization, plane polarized light; Brewster’s law, uses of plane polarized light and Polaroids.

Unit VII : Dual Nature of Matter and Radiation  Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation – particle nature of light. Matter waves – wave nature of particles, de Broglie relation. Davisson-Germer experiment (experimental

Unit VIII : Atoms and Nuclei

Radioactivity – alpha, beta, and gamma particles/rays, and their properties; radioactive decay law. Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission and fusion.

Unit IX : Electronic Devices Energy bands in solids (qualitative ideas only), conductors, insulators, and 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. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch.

Unit X: Communication Systems Elements of a communication system (block diagram only); bandwidth of signals (speech, TV, and digital data); bandwidth of transmission medium. Propagation of electromagnetic waves in the atmosphere, sky, and space wave propagation. Need for modulation. Production and detection of an amplitude-modulated wave. 

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Om Stationers, 7007326732

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0808

BARIPADA

UTTAR PRADESH DEHRADUN

( 10 )

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

PHYSICS

[This includes Questions pertaining to Domain Specific Subject only] Max. Marks: 200 

Time allowed: 45 Minutes

General Instructions: 1. The test is of 45 Minutes duration. 2. The test contains 50 questions out of which 40 questions need to be attempted. 3. Marking Scheme of the test: a. Correct answer or the most appropriate answer: Five marks (+5). b. Any incorrectly marked option will be given minus one mark (−1). c. Unanswered/Marked for Review will be given zero mark (0).

1. A beam of electron is used in Young's double slits

experiment. The slit width is d. When velocity of electron is increased, then: (1) No interference is observed (2) Fringe width increases (3) Fringe width decreases (4) Fringe width remains same 2. Match List - I with List - II. List - I (I)

(B) Compound Microscope

(II) Two convex lenses both of small focal length

(C) Refracting Telescope

(III) Two convex lenses, one of small focal length other of large focal length

A concave mirror is used as an objective

(D) Reflecting (IV) Single convex lens of Telescope small focal length Choose the correct answer from the options given below: (1) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) 3. Match List - I with List - II

(A)

(II)

Electric dipole

(C)

(III) q>0

(D)

(IV) Unequal charges

List - II

(A) Simple Microscope

List - I (Field lines)

(B)

Choose the correct answer from the options given below: (1) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (4) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

4. The emfs and resistances in the given circuit have the following values

List - II (Charge configuration) (I)

q (a) (C) (a) is used to make temporary magnets (D) (b) can be easily demagnetised

Which of the given statements are correct ? (1) (A), (C) and (D) (3) (B), (C) and (D)

(2) (A), (B) and (D) (4) (A) and (B) Only

47. The magnetic energy stored in a current carrying solenoid is (if B=magnetic field, l=length of solenoid, A =area of cross section of solenoid):

16

OSWAAL CUET (UG) Chapterwise Question Bank PHYSICS

(1)

B2 2µ0

(2)

µ0B2 2

(3)

µ0 B2 Al 2

(4)

B2 Al 2µ0

48. The

mutual inductance of a pair of solenoid

depends (A) on their separation (B) on their relative orientation (C) on the current flowing (D) on the number of turns in the solenoids

Choose the correct answer from the options given below: (1) (D) only (2) (A) and (B) only (3) (B), (C) and (D) only (4) (A), (B) and (D) only

49. An

electron is projected with uniform velocity

along the axis of a current carrying long solenoid. Which of the following is true ? (1) The electron will be accelerated along the axis (2) The electron path will be circular about the axis (3) The electron will have helical path (4) The electron will continue to move with uniform velocity along the axis of the solenoid

50. A

circular loop of area 1 cm2, carrying a current

of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is: (1) Zero (2) 10−4 Nm −2 (3) 10 Nm (4) 1 Nm



Study Time

CHAPTER

1

Max. Time: 1:50 Hours Max. Questions: 50

ELECTROSTATICS

  Revision Notes Charge:  Charge is an intrinsic property of the matter.  Charges developed on bodies, when they are rubbed against each other are called as frictional charges.  The charge on a body is expressed as q = ± ne, where n is an integer (i.e., no. of electrons present on the body) and e is the charge of an electron. Charge is additive in nature & always conserved. The charge of an electron is e– = – 1.6 × 10–19C and charge on proton is e + = + 1.6 × 10–19C.  Like charges repel each other and unlike charges attract each other. Conductors and Insulators:  Those substances which allow electricity to pass through them easily are called conductors. e.g., metals, human and animal bodies.  Those substances which do not let electricity pass through them easily are called insulators. e.g., glass, porcelain, plastic, nylon, wood. Basic Properties of Electric Charges:  Additivity of charges- If a system contains n charges q1, q2, q3,...,qn, then the total charge of the system is q1, + q2, + q3, + ... + qn,.  Conservation of charge- The total charge of the isolated system is always conserved.  Quantisation of charge- The charge on a body is given by, q = ne where, n is any integer, positive or negative and e is charge of an electron, i.e., e = 1.6 × 10–19C. Coulomb’s Law:  If two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by qq F = k 1 22 r  k=

1 = 9 × 109 Nm 2C −2 , ε0 = 8.85 × 10–12 C2/Nm2 4πε 0

Electric Field:  The space around a charge upto which electric force can be experienced is called the electric field.

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The electric field strength due to a  source charge ‘Q’ at a distance ‘r’ is given Electric charge 1 Q and electric by: = E = 2 fields 4πε0 r  If a test charge q is placed at a point where electric field is E, then the force on the test charge is F = qE Electric Field Lines:  Electric field lines are imaginary lines that originates from the positive charge and terminates at negative charges.

In a charge free region, electric field lines can be taken to be  continuous curves without any breaks.  The electric field lines never intersect each other.  Electric field lines do not form any closed loops.  The denser the electric field lines, the stronger the electric field. Electric Flux: Scan to know more about  Electric flux is the algebraic sum of this topic electric field lines passing through the surface normally.  Due to arbitrary arrangement of electric field lines, electric flux can be quantified as   Electric flux Φ E= E ⋅ ∆S= EDS cosq S E  E

S

(permittivity of free space.) If q1 = q2 = 1C, r = lm, then F = 9 × 109 N . Therefore,  1C is the charge that when placed at a distance of 1m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 109 N.  In a system of charges q1 , q2 , q3 ,…, qn , the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3 , q4 , q5 ,…, qn . The total force on the charge q1 is the sum of individual forces due to all the charges. This is called superposition principle.

In a non-uniform  electric field, the flux   is Φ E =∫ E  ds Electric Dipole:  An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space.

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Electric dipole

2 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

ELECTROSTATICS

3

4 Oswaal CUET (UG) Chapterwise Question Bank By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole. +q

–q

2a  p  The of an electric dipole is defined by  dipole moment p= q × 2a that is, it is a vector whose magnitude is charge q times the separation 2a and the direction is along the line from –q to q . Electric field due to a dipole:  At point P, at a distance r from the centre of the dipole, for 4qa 1 2p r  a , total field is = E = 4πε 0 4πε 0 r 3

At point P on the equatorial plane due to a dipole at a large  distance, 1 p E= 4πε 0 r 3 +q  P 2a r E+q E–q E –q Dipole in a uniform external field:  In a dipole, when the net force on dipole due to electric field is zero and centre of mass of dipole remains fixed, the forces on charged ends produce net torque τ about its centre of mass. = τ F 2a= sin θ qE 2a= sin θ pE sin θ    τ= p × E

If q = 0° or 360°, dipole exists in stable equilibrium state.   If q = 180°, dipole exists in an unstable equilibrium state. Continuous charge distribution: Q  Linear charge density, λ = l  where Q is the charge and l is the length over which it is distributed. Q  Surface charge density, σ = S  S is the area of the surface. Scan to know Q  Volume charge density, ρ = more about V this topic  V is the volume of distribution Gauss’ law:  The net outward normal electric flux through any closed surface of any shape is Gauss’ law equal to 1/e0 times of net charge enclosed by the surface.

PHYSICS

 The electric flux at all points on Gaussian surface is    q    E i ds  . s 0  If the flux is positive, then the net charge enclosed is positive.  If the flux is negative, then the net charge enclosed is negative.  If the flux is zero, then no net charge is enclosed.  The electric field due to a point charge on Gaussian surface 1 q is E = . 4πε 0 r 2 Application of Gauss’ law:  Electric field due to a uniformly charged infinitely long 2λ straight wire- E = 4πε0  Electric field due to a uniformly charged infinite plane sheetσ E= 2ε0  Electric field due to a uniformly charged thin spherical shellq (i) Field outside the shell- E = 4πε 0 r 2 (ii) Field inside the shell- E = 0 Electric potential:  Amount of work done by an external force in moving a unit positive charge from infinity to a point in an electrostatic field without producing an acceleration. V= 

W q

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Electric

potential and where, W = work done in moving a potential charge q through the field, q = charge being difference moved through the field. J Nm .  The SI units of electric potential are , volt, C C

Potential difference:  Amount of work done in moving a unit charge form one point to another in an electric field is called electric potential difference. Work ∆PE W Electric potential difference = = = Charge Charge q Between two points A and B, WAB = −VAB × q where, VAB is potential difference between A and B. = VB − VA  In an electric field, the work done by electric field to move a test charge q by a distance dl is dW.   dW = qE  dl  B   qE  dl B  WAB ∫ ∆V = VAB = VB − VA = − = − A = − ∫ E  dl A q q  The electric field at a point is related to the negative potential   dV  gradient as follows- E = −    dr  –1 Unit- Vm Electric potential due to a point charge:  Electric potential due to a point source charge q at a distance 1 q  r from it is given by- V = 4πε 0 r

5

ELECTROSTATICS

Electric potential due to a dipole:  Electric potential due to a dipole at a point at a distance r and making an angle q with the dipole moment pe is given by1 p cos θ V =  e 2 4πε0 r  On the axial line, the electric potential is given by1 p Vaxial =  e 4πε 0 r 2  On the equatorial line, the electric potential is- Vequitorial = 0

Electric potential due to system of charges:  Potential at a point due to system of charges is the sum of potentials due to individual charges. q 5 q q 1 4 r5P r4P r1P r3P

P

q 3

r2P

q 2 V= V1 + V2 + … + Vn Equipotential surfaces:  An equipotential surface is a surface with a constant value of potential at all points on the surface. It requires no work to move the charge on such surface.  Electric field is always perpendicular to the equipotential surface.  Spacing among equipotential surfaces allows to locate regions of strong and weak electric field.  Equipotential surfaces never intersect each other. If they intersect, then the intersecting point of two equipotential surfaces results in two values of electric potential, which is impossible. Electric potential energy of a system of charges:

 Potential energy of a system of two charges- U =

1 q1q2 4πε0 r12

 Potential energy of a system of three charges U =

1  q1q2 q1q3 q2 q3  + +   r13 r23  4πε0  r12

Electric potential energy in an external field:

  Potential energy due to a single charge- U = qV ( r )

 Potential energy due to two charges in an external field  1 q1q2 U = q1V r1 + q2V r2 + 4πε 0 r12

()

( )

Potential energy of a dipole in an external field U ( θ ) =− pE cos θ Where, p = 2aq and q is the angle between electric field and dipole. Conductors and insulators:  Conductors- Materials through which charges can move freely. Examples- metals, semi-metals such as carbon, graphite, antimony and arsenic.  Insulators- Materials in which the electrical current does not flow easily. Examples- Plastics and glass.

Dielectrics and polarisation:  Dielectrics- Materials in which induced dipole moment is linearly proportional to the applied electric field.  Polarisation- Electric polarisation is the difference between electric fields E (induced) and E (imposed) in a dielectric due to bound and free charges. P = ε0χe E Where, χe is electric susceptibility of the dielectric medium. Capacitors and capacitance:  A capacitor is a device that is used to store charge.  The ratio of the charge q and potential V of a conductor is called capacitance (C). Q V  The SI unit of capacitance is farad (F). Combination of capacitors:



C=

1 1 1 1  In series- = + +…+ Cs C1 C2 Cn

 In parallel- C= C1 + C2 + … + Cn p Capacitance of parallel plate capacitor:  Parallel plate capacitor is a capacitor with two identical plane parallel plates separated by a small distance where space between them is filled by dielectric medium. Capacitance of parallel-plate capacitor  with area A separated by a distance d is, ε A C= 0 d

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Combination of capacitors Scan to know more about this topic

Parallel plate capacitor

If we have a number of slabs of same area as the plates of  the capacitor with thickness t1 , t2 , t3 ,… and dielectric constants K1, K2, K3,... inserted between the plates, the capacitance of the capacitor is given by, ε0 A C= t1 t2 t + + 3 + .... K1 K 2 K 3  If a single slab of thickness ‘t’ and dielectric constant ‘K’ is introduced between the plates having separation ‘d’ then the capacitance of the capacitor is given by, ε0 A K ε0 A = C = t (d − t ) K (d − t ) + t + 1 K  If a single slab of conductor of thickness ‘t’ is introduced between the plates, having separation ‘d’ then the capacitance of the capacitors is given by, ε0 A ε0 A = C = t (d − t ) d − t + ∞ 1 Energy stored in a capacitor: = U

1 Q2 1 1 = = QV CV 2 2 C 2 2

 The energy density of in a region with electric field is, 1 U = ε0E 2 2

6 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Match List - I with List – II

A

List - I (Field lines)

List - II (Charge configuration)

(A)

(I)

q0

(D)

(IV)

Unequal charges

(1) (2) (3) (4)  2.

Choose the correct answer from the options given below: (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (CUET 2023, 26th May) A spherical metallic conductor of radius 6.0 cm is held in air. The maximum charge it can hold if dielectric strength of air is 20 kV/cm will be   1 = 9 × 109 Nm 2C −2   Take 4πε0  

(1) 0.8 mC (2) 8.0mC (3) 80 mC (4) 8 mC  (CUET 2023, 26th May) 3. Two capacitors of capacitances 8 mF and 20 mF are connected in series with a battery. The voltage across the 8 mF capacitor is 5 V. The total battery voltage is (1) 5 V (2) 10 V (3) 7 V (4) 125 V  (CUET 2023, 26th May) 4. Electrical capacity of earth is approximately (1) 57.6 mF (2) 1 F (3) 711 mF (4) 9 × 109 mF  (CUET 2023, 26th May) 5. Three points A, B and C lie in a uniform Electric field E of 2 × 103 N/C as shown in figure. The potential difference between points A and C is B A

C

(1) 100 V (2) 60 V (3) 80 V (4) Zero  (CUET 2023, 26th May) 6. For a given charged conducting hollow sphere arrange the electric field at the given points A, B and C in decreasing order

C

(1) EB > EC > EA (2) EA = EB > EC (3) EC > EB > EA (4) EA > EB > EC  (CUET 2023, 26th May) 7. Match List - I with List - II. List - I (Physical Quantity)

List - II (S.I. Unit)

(A) Linear charge density (B)

Electric dipole moment

(C) Polarisation vector

(I) newton (metre)2/ columb (II) coulomb/metre (III) coulomb/(metre)2

(D) Electric flux (IV) coulomb - metre Choose the correct answer from the options given below: (1) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (4) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)  (CUET 2023, 26th May) 8. Three charges + 4q, Q and q are placed along x-axis at points, x = 0, l/2 and l respectively. The value of Q to make the net force on q to be zero q (1) 4q (2) − (3) −2q (4) −q 2  (CUET 2023, 26th May) 9. Which one of the following correctly represents the variation of electric field and electric potential with a distance r from a point charge? E E, V E V E,V V (1)



(2)

r

r

VE

(3)

EV

E, V

E,V

r

E

B

(4) r

10. A circular lamina of radius ‘R’ is having surface charge density s cm–2. Electric field at axial distance ‘2R’ is (1) 

σ  2  σ σ σ (2) (4) 1 −  (3) 2ε 0  5 4πε0 R 2 2ε 0 4ε 0 (CUET 2022, 5th August)

11. Two-point charges +5 mC and –5 mC are placed at O (0 mm, 0 mm) and P (3 mm, 4 mm) respectively force on +5 mC is

7

ELECTROSTATICS

Y

X

( ) (3) 1.8 ( 3i + 4j ) × 10 N

(

) (4) 1.8 ( −3i + 4j ) × 10 N

(2) −1.8 3i + 4j × 10+3 N

(1) 1.8 3i − 4j × 10+3 N

3

3

 (CUET 2022, 5th August) 12. Three charges Q, +q and +q are placed at the vertices of an equilateral triangle of side l, as shown in the figure. If the net electrostatic energy of the system is zero, the Q is equal to Q l

l

+q

l

q 2



(1) Vo V (2) 0 K 8V0 8V0 (3) (4) 5K K  (CUET 2022, 5th August) 14. Electric field at the surface of a conducting shell of radius ‘r’ is measured as X. Electric field at a distance 3r from the centre of the shell is X (4) X 9  (CUET 2022, 6th August) 15. A charge + 10 μC is placed at (0 mm, 0 mm). Another charge –5 μC is moved from (3 mm, 0 mm) to (0 mm, 3 mm). Work done by the external agency is (1) 0 J (2) –150 J (3) + 150 J (4) –300 J  (CUET 2022, 6th August) 16. Two charges in air experiences a force of 20 N. If the space between them is filled with a medium of dielectric constant K = 4, the new force will be (1) 20 N (2) 5 N (3) 4 N (4) 2 N  (CUET 2022, 8th August) 17. Which one of the following statements is wrong? (1) Electric field lines start from positive charges and end at negative charges.

(1)

1 thickness   d and placed between plates of parallel 4 plate capacitor, where d is the separation of the plates. The ratio of capacitances of capacitor with dielectric and with

+q

q (2) Zero (3) (4) –q − 2  (CUET 2022, 5th August) 13. Consider a capacitor circuit charged by a battery to a potential difference V0 as shown in the figure. When a dielectric slab is inserted into both the capacitors, new potential difference will be

(1)

(2) In a charge free region, electric field lines can be taken to be continuous without any break. (3) Two field lines never cross each other. (4) Electrostatic field lines form closed loops. 18. Select the correct statement from the following A. Gauss’s law is true for any closed surface, no matter what its shape or size. B. If the net electric flux through a closed surface is zero, we can conclude that the total charge contained in the closed surface is also zero. C. Gauss’s law is not based on the inverse square dependence on distance contained in the Coulomb’s law. D. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface. Choose the correct answer from the options given below (1) A only (2) A and C only (3) A, B and D only (4) B, C and D only (CUET 2022, 8th August) 19. A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a

X 3

(2)

X 6

c air as medium   is  c0  (1) 20. (1) (2) (3) (4) 21.

K 4K 3K + 1 4 (2) (3) (4) 3K + 1 K +3 4K 3 +1 Charged bodies are considered as point charges if the linear size of the charged bodies is Much smaller than the separation between them Greater than the separation between them Equal to the separation between them Independent of separation between them (CUET 2022, 8th August) In the Fig. below, two positive charges q2 and q3 fixed along the y axis, exert a net electric force in the + x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1 y

y q2

q2

(3)

q1

q1 q3

(a)

(1) (2) (3) (4)

Q (x, o)

x O

O

x

q3   

(b)

shall increase along the positive x-axis. shall decrease along the positive x-axis. shall point along the negative x-axis. shall increase but the direction changes because of the intersection of Q with q2 and q3.  [NCERT Exemplar, Q.1.1 Page No. 1,] 22. A point positive charge is brought near an isolated conducting sphere (Fig.). The electric field is best given by

8 Oswaal CUET (UG) Chapterwise Question Bank



PHYSICS

(2) E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only. (3) E on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q1, q3 and q5 only. (4) Both E on the LHS and q on the RHS will have contributions from q2 and q4 only.  [NCERT Exemp. Q.1.4 Page No. 3] 25. A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is (1) directed perpendicular to the plane and away from the plane. (2) directed perpendicular to the plane but towards the plane. (3) directed radially away from the point charge. (4) directed radially towards the point charge.  [NCERT Exemp. Q.1.6 Page No. 3] 26. A capacitor of 4 µF is connected as shown in the circuit. The internal resistance of the battery is 0.5Ω. The amount of charge on the capacitor plates will be

(i)

(ii)

(iii)

4F 10 

(1) fig (i)

(2) fig (ii)

(iv)

(3) fig (iii) (4) fig (iv)

23. The Electric flux through the surface

2.5V

S

2

S

S

S +q

+q

+q

+q

(i)

(1) (2) (3) (4)  24.

(ii)

(iii)

(iv)

in Fig. (iv) is the largest. in Fig. (iii) is the least. in Fig. (ii) is same as Fig. (iii) but is smaller than Fig. (iv) is the same for all the figures. [NCERT Exemp. Q.1.3 Page No. 2] Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in Fig. S is a Gaussian surface. The Gauss’s law is given by

(1) 0 (2) 4 µC (3) 16 µC (4) 8 µC  [NCERT Exemp. Q.2.1 Page No. 10] 27. A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge (1) remains a constant because the electric field is uniform. (2) increases because the charge moves along the electric field. (3) decreases because the charge moves along the electric field. (4) decreases because the charge moves opposite to the electric field [NCERT Exemp. Q.2.2 Page No. 10] 28. Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B. 30V 40V

q

∮E  ds = ε

s

B

A

A

30V

B

B

A

0

Which of the following statements are correct? Gaussian Surface q1

S q2 q4

q5

q3

(1) E on the LHS of the above equation will have a contribution from q1, q5 and q3 while q on the RHS will have a contribution from q2 and q4 only.

10V 20V 30V 40V 50V

10V 20V 50V

Fig. I

Fig. II

(1) (2) (3) (4)

10V

20V

40V

50V

Fig. III

The work done in Fig. (i) is the greatest. The work done in Fig. (ii) is least. The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii). The work done in Fig. (iii) is greater than Fig. (ii)but equal to that in Fig. (i). [NCERT Exemp. Q.3 Page No. 11] 29. Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately (1) spheres. (2) planes. (3) paraboloids (4) ellipsoids.  [NCERT Exemp. Q.5 Page No. 11]

9

ELECTROSTATICS

30. A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant k1 and the other has thickness d2 and dielectric constant k2 as shown in Fig. This arrangement can be thought as a dielectric slab of thickness d ( = d1 + d2) and effective dielectric constant k. The k is

(1) (3)

d1

k11 k

d1

kk22

k1d1 + k2 d 2 d1 + d 2

k1k2 ( d1 + d 2 )

( k1d1 + k2 d 2 )

(2)

k1d1 + k2 d 2 k1 + k2

2k1k2 (4) k1 + k2

[NCERT Exemp. Q.6 Page No. 12]  [B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: (1) Both (A) and (R) are true and (R) is the correct explanation of (A). (2) Both (A) and (R) are true and (R) is not the correct explanation of (A). (3) (A) is true but (R) is false. (4) (A) is false but (R) is true. 1. Assertion (A): Circuit containing capacitors should be handled very carefully even when the power is off. Reason (R): The capacitors may break down at any time. 2. Assertion (A): In a non-uniform electric field, a dipole will have translatory as well as rotatory motion. Reason (R): In a non-uniform electric field, a dipole experiences a force as well as torque. 3. Assertion (A): Electric lines of force cross each other. Reason (R): The resultant electric field at a point is the superposition of the electric fields at that point. 4. Assertion (A): If two spherical conductors of different radii have the same surface charge densities, then their electric field intensities will be equal. Total charge Reason (R): Surface charge density = Area 5. Assertion (A): In a cavity of a conductor, the electric field is zero. Reason (R): Charges in a conductor reside only at its surface. 6. Assertion (A): When the inverse square law is not obeyed, Gauss’ law shows deviation. Reason (R): The conservation of charges leads to Gauss’ law. 7. Assertion (A): A negative charge in an electric field moves opposite direction of the electric field. Reason (R): On a negative charge a force acts in the opposite direction of the electric field. 8. Assertion (A): Electric field is always normal to equipotential surfaces and along the direction of decreasing order of potential.



Reason (R): Negative gradient of electric potential is electric field. 9. Assertion (A): Work done in moving a charge between any two points in a uniform electric field is independent of the path followed by the charge between these two points. Reason (R): Electrostatic forces are non-conservative. 10. Assertion (A): Two parallel metal plates having charge +Q and –Q are facing at a distance between them. The plates are now immersed in kerosene oil and the electric potential between the plates decreases. Reason (R): Dielectric constant of kerosene oil is less than 1.

[C] COMPETENCY BASED QUESTIONS

I. Read the following text and answer the following questions on the basis of the same: Faraday Cage: A Faraday cage or Faraday shield is an enclosure made of conducting mate rial. The fields within a conductor cancel out with any external fields, so the electric field within the enclosure is zero. These Faraday cages act as big hollow conductors you can put things in to shield them from electrical fields. Any electrical shocks the cage receives, pass harmlessly around the outside of the cage.

1. Which of the following material can be used to make a Faraday cage? (1) Plastic (2) Glass (3) Copper (4) Wood 2. Example of a real-world Faraday cage is (1) car (2) plastic box (3) lighting rod (4) metal rod 3. What is the electrical force inside a Faraday cage when it is struck by lightning? (1) The same as the lightning (2) Half that of the lightning (3) Zero (4) A quarter of the lightning 4. If isolated point charge +q is placed inside the Faraday cage. Its surface must have charge equal to (1) Zero (2) +q (3) –q (4) +2q 5. A point charge of 2mC is placed at centre of Faraday cage in the shape of cube with surface of 9 cm edge. The number of electric field lines passing through the cube normally will be (1) 1.9 × 105 Nm2/C entering the surface. (2) 1.9 × 105 Nm2/C leaving the surface. (3) 2.0 × 105 Nm2/C leaving the surface. (4) 2.0 × 105 Nm2/C entering the surface. II. Read the given text and answer any four of the following questions on the basis of the same: Super capacitor: Super capacitor is a high capacity capacitor with a capacitance value much higher than normal capacitors but with lower voltage limits. Such

10 Oswaal CUET (UG) Chapterwise Question Bank





6. (1) (3) 7. (1) (2)

capacitors bridges the gap between electrolytic capacitors and rechargeable batteries. In automobile, bus, train, crane, elevator such capacitors are used for regenerative braking, short term energy storage or burst-mode power delivery. Super capacitors have many advantages over batteries: they are very low weight and generally don’t contain harmful chemicals or toxic metal. They can be charged and discharged innumerable number of times without ever wearing out. The disadvantage is that super capacitors aren’t wellsuited for long-term energy storage. The discharge rate of super capacitors is significantly higher than lithium-ion batteries; they can lose as much as 10-20% of their charge per day due to self-discharge. Capacity of super capacitor is very low. (2) medium. very high. (4) may have any value. Super capacitor makes a bridge between electrolytic capacitor and rechargeable battery. single use battery and electrolytic capacitor.

(3) (4) 8. (1) (2) (3) (4)

PHYSICS

9. (1) (2) (3) (4)

electrolytic capacitor and dynamo. electrolytic and non-electrolytic capacitors. Super capacitors can be charged and discharged few number of times. once only. several number of times but less than rechargeable batteries. several number of times much more than rechargeable batteries. Self-discharge rate of Super capacitors 10-20% of their charge per day 1 - 2% of their charge per day 0% of their charge per day 100% of their charge per day

10. (1) (2) (3) (4)

Super capacitors are used for degenerative braking. regenerative braking. small appliances. long time charge storage.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (3)

3. (3)

4. (3)

5. (3)

6. (1)

7. (4)

8. (4)

9. (3)

10. (2)

11. (3)

12. (3)

13. (3)

14. (3)

15. (1)

16. (2)

17. (4)

18. (3)

19. (2)

20. (1)

21. (1)

22. (1)

23. (4)

24. (2)

25. (1)

26. (4)

27. (3)

28. (3)

29. (1)

30. (3)

8. (1)

9. (3)

10. (3)

8. (4)

9. (1)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (4)

4. (2)

5. (1)

6. (2)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (1)

3. (3)

4. (3)

5. (3)

6. (3)

7. (1)

ANSWERS WITH EXPLANATIONS [A] MULTIPLE CHOICE QUESTIONS

1. Option (3) is correct. Explanation: Electric field lines always point away from a positive charge and towards a negative charge. In fact, electric fields originate at a positive charge and terminate at a negative charge. For Unequal charges the field lines will not be proportionate at the two sides of the charge, that is the number of field lines will differ. 2. Option (3) is correct. Explanation: Consider that the charge is Q. The dielectric strength of air is given. The maximum charge which can be given without ionizing the air around it is given by, kQ E= 2 r Or,

9 × 109 × Q 20 ×103 = 2 6 20 × 103 × 36 9 × 109

Or,

Q=

∴

Q= 80 ×10−6 C = 80 µC

3. Option (3) is correct. 8 × 20 Explanation: Ctotal = = 5.71µF 8 + 20 Q = CV = 8 × 5 = 40µF Vtotal=

Q 40 = = 7V C 5.71

4. Option (3) is correct. Explanation: C = 4pe0R 1 Here, 4pe0 = ; 9 × 109 R = 6400 km = 6.4 × 106m 6.4 × 106 C= = 0.711 × 10−3 F = 711µF 9 × 109 5. Option (3) is correct. Explanation: The line joining B to C is perpendicular to electric field, so potential at B = potential at C Distance AC = 5 cm Potential difference between A and C = E × (AB) 2 × 103 × (4 × 10–2) = 80 volt.

11

ELECTROSTATICS

6. Option (1) is correct. Explanation: There are no charges inside the hollow conducting sphere, all charges reside on its surface. So, electric field inside the hollow conducting sphere is zero. Electric field decreases exponentially outside the conductor. 7. Option (4) is correct. Explanation: Linear charge density (l) is the quantity of charge per unit length, measured in coulomb per meter (Cm–1) at any point on a line charge distribution. The SI unit for electric dipole moment is the coulomb-meter (Cm).  Polarization vector, P is equal to the bound charge per unit area or equal to the surface density of bound charges (because surface charge density is charge per unit area), S.I. unit is Cm–2. The Sl unit of electric flux is volt meter (V-m), this is also equal to newton-meter squared per coulomb (Nm2Cm–1). 8. Option (4) is correct. Explanation: The force exerted on the charge q by charge Q is given as:  qQ  4kqQ = F ( q, Q ) k= l2 / 4 l2   The force exerted on the charge q by charge 4q is given as:  4q 2  4kq 2 F= ( q, 4q ) k=  2  l2  l  The net force is zero, so, F ( q, Q ) + F ( q, 4q ) = 0 4q q Q 4kqQ 4kq 2 + 2 = 0 2 l/2 l/2 l l

Q+q= 0 \ Q=–q Therefore, the charge of Q should be – q for the net force on q to remain zero. 9. Option (3) is correct. 1 q Explanation: E  40 r 2

V

1 q 4 0 r

12. Option (3) is correct. Explanation:

Given, Or,

σ (1 − cos θ) , where cos q  = 2ε 0

Putting the value of cos q E=

Q=

Charge = 4C0V0 3C K   =  KC0 + 0 V1 2   8V0 5K 14. Sol. Option (3) is correct. Explanation: For shell,

So, V1 =



(

)

(

)

9 × 109 × 5 × 10−6 × 5 × 10−6 4j + 3i × 10−3  F = 42 + 32 × 10−9

(

 F = 1.8 3i + 4j × 103 N

(

)

)

E

1 r2

Or,

E1 r2 = 22 E2 r1

Or,

X (3r ) 2 = E2 (r )2

X 9 15. Option (1) is correct. Explanation:

\

E2 =

X (0mm, 3mm)

B

(0mm, 0mm) O

4R2 + R2

11. Option (3) is correct. Explanation: Force between two-point charges Q1 and Q2   kQ1Q2 r2 − r1  F = | r2 − r1 |3

−q 2

13. Option (3) is correct. Explanation: On inserting dielectric, charge on the combination of capacitor does not change. Equivalent capacitance = C0 + 3C0 = 4C0 New potential difference be V1

2R

σ  2  1 −  2ε 0  5

U=0 kq 2 2kqQ + =0 l l

So,

So, E will come down faster then V with an increase of r as E = r2 10. Option (2) is correct. Explanation: E=

kqQ kqQ kq 2 + + l l l

Net electrostatic energy = U =

+10 µC

A (3mm, 0mm)

X

Z Since, OA = OB, potentials at A and B are same and potential difference between A and B is 0. So, work done is also 0. 16. Option (2) is correct.  1   Q1Q2  Explanation: Fair =   2   4πε 0   r 



 1  Q1Q2  Fmed =     4πε  r 2  Fair   =K=4 Fmed  0

12 Oswaal CUET (UG) Chapterwise Question Bank

Now, as a positive charge is placed on the x axis, the force between q1 and Q is also attractive and is in the + x direction. y

20 =4 Fmed

Or,

∴ Fmed = 5 N 17. Option (4) is correct. Explanation: Electrostatic field lines don’t form closed loops since it is conservative in nature. 18. Option (3) is correct. Explanation: Gauss’s law is true for any closed surface, no matter what its shape or size. So, option (C) is correct. Gauss’s law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. So option (B) is correct. Flux is due to charge enclosed. The electric field due to a charge outside the Gaussian surface contributes nothing through the surface. So, option (D) is correct. Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. So, option (C) is incorrect. 19. Option (2) is correct. Explanation: C0 =

C=



=



=



=

Aε 0 d

ε0 A d −t +

t K

ε0 A d−

d d + 4 4K

ε0 A d 3d + 4 4K ε0 A d 1 3 +  4 K

ε0 A d C × = d 1  Aε 0 C0 3 +  4 K



=

4K 3K + 1

20. Option (1) is correct. Explanation: When the linear size of charged bodies is much smaller than the separation between them, the size may be ignored and the charged bodies are considered as point charges. 21. Option (1) is correct. Explanation: As the net electric force on charge q1 is in + x direction, hence the nature of force between q1, q2 and q2, q3 should be attractive which means q1should be negative. y q2 F1, 2 –q1

PHYSICS

Fnet

x

F1, 3 q3

q

2

F1, 2 –q

1 F1, 3

Q

Fnet

x (x, 0)

Fq.Q q

3



Hence, the net force on charge q1 shall increase along the + x direction. 22. Option (1) is correct. Explanation: The electric field lines come out of positive charge and go into negative charge. The field lines are always normal to the conducting surface. Now, when a positive charge is brought near an isolated conducting sphere, without touching it, then the free electrons in the sphere gets attracted towards the positive charge. As a result, the left surface of the sphere becomes charged with negative charge, whereas the right surface becomes charged positively. It should be emphasised that neither type of charge is free to exit the metal sphere. Therefore, they are located on the sphere’s exterior. 23. Option (4) is correct. Explanation: According to Gauss law of electrostatics, the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Hence, the electric flux through a surface does not depend on its shape, size or area. Therefore, the electric flux through all the surfaces is same as the charge enclosed is same. 24. Option (2) is correct. Explanation: On the LHS of the equation, the electric field is due to all the charges present both inside as well as outside the Gaussian surface. On the RHS of equation, the flux is due to all the charges enclosed in the Gaussian surface, i.e., q2 and q4. 25. Option (1) is correct. Explanation: Free electrons are drawn to a point positive charge when it is positioned close to an isolated conducting plane. As a result, a little amount of negative charge forms on the plane’s surface in the direction of the positive charge, and a similar positive charge forms on the other side of the plane. The electric field lines are parallel to the surface and away from positive charge. As a result, option (1) is true since the field at point P on the other side of the plane is directed away from and perpendicular to the plane. 26. Option (4) is correct. Explanation: At steady state capacitor gets fully charged and offers infinite resistance in DC circuit and acts as open circuit, hence no current flows through the capacitor and 10 W resistance. So, the potential difference across the capacitor is equal to potential difference across 2 W resistance. Current through 2 W resistance, V 2.5 = I = = 1A R + r 2 + 0.5 Potential difference across 2 W resistance, V = IR =1 × 2 =2V

13

ELECTROSTATICS

So, potential difference across capacitor is also 2V. Therefore, the charge on the capacitor, Q = CV = 2µF × 2V = 8µC 27. Option (3) is correct. Explanation: Electric potential energy decreases in the direction of the field. And as the particle is positively charged it will move along the direction of the electric field hence its electric potential energy will decrease. 28. Option (3) is correct. Explanation: In all the three figures, VA = 20V and VB = 40V Work done in carrying a charge q form A to B is W = q(VB – VA) Hence, work done is same in all figures. 29. Option (1) is correct. Explanation: From a large distance, a collection of charges whose sum is not zero can be considered as a point charge. The equipotential surfaces due to a point charge are spherical. 30. Option (3) is correct. Explanation: This system as shown in the figure can be considered as two capacitors say C1 and C2 connected in series. k Aε k Aε We know, C1 = 1 0 and C2 = 2 0 d1 d2 Now,

1 1 1   Ceq C1 C2

k1 Aε0 k2 Aε0 × k1k2ε0 A d1 d2  Ceq = = k2 Aε0 k2 Aε0 k1d 2 + k2 d1 + d1 d2

…(i)

We can write equivalent capacitance as C=

k ε0 A d1 + d 2

 On comparing (i) and (ii), we get, k k ( d + d2 ) k= 1 2 1 k1d 2 + k2 d1

[B] ASSERTION REASON QUESTIONS

…(ii)

1. Option (1) is correct. Explanation: Even when power is off capacitor may have stored charge which may discharge through human body and thus one may get a shock. 2. Option (1) is correct. Explanation: When an electric dipole is placed in a uniform electric field at an angle q with the field, the dipole experiences a torque. The torque produced by two parallel forces qE acting as couple t = qE (2a sin q) In case of non-uniform field, force acting on both the ends of the dipole will not be equal. So, there will be a combination of couple and a net force. In this way, dipole will have both rotational as well as linear motion. So, both assertion and reason are true. Reason also explains the assertion. 3. Option (4) is correct. Explanation: Electric lines of force never cross each other. If they cross each other, then at that point, we get two directions of electric field, which is not possible. So, the assertion is false. The resultant electric field at a point is a vector sum of the electric fields at that point. So, the reason is true.

4. Option (2) is correct. Explanation: If s be the surface charge density of the two spheres of radius r and R, then electric fields for the two spheres are respectively: K 4r 2 E1 =  K 4 r2 E2 =

K 4πR 2σ = K 4πσ R2

o, electric field intensities are equal. The assertion is true. S Surface charge density is charge per unit area = Total charge/area. So, reason is also true. But the reason does not explain the assertion. 5. Option (1) is correct. Explanation: The charge enclosed by the Gaussian surface surrounding the cavity is zero. Hence, the electric field is also zero. So, the assertion is true. Charges in a conductor reside only at its surface. So, in cavity there is no charge. So, the reason is also true and properly explains the assertion. 6. Option (2) is correct. Explanation: The total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity, according to Gauss law. Inverse square dependency of the electric field on distance is the basis of Gauss theorem. Therefore, both assertion and reason are correct but reason is not correct explanation of assertion. 7. Option (1) is correct. Explanation: The direction of electric field is from equipotential surface with high electric potential energy to low electric potential energy. So, the negative charge gets attracted towards the equipotential surface with high electrostatic energy because of strong Coulomb attraction and hence, moves in direction opposite to the electric field. 8. Option (1) is correct.  Explanation: E   dV dr The electric field is always perpendicular to the equipotential surface. Therefore, the assertion is true. Negative gradient of electric potential is electric field. So, direction of electric field must be in the direction of the decreasing order of electric potential. Therefore, the reason is also true and is the correct explanation of assertion too. 9. Option (3) is correct. Explanation: Work done in moving a charge between any two points in a uniform electric field = charge × potential difference. So, it is independent of the path followed by the charge. Hence, the assertion is true. Electrostatic forces are conservative type. Hence, the reason is false. 10. Option (3) is correct. Explanation: Electric field for parallel plate capacitor in vacuum = E = s/e0 Electric field in dielectric = E′ = s/ke0. Since, the value of k for Kerosene oil is greater than 1, then E ′ < E and hence V ′ < V. Hence, the assertion is true. Dielectric constant of Kerosene oil is greater than 1. Hence, the reason is false.

14 Oswaal CUET (UG) Chapterwise Question Bank [C] COMPETENCY BASED QUESTIONS

1. Option (3) is correct. Explanation: A Faraday cage or Faraday shield is an enclosure made of a conducting material. Since, copper is the only metal given in the list of options, copper is the correct answer. 2. Option (1) is correct. Explanation: Cars are example of Faraday Cages in the real world. Cars can help keep us safe from lightning. Its metal body acts as a Faraday Cage. 3. Option (3) is correct. Explanation: The field within a conductor cancel out with any external fields, so the electric field within the enclosure is zero. 4. Option (3) is correct. Explanation: If a charge is placed inside an Faraday shield without touching the walls of the internal face of the shield, becomes charged with – q, and + q charge accumulates on the outer face of the shield. If the cage is grounded, the excess charges will be neutralized by the ground connection. 5. Option (3) is correct. Explanation: The number of electric field lines passing through the cube normally and leaving the surface = Q/e0 Q = 2 mC = 2 × 10–6C

24

PHYSICS

e0 = 8.85 × 10–12 C2/Nm2 \ Q/e0 = 2.2 × 105 Nm2/C 6. Option (3) is correct. Explanation: Super capacitor is a high-capacity capacitor with a capacitance value much higher than normal capacitors but with lower voltage limits. 7. Option (1) is correct. Explanation: Such capacitors bridge the gap between electrolytic capacitors and rechargeable batteries. 8. Option (4) is correct. Explanation: Super capacitors can be charged and discharged innumerable number of times without ever wearing out. 9. Option (1) is correct. Explanation: The disadvantage is that super capacitors aren’t well-suited for long-term energy storage. The discharge rate of super capacitors is significantly higher than lithium-ion batteries; they can lose as much as 10-20% of their charge per day due to self-discharge. 10. Option (2) is correct. Explanation: In automobile, bus, train, crane, elevator such capacitors are used for regenerative braking, short term energy storage or burst-mode power delivery.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

2

CURRENT ELECTRICITY Electrical resistance:

  Revision Notes

Electric current:  The time rate of flow of electric charges through any cross-section of a conductor is called electric current. If ∆q charge pass through a cross section in time ∆t, then ∆q Average current Iav = . ∆t Instantaneous current I = ∆lim t →0

∆q dq = ∆t dt

Scan to know more about this topic

Electric current

Drift velocity:  Drift velocity is an average velocity which is obtained by certain particles like electrons due to the presence of an electric field.   eE eV  Drift velocity, vd = − τ = τ m mL λ Where, relaxation time, τ = v

 R =

ρl A

Where, r = resistivity of the material l = length of the wire A = Area of cross-section of the wire  The SI unit of resistance is ohm (W).  r is called resistivity of the material which depend on the nature of the material of the conductor but not on its dimensions. I-V characteristics:  Ohm’s law and their I–V characteristic curve are straight lines passing through the origin.

Here, e = charge of an electron, m = mass of an electron, λ = mean free path  When electric current is set up in a conductor, electrons drift through the conductor with velocity vd, given by I or I neAvd = vd = neA Where, I = Electric current through the conductor, n = Number of free electrons per unit volume A = Area of cross-section of the wire e = Charge of an electron Mobility:  Mobility is the drift velocity of an electron when applied electric field in unity. v µ= d E = µ

Semiconductors, p-n junction diodes do not obey Ohm’s law  and their I–V curves are non-linear.

eτE / m eτ = E m

Ohm’s law:  Current I through a conductor is directly proportional to the potential difference V applied across the ends of the conductor provided the physical conditions such as temperature, mechanical strain, etc. remain unchanged. I ∝V V = IR Or, Where, R is a constant called resistance of the conductor.

Electric energy and power:  Electrical energy due to conduction of charged particles in a conductor causing electric current (I) is E = V × I ×t = I2 × R×t =

V2 ×t R

16 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

17

CURRENT ELECTRICITY

Where, E = electrical energy V = Potential difference across the ends of the conductor I = Current through the conductor R = Resistance of the wire  The SI unit of energy is joule (J).  Power is the work done per unit time. W P= t 2 = P I= R

V2 R

The unit of power is joule per second (J/s) or watt (W).  Electrical resistivity and conductivity:  Resistivity is the resistance of a conductor having unit length and unit area of cross-section. m ρ= 2 ne τ  Conductivity is the reciprocal of resistivity. σ=

1 ne 2 τ = ρ m

Temperature dependence of resistance:  With small change in temperature, resistivity varies with temperature as: = R R0 (1 + α∆t ) Where, a = temperature coefficient of resistance.  The temperature coefficient for conductors is positive i.e., resistance increases as the temperature rises.  The temperature coefficient for the insulators and semiconductors is negative i.e., their resistance decreases as the temperature increases. Internal resistance of a cell:  Cell is a device that maintains the potential difference between the two electrodes as a result of chemical reaction.  Internal resistance is the resistance of electrolyte which resists the flow of current when connected to a circuit.  E −V r =  V

When current is drawn from a cell, its terminal potential difference is less than the emf. Combination of cells in series and parallel:  Series combination of cells- Let n cells, Scan to know each of emf E and internal resistance r more about this topic be connected in series across an external resistance R, then the current in the circuit will be nE Is = Combination R + nr of cells

Parallel combination of cells- When  m cells are connected in parallel across a resistance R, then current through the resistance is given by mE IP = mR + r  If m cells having internal resistances r1 , r2 , r3 ,…, rm and emf E1 , E2 , E3 ,…, Em respectively are connected in parallel across resistance R, then the current through the external resistance is given by

 E1 E2 E3 E  + +…+ m   + r r r rm  2 3 IP =  1 1 1 1 1 R +  + + +…+  r r r r m   1 2 3 Kirchhoff’s Rule: Scan to know  First rule- The algebraic sum of current more about at a junction is zero i.e., ΣI = 0. This implies this topic that the total current entering a junction is equal to the total current leaving the junction. This follows law of conservation of charge. Kirchhoff’s  Second rule- Around any closed loop in laws a circuit, sum of the emfs and the potential differences across all elements is zero, i.e., ΣV = 0 or ΣV = ΣIR. This follows law of conservation of energy. Wheatstone bridge:  It is a circuit having four resistances P, Q, R and S, a galvanometer (G) and a battery connected as shown.

 R 

Potential difference and emf of a cell:  The emf and terminal potential difference of a cell- Let the emf of a cell is ‘E’ and its internal resistance is r. If an external resistance R is connected across the cell through a key, then IR = V = potential difference across the external resistance R. This is equal to the terminal potential difference across the cell.

E= V + Ir I=

E −V r

V= E − Ir V ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in fig.  r1 B

A 2

r2

(1) The equivalent emf eeq of the two cells is between e1 and e2, i.e., e1 < eeq < e2. (2) The equivalent emf eeq is smaller than e1. (3) The eeq is given by eeq = e1 + e2 always. (4) eeq is independent of internal resistances r1 and r2.  [NCERT Exemp. Q.3.2, Page No. 16] 23. A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100 W. He finds the null point at l1 = 2.9 cm. He told to attempt to improve the accuracy. Which of the following is a useful way? (1) He should measure l1 more accurately. (2) He should change S to 1000 W and repeat the experiment. (3) He should change S to 3 W and repeat the experiment. (4) He should give up hope of a more accurate measurement with a meter bridge.  [NCERT Exemp. Q.3.3, Page No. 17] 24. Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm. (1) The battery that runs the potentiometer should have voltage of 8V.

PHYSICS

(2) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V. (3) The first portion of 50 cm of wire itself should have a potential drop of 10V. (4) Potentiometer is usually used for comparing resistances and not voltages. [NCERT Exemp. Q.3.4, Page No. 17] 25. A metal rod of length 10 cm and a rectangular cross1 section of 1cm × cm is connected to a battery across 2 opposite faces. The resistance will be (1) Maximum when the battery is connected across 1 1 cm × cm faces. 2 (2) Maximum when the battery is connected across 10 cm × 1 cm faces. (3) Maximum when the battery is connected across 1 10 cm × cm faces. 2 (4) Same irrespective of the three faces.  [NCERT Exemp. Q.3.5, Page No. 17] 26. Which of the following characteristics of electrons determines the current in a conductor? (1) Drift velocity alone. (2) Thermal velocity alone. (3) Both drift velocity and thermal velocity. (4) Neither drift nor thermal velocity.  [NCERT Exemp. Q.3.6, Page No. 17] 27. Kirchoff’s junction rule is a reflection of (1) Conservation of current density vector. (2) Conservation of charge. (3) The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction. (4) The fact that there is accumulation of charges at a junction.  [NCERT Exemp. Q.3.7, Page No. 18] 28. Consider a simple circuit shown in fig.

stands for a

variable resistance R′. R′ can vary from R0 to infinity. r is internal resistance of the battery (r resistance of wire so, more heat is produced in filament. Therefore, Assertion is true. Filament is made of material having high resistance like tungsten so that heat produced is more. Melting point of the material also should be high so that it can sustain more heat. Hence, reason is also true. Reason properly explains the assertion. 2. Option (1) is correct. Explanation: The metallic body of electrical appliance is connected to the 3rd pin which is an earth pin. By any chance if the metallic body gets connected to the LIVE line, current flows to earth through this pin without giving any shock to the user. Hence, assertion is true. Electric bulb does not have any metallic enclosure and hence there is no requirement of earth pin. So, the reason is also true and properly explains the assertion. 3. Option (2) is correct. Explanation: Resistance of superconductor falls to zero at critical temperature. This property is very useful for power transmission without any loss. Assertion and reason both are true but reason does not explain the assertion. 4. Option (1) is correct. Explanation: Kirchhoff’s voltage law says that the sum of the voltages around any closed loop is zero. A closed loop starts from a node, traces a path through the circuit and returns to the same node. Since the total work done in moving a charge around this close path is zero, hence the electric field is conservative. So, the assertion is true. Potential difference between two points in a circuit does not depend on the path. This is true for conservative field. Hence, the reason is also true and it explains the assertion. 5. Option (3) is correct. Explanation: In a Wheatstone bridge, in balanced condition, if the galvanometer and the voltage source are interchanged, the balanced condition remains same since in both the cases P : Q, remains equal to R : S,. The assertion is true. The balanced condition of Wheatstone bridge depends on the value of the resistances. P : Q, should be equal to R : S,. So, if the resistance values are changed, the balanced condition also gets disturbed. So, the reason is false. 6. Option (2) is correct. Explanation: Voltmeter is connected in parallel with the circuit so as to measure the potential difference between two points across which it is applied. Resistance of voltmeter is very large but this does not explain why voltmeter is connected in parallel with the circuit. 7. Option (2) is correct. Explanation: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on because it draws more current due to its less resistance, consequently, the current through the bulb decreases and so it becomes dim. When the heater coil becomes sufficiently hot, its resistance becomes more and then its draws a little lesser current. Hence, the dimness decreases after sometimes.

26 Oswaal CUET (UG) Chapterwise Question Bank 8. Option (3) is correct. Explanation: Electrons that constitute current are negatively charged. A wire carrying current is not charged. Inside the wire the number of electrons equals to the number of protons, at any instant. Therefore, the conductor is neutral. The current in the conductor is due to the flow of electrons from the negative terminal to the positive terminal. 9. Option (1) is correct. Explanation: In the balance of electric field, the electrons move randomly in all directions. Hence, there is no net motion of electrons and no current flow in the conductor. But in the presence of electric field, electrons move in a particular direction. Each electron experiences a force in the direction opposite to that of the electric field and moves from the negative end to the positive end of the conductor. Thus, there is a current flow. e The drift velocity, vd = − E τ,i.e., vd ∝ E . m 10. Option (2) is correct. Explanation: Drift velocity is the average velocity of electrons in presence of electric field, which is independent of time.

[C] COMPETENCY BASED QUESTIONS 1. Option (4) is correct. Explanation: Current flowing the circuit. V 2 2 1 = = = I= A R + RAB 2 + 8 10 5 Potential gradient of wire AB V IRAB 1 8 = × k= = = 0.16V /m l l 5 10 2. Option (1) is correct. Explanation: Given, l = 225 cm = 2.25 m As V = kl V = 0.16 × 2.25 V = 0.36 volt

24

PHYSICS

3. Option (2) is correct. Explanation: With increase in Resistance of circuit, current in the circuit will decrease. To obtain same balancing length, more Resistive wire would be required. Hence, balancing length will increase. 4. Option (2) is correct. Explanation: The value of X has no effect on the balancing length as it does not change the current of circuit. 5. Option (2) is correct. Explanation: As Terminal Potential drop across battery E is V = E – IX E V = E − .X S+X With rise in value of S, the terminal Potential difference will increase. Hence, balancing length will also increase. 6. Option (2) is correct. Explanation: The heating element is made up of nichrome 80/20 (80% nickel, 20% chromium). 7. Option (2) is correct. Explanation: Nichrome 80/20 means an alloy of 80% nickel, 20% chromium. 8. Option (4) is correct. Explanation: Electricity consumption is measured by kWh. So, 1200 W toaster will consume more electricity. 9. Option (1) is correct. Explanation: The designed electric toaster is operated at 220 volts AC, single phase. 10. Option (4) is correct. Explanation: The element is wound separately on Mica sheets and fitted with body of toaster with the help of ceramic terminals.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

3

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

  Revision Notes

Concept of magnetic field: Magnetic field is the region around  a magnet or a moving electric charge where the force of magnetism is felt. SI unit of magnetic field is Tesla. 1 Tesla = 104 Gauss.  Oersted’s experiment: In 1820, Hans Scan to know Christian Oersted discovered that there more about is a relationship between electricity and this topic magnetism. A magnetic needle was placed near a current carrying wire and it was shown that moving electrons can create a magnetic field. Oersted’s  Biot - Savart law and its application Experiment to current carrying circular loop: The magnetic field due to a current element is given by the following relation:   µ 0 I dl ×  r = ⋅ B 2 4π r µ 0 Idl B = ⋅ 2 sinθ 4π r  r) (θ be the angle between Idl and  Magnetic field at a point due to a current carrying circular loop: 

B

 0 IR 2

[Here, r2 = R2 + x2]



2  R2  x2 

3/ 2

r

 Magnetic field at the centre of the coil with N turns,   NI  r [x = 0] B  0 2R Magnetic field at very large distance from the centre: 2 NiA B 0 3  [Here, R2 v where, R = Radius of curvature of spherical surface and  If light ray bends away from normal after refraction, then object is placed at rarer medium. second medium is optically rarer. If the light ray bends towards u = Object distance from spherical surface normal, then second medium is optically denser. v = Image distance from spherical surface Total Internal reflection and its applications, optical fibres: Lenses:  When light travels from an optically denser medium to a rarer medium at the interface, it is reflected back into the same  A lens is a piece of transparent glass bounded by two medium when angle of incidence is greater than critical angle. surfaces out of which at least one surface is spherical. This reflection is called total internal reflection.  There are two types of lenses:  Critical angle is the angle of incidence for which angle of  Convex lens: A convex lens is one which is thinner at refraction is 90°. edges and thick at centre.  Concave lens: A concave lens is one which is thick at edges and thin at centre.  Image formation in convex lens for different positions of object:

µ2

Position of the object

Position of the image

Relative size of the image

Nature of the image

At infinity

at focus F2

Highly diminished, point sized

Real and inverted

Beyond 2F1

Between F2 and 2F2

Diminished

Real and inverted

At 2 F1

at 2F2

Same sized

Real and inverted

Between 2F1 and 2F2

Beyond 2F2

Enlarged

Real and inverted

At Focus F1

At infinity

Infinitely enlarged

Real and inverted

Between focus F1 and optical On the same side of the Enlarged entre lens as object  Image formation in concave lens for different positions of object:

Virtual and erect

Position of the object

Position of the image

Relative size of the image

Nature of the image

At infinity

At focus F1

Highly diminished point sized

Virtual and erect

Between infinity and the Between focus F1 and optical Diminished Virtual and erect optical centre O of the lens centre O Thin lens formula: Magnification: v h  Thin lens is a lens whose thickness is negligible compared m= = i to the radii of curvature of the lens surfaces. Relation between u ho object distance (u), image distance (v) and focal length (  f  ) of Power of lens: a thin lens:  The power of a lens is defined as the reciprocal of its focal 1 1 1 − = length, represented by the letter P and expressed as v u f 1 P= Lens maker’s formula: f 2 1 1 1  The SI unit of power is dioptre when focal length is in metre.  =( µ 21 − 1)  −  where μ21 f 1  R1 R2 

66 Oswaal CUET (UG) Chapterwise Question Bank Combination of thin lenses in contact and combination of a lens and a mirror:  When two or more lenses are combined, the equivalent focal length and power of the combination is 1 1 1 1     ..... f equ f1 f 2 f 3 P = P1 + P2 + P3 + …… equ  When lens and mirror are used coaxially, the image formation is considered one after another in steps. The image formed by the 1st lens facing the object serves as the object (may be real or virtual) for the mirror on the 2nd lens. In each case, lens appropriate lens or mirror formula to be applied with proper sign convention. Refraction and dispersion of light through a prism:  Refraction through Prism: i+e=d+A Here, i = e

Refracted ray always bends towards the base. Angle of deviation, δ = (i – r1) + (e – r2)  At minimum deviation, i = e (= i) and r1 = r2 (= r), then

µ 21 =

PHYSICS

Scattering of light - blue colour of the sky and reddish  appearance of the sun at sunrise and sunset: Scan to know Scattering: Scattering of light is the more about this topic phenomenon by which a beam of light is redirected in many different directions when it interacts with particle of size comparable to the wavelength of the incident light. Blue colour of sky: Sunlight in Why sun earth’s atmosphere is scattered by the gas appears red during sunrise molecules and the tiny particles floating and sunset? in air. Blue light is scattered more than the other colours because it is the visible light of smallest wavelength which is comparable to the gas molecules and floating tiny particles. So, the sky appears blue. Reddish appearance of the sun at sunrise and sunset: During sunrise and sunset sun stays near the horizon. So the rays travel longer distance through the atmosphere. On the way most of the blue light is scattered away since its wavelength is the least in the visible range and comparable to the gas molecules and floating tiny particles in the atmosphere. The red light is scattered the least since its wavelength is largest in the visible range and hence not comparable to the gas molecules and floating tiny particles in the atmosphere. So, the red light reaches our eyes predominantly and therefore appears red at sunrise and sunset. Optical instruments: Human eye, image formation, and accommodation:  Human eye: Vision is the most precious sense of human being. Our visual system is composed of highly sophisticated brain and modest optical instrument - eye.

sin ( A + δ m ) / 2  sin [ A / 2]

For thin prism, δm = (μ21 – 1)A  Angle of incidence vs. angle of deviation graph:

Dispersion through prism:  In a medium, lights of different colours travel with different velocities and hence refracted in different angles. Thus white light passing through a prism gets split into its constituent colours and form a spectrum. This phenomenon is known as dispersion.  Angular dispersion: The angle subtended between the directions of emergent violet and red rays of light is known as the angular dispersion. Angular dispersion = θ = δv – δR = (μv – μR)A  Dispersive power: Dispersive power is defined as the ratio of angular dispersion to mean deviation (deviation of yellow light) by the prism. 

v   R   v   R    1 

Different parts of human eye:   Cornea: Clear dome shaped front portion of eye through which light enters into eye.  Iris: The coloured portion behind the cornea.  Pupil: A dark hole at the middle of Iris.  Pupil: A dark hole at the middle of Iris.  Aqueous humour: Fluid present in between cornea and iris.  Lens: Just behind the iris; focus the incoming light on the retina.  Ciliary Muscle: Changes the curvature of the lens to focus the incoming rays on retina.  Retina: Photosensitive screen on which the incoming light is focused and image is formed. The image is real, inverted, smaller than object. Retina is made of photoreceptor cells - rods and cones which converts light into electrical energy. Rods perceives black and while and enable night vision. Cones perceive colour and provides details of the vision.  Optic nerve: Carries the electrical signal to brain. Range of clear vision of human eye: 25 cm (near point) to infinity (far point).

67

OPTICS

Image formation by human eye: With the help of convex  eye lens an inverted real and smaller in size image is formed on the retina. Photoreceptor cells convert the light signal into electrical signal. The signal is carried to brain by the optical nerves where image is again inverted and we see an erect image.

Hypermetropia:  Hypermetropia is far-sightedness. In this condition, the distant objects appear clear whereas near objects appear blurred. Causes: (i) Reduction of eyeball size, (ii) Decrease of curvature of eye lens. Correction: Use of a convex corrective lens of appropriate power.

Accommodation:  The ability to change the curvature of the eye lens with the help of ciliary muscle to focus far and distant object. Correction of eye defects (myopia and hypermetropia) using lenses:  Defects of human eye: (i) Myopia, (ii) Hypermetropia  Myopia: Scan to know Myopia is near-sightedness. In this more about condition, the near objects appear clear this topic whereas distant objects appear blurred. Causes: (i) Elongation of eyeball size, (ii) Increase of curvature of eye lens. Correction: Use of a concave corrective Defects of eye lens of appropriate power.

Myopic eye without correction: Image of object at infinity forms in front of retina.

Hypermetropic eye without correction: Image of object at near point forms behind retina.

Hypermetropic eye without correction: Near point shifts away from eye.

Hypermetropic eye after correction. Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers: Microscope: An optical instrument which helps us to see  and study micro objects or organisms.  Simple microscope:

Myopic eye without correction: Far point shifts close to eye.

Image at near point of distinct vision Myopic eye after correction

68 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

Magnification:   When image forms at near point of distinct vision: D m=1+ f D  When image forms at infinity: ⇒ m = f

Image at infinity  Compound microscope:

Magnification:   When image forms at near point of distinct vision:  L  D m    1   fe   f0    L  D   When image forms at infinity: m       f0   fe   Telescope: It is an optical instrument which helps us to see and study far off objects magnified and resolved (with clarity).

 Reflecting telescope:

 f  Magnification m   0   fe 

69

OPTICS

 Refracting telescope:

Normal Adjustment: Image at Infinity

Image at near point of distinct vision Magnification:   When image forms at near point of distinct vision:  f  f  m = –  0  1  e  f D    e  f   When image forms at infinity: m = –  0   fe  Wave optics: Wave front and Huygens’ Principle Wavefront: Wavefront is locus of all points in which light  waves are in same phase. Propagation of wave energy is perpendicular to the wavefront.  Shapes of wavefronts: Source

Wavefront

Point source

Spherical wavefront

Line source

Cylindrical wavefront

Plane source

Plane wavefront

Point source very far away

Plane wavefront

 Huygens’ Principle:  Every point of a wavefront becomes secondary source of light.  These secondary sources give their own light waves. Within small time, they produce their own wave called secondary wavelets. These secondary wavelets have same speed and wavelengths as waves by primary sources.  At any instant, a common tangential surface on all these wavelets gives new wavefront in forward direction. Reflection, and refraction of plane wave at a plane surface using wavefronts: Reflection of plane wavefront: XY is a reflecting surface.  A plane wavefront AP is incident on it. The lines LA and MP which are perpendicular to the incident wavefront AP, represent incident rays. The wavefront arrives at point A first and becomes the source of secondary wavelets. Time required by the P end of the wavefront to reach the reflecting surface is t and PP′ = vt (v = velocity of light in the medium). Arc with radius PP′ is drawn. Tangent A′P′ is the reflected wavefront. The lines A′L′ and P′M′ perpendicular to the reflected wavefront are reflected rays.

L

N

N'

L' M P

A' i

r i

X

M'

A

r

Y P'

 Refraction of plane wavefront: PQ is a line of separation of two media. A plane wavefront AB is incident on it. The lines perpendicular to the incident wavefront represent incident rays. The wavefront arrives at point A first and becomes the source of secondary wavelets. Time required by the B end of the wavefront to reach the line of separation is t and BB′ = v1t (v1 = velocity of light in the medium of refractive index m1). By that time wavelets from A reaches A′ and AA′ = v2t (v2 = velocity of light in the medium of refractive index m2). Arc with radius AA′ is drawn. Tangent A′B′ is the refracted wavefront. The lines perpendicular to the refracted wavefront are refracted rays.

70 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

Proof of laws of reflection and refraction using Huygens’ Principle:  Proof of law of reflection:

The figure drawn here shows the reflected wavefront CE corresponding to the given incident wavefront AB. It is seen that DEAC and DBAC are congruent. ∠i = ∠r

\

This is law of reflection.  Proof of law of refraction:

Central maxima forms at O. Here, path difference = 0   At P, which is at x distance from O, path difference = S2 P – xd S1 P = D xd  Condition for P to be a bright spot: = 0, λ. 2λ, ….., D nDλ nλ, i.e., xnth bright = d xd 3λ  Condition for P to be a dark spot: = 0, , ….., D 2  2n  1 D  2n  1  , i.e., x nth dark = 2d 2 Dλ  Fringe width = β = d Coherent sources, and sustained interference of light:

The figure drawn here shows the refracted wavefront CE corresponding to the given incident wavefront AB. It is seen that

sin i =

BC t = v1 AC AC



sin r =

AE t = v2 AC AC

\

v sin t = 1 = μ21 v sin r 2

This is Snell’s law of refraction. Interference:  Interference is a phenomenon in which waves from two coherent sources combined by adding their intensities with due consideration of their phase difference. The resultant wave will have greater intensity (constructive interference) or lower intensity (destructive interference) if the two waves are in phase or out of phase respectively. Condition for constructive Interference: Phase difference  between the waves should be even multiple of p. Hence Path difference = 0, λ,…….nλ. Here, n = 0, 1, 2, 3……. Condition for destructive interference: Phase difference  between the waves should be odd multiple of p. Hence Path

 2n  1  . Here, n = 1,2,3…… λ 3λ , , ……. 2 2 2 Young’s double hole experiment and expression for fringe width: difference =

 Coherent sources: Two sources are said to be coherent sources if they produce two waves having same frequency, same waveform and have time independent constant phase different between them. Sustained interference of light: Sustained interference of  light indicates that the positions of the maxima and minima of light intensity remains constant throughout the screen. Conditions for sustained interference:  (i) The two interfering waves must be coherent which means the two light waves must have a constant phase difference or must be in the same phase. (ii) The two waves must have the same wavelength. There may be little difference in amplitudes. (iii) The two sources should be very narrow. (iv) The two sources must lie very close to each other.  Intensity distribution pattern:

71

OPTICS

Diffraction due to a single slit, width of central maximum:  Diffraction: It is the bending of light around the corners of an obstacle or aperture into the region where we expect shadow of the obstacle.

Path difference between ray from L and ray from N = NQ  = a sinθ (where width of the opening = a and q = angle of elevation of point P from principal axis.) λ For first maxima: a sinθ = λ, i.e., θ =  (sinθ ≈ θ for small a value of θ )  P is a dark point when path difference = λ, 2λ, …………, nλ

a 1.22λ 2 µsinθ  Resolving power of microscope =  Resolving power of telescope =

λ

Polarization, plane polarized light: Scan to know  Polarization: Polarization is a more about phenomenon by which vibrations of light this topic waves are restricted in a particular plane.  Plane polarized light: A polarized light which vibrates in single plane, perpendicular to direction of Polarization of propagation is known as plane polarized light light. Light waves are electromagnetic in nature i.e., they are composed of electric as well as magnetic waves. In unpolarized light, the orientation of these fields is along various planes. Passing unpolarized light through a polarizing filter the vibrations may be restricted in a single plane and such light is known as plane polarized light.

3λ  P is a bright point when path difference = , ………, 2  2n  1  2  Fringe width =

Dλ a

 Width of central bright fringe =  Intensity distribution pattern:

2Dλ a

There is no gain or loss of energy in interference or diffraction, which is consistent with the principle of conservation of energy. Energy is only redistributed in these phenomena. Resolving the power of microscopes and astronomical telescopes:  Resolving power: The resolving power of an optical instrument is measured by its ability to differentiate two lines or points in an object. The greater the resolving power, the smaller the minimum distance between two lines or points that can still be distinguished.

Brewster’s law:  Brewster’s law gives the exact angle of incidence for different materials and wavelengths of light for which reflected ray is completely polarized.  The law states that maximum polarization of an unpolarized light ray occurs when the refractive index of the reflector material is equal to the tangent of the angle of incidence of the ray. Uses of plane polarized light and Polaroids: Uses of plane polarized light:  Used to reduce glare of bright light - sunglass, windows of aeroplane etc.  In number of digital applications, digit, letters, figures are formed by LCD through polarization.  Polarized LASER beam is used for playing CD.  Used for photography.  Used for recording and reproducing 3D pictures.  Used for study of asymmetries in molecules and crystals.  Used for determination of refractive index using Brewster’s law. Polaroids: Polaroid is a large sheet of synthetic material packed with tiny herapathite crystals in cellulose acetate with their optic axis parallel so that it transmits light vibrations only in one direction. The layers of the crystal are mounted between two glass sheets for protection. This acts as a sheet of the polarizer and known as polaroids. It can be used as a polarizer as well as an analyzer.

72 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. A converging lens is used to form image on a screen. When the lower half of the lens is covered by an opaque screen then, the correct statements are: (1) Intensity of Image will decrease. (2) Intensity of Image will increase. (3) Complete Image will be formed. (4) Half the Image will disappear. Choose the correct answer from the options given below. (1) (A), (B) and (C) Only (2) (A), (C) and (D) Only (3) (A) and (C) Only (4) (B) and (C) Only  (CUET 2022, 7th June) 2. In Young’s double slit experiment, the ratio of slit widths is 4 : 1. The intensity ratio in the interference pattern would be: (1) 1 : 2 (2) 1 : 3 (3) 9 : 1 (4) 1 : 9  (CUET 2022, 7th June) 3. Huygen’s wave theory of light could not explain the phenomenon of: (1) interference (2) diffraction (3) rectilinear propagation of light (4) photoelectric effect (CUET 2022, 7th June) 4. Two sources are producing waves as given in the options below. Which two sources are called coherent? (1) of same velocity (2) of equal wavelength (3) having a constant phase difference (4) having wavefront of same shape (CUET 2022, 7th June) 5. The radii of curvature of the faces of a double convex lens of focal length 12 cm made up of glass (m = 1.5) are 10 cm and ‘p’ cm respectively. The value of p is: (1) 10 cm (2) 15 cm (3) 30 cm (4) 20 cm 6. The essential condition for total internal reflection to occur is: (1) angle of incidence i > critical angle ic (2) angle of incidence i > critical angle ic (3) angle of incidence i = critical angle ic (4) angle of incidence i = 90° (CUET 2022, 7th June) 7. A student has drawn the following courses of rays through a glass prism. The one which represents the position of minimum derivation is: (1)

(2)

(3)

(4)

 (CUET 2022, 7th June) 8. The Brewster’s angle for air to glass interface when an unpolarised light is incident on a glass (n = 1.5), such that reflected and refracted rays are perpendicular to each other, is nearly: (1) iB = 30° (2) iB = 45° (3) iB = 57° (4) iB = 47°  (CUET 2022, 7th August) 9. Nowadays optical fibres are extensively used for transmitting audio and video signals through long distances. The optical fibres work on (1) Double refraction (2) Refraction (3) Total internal reflection (4) Reflection  (CUET 2022, 5th August) 10. A star is seen using a telescope whose objective lens has a diameter of 250 cm. The wavelength of light coming from the star is 500 nm. The limit of resolution of telescope is (1) 1.2 × 10–7 radians (2) 2.4 × 10–7 radians –7 (3) 1.5 × 10 radians (4) 3.9 × 10–7 radians  (CUET 2022, 6th August) 11. Match List-I and List-II List-I

List-II

A. Simple microscope

I. Parabolic mirrors are used

B. Compound microscope

II. Only one convex lens is used

C. Telescope (reflecting)

III. Objective of large focal length and aperture is used

D. Telescope IV. Objective of small focal (refracting) length and aperture is used Choose the correct answer from the options given below: (1) A-III, B-II, C-I, D-IV (2) A-II, B-III, C-IV, D-I (3) A-I, B-III, C-II, D-IV (4) A-II, B-IV, C-I, D-III  (CUET 2022, 6th August) 12. A convex mirror of focal length f1 is placed at a distance d from a convex lens of focal length f2. A beam of light rays coming from infinity falling on convex lens-convex mirror combination and returns back to infinity. Distance d is (1) (f1 + f2) (2) (f1 – f2) (3) (f2 – 2f1) (4) (f2 + 2f1) 13. A short pulse of white light is incident from air to a glass slab at normal incidence, after travelling through the slab, the first colour to emerge is (1) Blue (2) Green (3) Violet (4) Red  (CUET 2022, 6th August) 14.

73

OPTICS



There are four light rays incident on a right angled prism. The refractive index of prism material for the rays is 1.43, 1.39, 1.38, 1.21 respectively. The ray that suffers total internal reflection is (1) I (2) II (3) III (4) IV  (CUET 2022, 6th August) 15. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of the two liquids L1 or L2 having refractive indices n1 and n2 respectively (n2 > n1 > 1). The lens will diverge a parallel beam of light if it is filled with (1) Air inside and placed in air (2) Air inside and immersed in L1 (3) L1 inside and immersed in L2 (4) L2 inside and immersed in L1  (CUET 2022, 6th August) 16. Two prisms made of material of same refractive index and having angles of the prism 65° and 60° respectively are combined to form a glass block as shown. If for any ray passing across the block the minimum angle of deviation in 3° then refractive index for the material of the prism is

(1) 1.5 (2) 2.0 (3) 1.6 (4) 1.35 17. In Young’s double slit experiment when light of wavelength 600 nm is used the fringe width is 0.1 mm. Calculate the fringe width when the entire apparatus is immersed in a 4 medium of refractive index by keeping the source of 3 light in air (1) 0.01 mm (2) 1.05 mm (3) 0.075 mm (4) 0.0075 mm  (CUET 2022, 8th August) 18. Among the following which one is not due to the phenomenon of total internal reflection? (1) Mirage (2) Spectacular brilliance of diamond (3) Optical fibres (4) Dispersion of light 19. An object is placed 12 cm in front of a convex mirror of focal length 18 cm, the value of the image distance is (1) 18 cm (2) 12 cm (3) 7.2 cm (4) 6 cm  (CUET 2022, 8th August) 20. A screen is placed 90 cm from an object. The image of the object on the screen formed by a convex lens at two different locations separated by 20 cm. The focal length of the lens is (1) 10.5 cm (2) 15.4 cm (3) 18.5 cm (4) 21.4 cm  (CUET 2022, 8th August) 21. The resolution of a telescope where objective has a diameter of 200 inches when the light of wavelength 4000 Å is coming from the star is (1) 9.606 × 10–8 rad (2) 96.06 × 10–8 rad –8 (3) 0.9606 × 10 rad (4) 9.006 × 10–4 rad  (CUET 2022, 8th August)

22. The Brewster’s angle for glass of refractive index 1.5 is (1) 57.3° (2) 45° (3) 60° (4) 67.3°  (CUET 2022, 8th August) 23. A plane wavefront is incident on a thin convex lens. The nature of wavefront immediately after refraction is (1) Plane wavefront (2) Spherical and converging (3) Spherical and diverging (4) Cylindrical and diverging  (CUET 2022, 8th August) 24. The reddish appearance of the sun during sunrise and sunset is due to (1) Wavelength of red light is more and hence its scattering is more. (2) Red light has higher energy and hence reaches our eyes even after travelling long distances. (3) Due to its longer wavelength, its scattering is least hence travels longer and reach our eyes. (4) Because it does not follow Rayleigh scattering.  (CUET 2022, 8th August) 25. A parallel beam of light ray parallel to the x-axis is incident on a parabolic reflecting surface x = 2by2 as shown in the figure. After reflecting it passes through focal point F. What is the focal length of the reflecting surface?

(1)

1 2b

(2)

1 8b

(3)

1 4b

(4)

1 b

26. A beam of light consisting of two wavelengths 5000 Å and 6000 Å is used to obtain interference fringes in Young’s double slit experiment. The least distance from the central maxima, here the bright fringes due to both wavelengths coincide, will be (If separation between slits = 1 mm and separation between slits and screen is 1 m) (1) 4 mm (2) 3 mm (3) 2 mm (4) 1mm 27. Which of the following statements are true for refraction of white light through a glass prism at minimum deviation position of prism. (i) The angle of prism becomes zero. (ii) Angle, of refraction at first refracting surface r1 is equal to angle of refraction at second refracting surface r2. (iii) The refracted ray inside the prism is parallel to the base of the prism. (iv) Angle of emergence becomes 90° (v) Angle of incidence is equal to angle of emergence. Choose the correct answer from the options given below: (1) (ii), (iii) and (v) only (b) (i), (iii) and (iv) only (3) (ii), (iii) and (iv) only (4) (i), (iv) and (v) only 28. Match List-I with List-II List-I

List-II

(A) Convex mirror

(I) Accommodation

(B)  Total internal reflection

(II) Reflecting type

74 Oswaal CUET (UG) Chapterwise Question Bank (C) Ciliary muscles

(III) Optical fiber

(D) Cassegrain telescope

(IV) Used as a rear view mirror

Choose the correct answer from the options given below: (1) (A) - (IV), (B) – (I), (C) – (III), (D) – (II) (2) (A) - (IV), (B) – (III), (C) – (I), (D) – (II) (3) (A) - (II), (B) – (III), (C) – (I), (D) – (IV) (4) (A) - (I), (B) – (III), (C) – (IV), (D) – (II)  (CUET 2022, 23rd August) 29. A converging beam of rays is incident on a concave lens. After passing through the lens, the rays converge at a distance of 15 cm from the lens on the other side. If the lens is removed, the converging point of rays decreases by 5 cm. The focal length of lens is (1) –10 cm (2) –20 cm (3) –30 cm (4) –5 cm 30. A beam of light converges at a point M. Now a concave lens of focal length 32 cm is placed in the path of convergent beam 24 cm from M. Now the beam will converge at: (1) 56 cm from lens (2) 8 cm from lens (3) +96 cm from lens (4) –32 cm from lens  (CUET 2022, 23rd August)

[B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): A convex mirror cannot form real images. Reason (R): Convex mirror converges the parallel rays that are incident on it. 2. Assertion (A): The focal length of a concave mirror is f and an object is placed at a distance x from the focus. The f magnification produced by the mirror is . x Reason (R): Magnification = size of image / size of object. 1 1 1 3. Assertion (A): The mirror formula + = is valid for v u f mirrors of small aperture. Reason (R): Laws of reflection of light is valid for only plane surface and not for large spherical surface. 4. Assertion (A): A diamond of refractive index

6 is

immersed in a liquid of refractive index 3 . If light travels from diamond to liquid, total internal reflection will take place when angle of incidence is 30°. Reason (R): m = 1sin C, where m is the refractive index of diamond with respect to the liquid. 5. Assertion (A): A double convex air bubble is formed within a glass slab. The air bubble behaves like a converging lens. Reason (R): Refractive index of glass is more that the refractive index of air.

PHYSICS

6. Assertion (A): According to Huygen’s theory no backward wavefront is possible. Reason (R): Amplitude of secondary wavelets is proportional to (1 + cosq), where q is the angle between the ray at the point of consideration and direction of secondary wavelet. 7. Assertion (A): Wavefront emitted by a point source of light in an isotropic medium is spherical. Reason (R): Isotropic medium has same refractive index in all directions. 8. Assertion (A): When a light wave travels from rarer to denser medium, its speed decreases. Due to this reduction of speed the energy carried by the light wave reduces. Reason (R): Energy of wave is proportional to the frequency. 9. Assertion (A): No interference pattern is detected when two coherent sources are too close to each other. Reason (R): The fringe width is inversely proportional to the distance between the two slits. 10. Assertion (A): Diffraction takes place with all types of waves. Reason (R): Diffraction is perceptible when the wavelength of the wave is comparable to the dimension of the diffracting device.

[C] COMPETENCY BASED QUESTIONS

I. Based on following passage answer questions from 1-5. It is useful to design lenses of derived focal length using surface of suitable radii of curvature. Lens maker formula helps in finding various unknowns and is widely applied.  (CUET 2022, 5th August) 1. The radii of curvature of faces of a double concave lens are 3 20 cm and 30 cm. If refractive index of glass is , then 2 the focal length of the lens will be: (1) 24 cm (2) –24 cm (3) 120 cm (4) –120 cm 2. Find the focal length of an equiconvex lens whose each face has radius of curvature 12 cm. Given refractive index 3 of glass is . 2 (1) 0 (2) Infinite (3) –12 cm (4) 12 cm 3. A convex lens has 30 cm focal length in air. What is its 4 focal length in a liquid whose refractive index is ? 3 3 (Given refractive index glass is ) 2 (1) –120 cm (2) +120 cm (3) 60 cm (4) –60 cm 4. If a convex lens made of a material of refractive index n2 is immersed in a liquid of refractive index n1 such that n1 = n2, the incident beam of light will emerge as shown (A) (B)



75

OPTICS

(3)

7. Path difference between the waves meeting at point P is given by xd 2xD 1xd (3) (4) 2D D d th 8. In YDSE 20 bright fringe is obtained 20 mm distance from central bright point. The distance between the 5th bright fringe and 3rd dark fringe is

(1) (4)

5. (1) (3) II.



Find the power of a concave lens whose focal length is 40 cm. –2.5D (2) +2.5D +2.5 × 10–2D (4) –2.5 ×10–2D Based on following passage answer questions from 6-10. In Young’s double slit experiment as shown in figure, interferencse of light waves was observed on the screen. Thomas Young made two pinholes S1 and S2 (very close to each other) on an opaque screen in front of parent source S. S1 and S2 behaved as coherent sources, and produced interference pattern on the screen which has alternate bright fringe and dark fringes. This experiment proved Huygen’s wave theory of Light.

6. In YDSE experiment position of bright fringes and dark fringes is given by λD λd (1) xn = n (2) xn = n d D (3) xn = nλ

(4) xn = n

λ

xD d

(2)

(1) 1 mm (2) 2 mm (3) 2.5 mm (4) 8 mm 9. In YDSE the distance between two consecutive bright fringes is given by 3 mm. What would be the fringe width if YDSE is performed in a medium of refractive index 1.2? (1) 3 mm (2) 1.5 mm (3) 2.5 mm (4) 0.5 mm λ . 10. In YDSE as shown, the path difference SS1 − SS 2 = 3 This causes the shift in position of central bright point. The new position of central bright point is

(1) −

λD d

(3) −3

(2) −2

λD d

(4) −

λD d

1 λD 3 d

d

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (3)

3. (4)

4. (3)

5. (2)

6. (1)

7. (3)

8. (3)

9. (3)

10. (2)

11. (4)

12. (3)

13. (4)

14. (1)

15. (4)

16. (2)

17. (3)

18. (4)

19. (3)

20. (4)

21. (1)

22. (1)

23. (2)

24. (3)

25. (2)

26. (2)

27. (1)

28. (2)

29. (3)

30. (3)

8. (4)

9. (1)

10. (2)

8. (3)

9. (3)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (3)

2. (1)

3. (3)

4. (4)

5. (4)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (4)

3. (2)

4. (4)

5. (1)

6. (1)

7. (2)

ANSWERS WITH EXPLANATIONS [A] MULTIPLE CHOICE QUESTIONS

1. Option (3) is correct. Explanation: The complete image will be formed. Since one half of the lens is covered, the intensity will be lower.

2. Option (3) is correct. Explanation: We know, I ∝ a I1 a \ = 1 I2 a2

76 Oswaal CUET (UG) Chapterwise Question Bank

 I + I2 I max =  1  I − I I min 2  1   =    

 I1 +1 I2   I1 − 1  I2 

Or,

I max I min

Or,

 4 +1 I max =   I min  4 −1

Or,

I max = (3)2 I min

\

I max =9:1 I min

   

2

2

2

3. Option (4) is correct. Explanation: Only particle nature of light can explain photoelectric effect. 4. Option (3) is correct. Explanation: It is essential that two coherent waves should have constant phase difference. 5. Option (2) is correct. Explanation: 1 1  1 = ( µ − 1)  −  f  R1 R2  Or,

 1 1 1 = (1.5 − 1)  +  12  10 p 

Or,

 1 1 1 = 0.5 ×  +  12  10 p 

Or, Or,

1 1 1 − = 10 p 6 1 1 1 = − p 6 10

1 1 = p 15 \ p = 15 cm 6. Option (1) is correct. Explanation: When i > ic, then there is no refraction. Total light is reflected back. 7. Option (3) is correct. Explanation: For minimum deviation, angle of incidence = angle of emergence and the ray inside the prism is parallel to the base. Or,

8. Option (3) is correct. Explanation: From Brewster’s law tan(iB) = m iB = tan–1 = tan–1(1.5)  57° 9. Option (3) is correct. Explanation: Refractive Index of inner core of optical fibre is optically denser than the outer cladding. Hence repeated total internal reflection takes place between these two media and thus the signal is sent through long distance. 10. Option (2) is correct. Explanation: Limit of resolution of telescope is given by,



dq =

PHYSICS

1.22λ D

1.22 × 500 × 10−9 250 × 10−2 \ dq = 2.4 × 10–7 radians 11. Option (4) is correct. Explanation: In simple microscope, microscope is made of only one convex lens. In compound microscope, objective has small focal length and aperture since smaller the focal length higher the magnification. In reflecting type telescope parabolic mirrors are used to avoid aberration and to make the instrument light. Image formed is also brighter. In refracting type telescope objective has large focal length and aperture since large focal length increases the magnifying power and large aperture help in collecting large amount of light coming from the object so that a bright image is formed. 12. Option (3) is correct. Explanation: Or,

dq =

So, d = f2 – R = f2 – 2f1 13. Option (4) is correct. Explanation: Since red has the longest wavelength, its velocity will be fastest. So red will emerge first. 14. Option (1) is correct. Explanation: For total internal reflection, the angle of incidence should be greater than critical angle. Angle of incidence for all rays on the face opposite to the 900 angle of the prism is 450. 1 Critical angle = ic = sin −1   µ

For ray I,

 1  ic = sin −1  =  44.37°  1.43 

For ray II,

 1  ic = sin −1  =  46°  1.39 

For ray III,

 1  ic = sin −1  =  46.43°  1.38 

 1  ic = sin −1  =  55.73°  1.21  So, only for ray I, the angle of incidence is greater than critical angle. So, only this ray will suffer total internal reflection. 15. Option (4) is correct. Explanation: For ray IV,

77

OPTICS

A concave lens will behave like a diverging lens if the refractive index of the material of the lens is more than that of the surrounding medium. So, here, the liquid L2, having comparatively higher refractive index should be inside. The lens should be immersed in liquid L1, having comparatively lower refractive index. 16. Option (2) is correct. Explanation: Minimum angle of deviation = (μ – 1)A1 – (μ – 1)A2 or, (μ – 1)65 – (μ – 1) 60 = 3 or, 5μ – 5 = 3  5μ = 8  μ = 1.6 17. Option (3) is correct. Explanation: λD bair = d

bmed =

So,

bmed =



λD βair = µd µ 0.1 = 0.075 mm 4 3

18. Option (4) is correct. Explanation: Dispersion is the phenomenon of splitting of visible light into its component colours. Dispersion of light is caused due the change of speed of light ray of each wavelength by a different amount. This is not due to total internal reflection. 19. Option (3) is correct. 1 1 1 Explanation: + = v u f

1 1 1 = − 18 v 12

or, \ 20. Option (4) is correct.

1 30 = v 12 × 18 v = 7.2 cm

D2 − x2 Explanation: f= 4D Where, D = distance between object and screen = 90 cm x = Distance between two positions of lens = 20 cm \

f=

[902 − 202 ] [4 × 90]

 21.4 cm 21. Option (1) is correct. 1.22λ a l = 4000 Å = 4000 × 10–10 m a = 200 inch 200 × 2.54 = m 100

Explanation: Resolution = 

Resolution =

22. Option (1) is correct. Explanation: Brewster’s angle or,  The value is close to 57.3°. 23. Option (2) is correct. Explanation:

24. Option (3) is correct. Explanation: During sunrise and sunset, the sunrays have to pass through greater distance through the atmosphere. The blue light gets scattered the most; the red colour is less scattered and reaches us. Thus, we find reddish colour of the sun during sunrise and sunset. 25. Option (2) is correct. Explanation: Given, x = 2by 2 So,

y2 =

x 2b

Comparing with the general equation of parabola y 2 = 4ax, (where a = focal length) x = 4ax 2b So,

a=

1 8b

Focal length of the reflecting surface is

1 . 8b

26. Option (2) is correct. Explanation: If nth bright fringe of 5000 Å wavelength coincides with (n – 1)th bright fringe of 6000 Å then n × 5000 = (n – 1)6000 or, 5n = 6n – 6  n=6 Least distance from central maxima D 1 = nλ1 = 6 × 5000 × 10−10 × d 1 × 10−3 –4 = 30 × 10 m = 3 mm 27. Option (1) is correct. Explanation: For minimum deviation,

1.22λ a

1.22 × 4000 × 10−10 200 × 2.54 100 = 9.6 × 10–8 rad

iB = tan–1(m) iB = tan–1(1.5) iB = 56.3°

=

i=e r1 = r 2

78 Oswaal CUET (UG) Chapterwise Question Bank The refracted ray inside the prism. is parallel to the base of the prism. 28. Option (2) is correct. Explanation: Parabolic reflector is used in Cassegrain telescope. Optical fibre utilises the total internal reflection phenomenon. Ciliary muscles control the curvature of eye lens. This is known as accommodation. Convex mirror is used as rear-view mirror since it has large field of view and produces erect image. 29. Option (3) is correct. Explanation:

O is the virtual object. I is the real image. Using lens formula, 1 1 1 − = v u f 1 1 1 = − 15 10 f  f = –30 cm 30. Option (3) is correct. Explanation: Given focal length of concave lens = 32 cm object is 24 from point M; U = 24 cm Or,





1 1 1 = − f v u



−1 1 1 = − 32 v 24



1 1 −1 1 = − + 32 24 v 12



 

1 −3 + 4 1 = = 96 96 v v = 96 cm from lens

[B] ASSERTION REASON QUESTIONS

1. Option (3) is correct. Explanation: Convex mirror always form virtual image. So, the assertion is true. Parallel rays incident on convex mirror do not actually meet. They get reflected in such a manner that their extensions meet at a point. So, the reason is false. 2. Option (1) is correct. Explanation: u= f + x Using mirror formula,

1 1 1 + = v u f

PHYSICS

1 1 1 − = − v f +x f Or,

v= −

f ( f + x)

x size of image v  Magnification (m) = u size of object = m

f ( f + x) x

×

1 = f +x

f x

So, both the assertion and reason are true and reason is the correct explanation of assertion. 3. Option (3) is correct. Explanation: The mirror formula is derived under the consideration that the incident rays are paraxial which means that the rays lie very close to the principal axis. Hence the mirror aperture is considered to be small. So, the assertion is true. Laws of reflection are valid for any surface plane or spherical. Hence the reason is false. 4. Option (4) is correct. Explanation: Refractive index of diamond with respect to the liquid is 6 = 2 So, critical angle for the diamond-liquid 3  1   pair of media is sin −1   = 45 . For total internal reflection,  2 angle of incidence should be greater than critical angle. Since, angle of incidence is 30°, total internal reflection cannot take place. So, the assertion is false but the reason is true. 5. Option (4) is correct. Explanation: Speed of light is slower in glass compared to that in air. Hence, the refractive index of glass is more than that of air. So, the reason is true. When a double convex air bubble is formed within a glass slab, the refractive index of the medium of the bubble is less than the refractive index of the surrounding medium. Hence, the lens will not behave like a converging lens. It will behave like a diverging lens. So, the assertion is false. 6. Option (1) is correct. Explanation: According to Huygen’s theory each and every point on a wavefront is the source of secondary wavelets. Secondary wavelets do not proceed backward. So, the assertion is true. Kirchhoff explained that amplitude of secondary wavelets is proportional to (1 + cosq), where q is the angle between the ray at the point of consideration and direction of secondary wavelets. In the backward direction q = 180°; so 1 + cosq = 0; so, the secondary wavelets do not proceed backward. Hence, reason is also true and is correct explanation of assertion. 7. Option (1) is correct. Explanation: If a medium has same refractive index at every point in all directions, then the wavefront obtained from a point source in such a medium is spherical since wave travels in all direction with same speed. Such a medium is known as isotropic medium. 8. Option (4) is correct. Explanation: When a light wave travels from rarer to denser medium, its speed decreases. But this reduction of speed does not imply the loss of energy carried by the light wave. So, the assertion is false. Energy of wave is proportional to the frequency of the wave which remains same in very medium. Hence there is no loss of energy. So, the reason is true.

79

OPTICS

9. Option (1) is correct. Explanation: No interference pattern is detected when two coherent sources are too close to each other. The assertion is true. 1 Fringe width is proportional to . When d becomes too small, d the fringe width becomes too large. So, the reason is also true and is correct explanation of assertion. 10. Option (2) is correct. Explanation: Diffraction is spreading of waves around obstacle. It takes place with all types of waves (mechanical, non-mechanical, transverse, longitudinal) and with very small moving particles (atom, neutron, electron etc.) which show wave like property. So, the assertion is true. Diffraction is perceptible when the wavelength of the wave is comparable to the dimension of the diffracting device. The reason is also true but reason is not the correct explanation of reason.

[C] COMPETENCY BASED QUESTIONS 1. Option (2) is correct. Explanation: Using lens maker’s formula Or,

1 1  1 = ( µ − 1)  −  R R f 2   1

no reflection, no refraction. The light rays will pass through without any deviation. 5. Option (1) is correct. Explanation: P =

1 1 (in meter) = f (−0.4)

So, f = –2.5 D 6. Option (1) is correct. Explanation: For bright fringe to be formed nλ D d (n = 0, ± 1, ± 2, . . . .) For dark fringe to be formed



x=



x=

(2n + 1)λ D 2d

1   n + λD 2  = d (n = 0, ± 1, ± 2, . . . .) 7. Option (2) is correct. Explanation:

1  1  3  1 =  − 1 − −  f  2  20 30 

1  −3 − 2  1 1 =  = − f 2  60  24 So, f  = –24 cm 2. Option (4) is correct. Explanation: Using lens maker’s formula

Or,

Or,

1 1  1 = ( µ − 1)  −  f  R1 R2 

1  3   1  1  1 2 =  − 1  −  −   = × f  2  12  12   2 12

1 1  2  =  ×  f  2   12  So, f  = 12 cm 3. Option (2) is correct. Explanation: Using Lens Maker’s Formula,

Or,

(When the lens is in air)

1 1  1 = ( µ g − 1)  −  R R f air 2   1

 µg  1 1  − 1 −  =  f liquid  µl   R1 R2  (When the lens is in the liquid) Taking ratio, 1 f liquid 3 4  = = µl and = 2 4=  µg  1 2 3 f air  8 So, fliquid = 4 × 30 = 120 cm 4. Option (4) is correct. Explanation: The refractive index of the material of the lens and that of the surrounding medium being same, there will be

1

Path difference = S2N In DS1S2N,

sinq =

In KPO,

tanq =

θ being small, tanq ≈ sinq

S2 N d x D

x S N = 2 D d





S2N =

xd D

8. Option (3) is correct. Explanation: For 20th bright fringe 



So,

 λD  20   = 20 mm  d  λD = 1 mm d

For 5th bright fringe, For 3rd dark fringe,

 λD  X5 = 5    d  = 5 × 1 = 5mm  λD  X’3 = 5    2d 

80 Oswaal CUET (UG) Chapterwise Question Bank  5  λ D  5 =    =  2  d  2 = 2.5 mm So, the distance between the 5th bright fringe and 3rd dark fringe is 5 – 2.5 = 2.5 mm 9. Option (3) is correct. Explanation: Fringe width = Distance between two consecutive bright fringe = 3mm (given) When the system is immersed in a medium



Then b′ =

β air 3 = = 2.5 mm µ 1.22

10. Option (4) is correct. Explanation:

SS1 – SS2 =



S2P – S1P =

λ

(given)

3

xd D

Total path difference xd λ nλ (For maxima) + = D 3 For central maxima, putting n = 0





24

PHYSICS

=

1 1 =0 − 15 10 x0 = −

λD 3d

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

7

  Revision Notes

DUAL NATURE OF MATTER AND RADIATION Scan to know more about this topic

CRUX POINTS: Photoelectric effect Photoelectric effect: When electron emission occurs by illumination of metal by the light of suitable frequency, it is known as photoelectric emission. Here, emitted Photoelectric electrons are called photoelectrons. When effect light falls on the metal surface, free electrons absorb energy from light and if this energy Scan to know is more than the work function of metal, the more about electron escapes from the surface. this topic Hertz and Lenard’s observations:  In 1888, Lenard observed that when ultraviolet light falls on zinc metal, metal becomes positively charged. With the discovery of electrons, it was established Photoelectric effect that this is due to emission of electrons. The current produced by these photoelectrons is called photoelectric current.  The frequency of the incident light below which emission of electrons does not take place, is called its threshold frequency.  Intensity of light has linear relationship with photoelectric current at a potential higher than the stopping potential.  For a given frequency of the incident radiation, the stopping potential is independent of its intensity.

Einstein’s photoelectric equation - Particle nature of light  Wave theory failed to explain the photoelectric effect:  According to wave theory, higher amplitude means higher energy but experiments show that even larger amplitude (higher intensity) of light below threshold frequency does not show photoelectric effect.  According to wave theory, same intensity of different colour should have same energy but experiments show that energy depends upon frequency, not on amplitude.  According to wave theory, wavefront should take some time to give energy to electron but experimentally, it was found that ejection of electron is instantaneous.  In 1900, Max Planck stated that electromagnetic energy can be emitted only in quantized form. E = hv (where, k is Planck’s constant).  Based upon this postulate, Einstein established quantum theory of radiation and was able to explain photoelectric phenomenon.  It states that light energy packets are known as photons (Particle nature of light).  In photoelectric effect, an electron absorbs a quantum of energy (E = hv) of radiation. If this absorbed quantum of energy exceeds, the minimum energy needed for the electron to escape from the metal surface (work function ϕ0), the electron is emitted with maximum kinetic energy. K.E. = hv – ϕ0 (where, ϕ0 is the work function of the metal).  At stopping potential, the kinetic energy of the ejected electron is zero. Below this potential, the electrons cannot be ejected. Hence, maximum kinetic energy of an electron is calculated by K ⋅ Emax = eV0 where, V0 is the stopping potential.  Work function of metal, ϕ0 =hv0 where, v0 is the cut-off frequency or threshold frequency.  Maximum speed of emitted photoelectrons can be calculated as 2KE max m Matter waves - Wave nature of particles  de-Broglie’s postulate is based upon the symmetry of nature. If radiation has dual nature, then matter should also have dual nature.  According to his hypothesis, moving particles of matter should display wave nature under suitable conditions. He named the wave as matter wave. It is the third type of wave. It is different from mechanical wave and electromagnetic wave.  de-Broglie proposed that the wavelength l known as deBroglie wavelength is associated with momentum of particle p h as l = p h Hence, de-Broglie’s wavelength of particle, l = mv  Wavelength of electron-wave: Kinetic energy of electron at potential V is K.E. = eV. vmax =

 Photoemission current starts only at certain minimum frequency of light known as threshold frequency. Below this frequency, photoemission does not take place in spite of the increase in the intensity of light.

0, i.e.,

onto

Here,

c

ejected hc

h

eV0

Second Level

Trace the Mind Map

First Level

V0



E



hc

Electrons come out with definite KE.

i.e., E

the surface

Kmax

hc

Third Level

Dual Nature of Radiation

Ra

D

nature Interference and diffraction can be explained by wave nature is small wavelength, it . s linear momentum are called ''photons'' . hc h Energy of each h E/c



n atio r el

Davisson Germer Experiment

Hertz and Lenard's Observation

Photoelectric Effect

de-B ro g

Kmax

h mv

c

m

h

Kmax

hc f=tan –1 e

• The wave nature of electron was verified by observing the diffraction pattern. • Inter-atomic space of Nickel was used as single slit. Intensity of scattered electron beam was plotted for different accelerating voltage. Intensity peak was observed at 54V at a scattering angle  = 50°. • Electron wavelength calculated 0.165 nm. It tally with the theoretically calculated wavelength.

Matter waves

l Nature ua n & of Ma tio tt dia

Contribution

a

e.g.,

.

l ie

h p

82 Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

er

83

DUAL NATURE OF MATTER AND RADIATION

Putting this value of kinetic energy in de-Broglie wavelength equation, h λe = 2meV By putting the value of mass of electron (m = 9.1093837 × 10–31 kg), its charge (e = 1.60217663 × 10–19C, and Planck’s Davisson-Germer experiment: Experimental arrangement:

constant (h = 6.62607015 × 10–34 J/Hz, it becomes

λe =

1.227 nm V

From this equation, we can infer that the wavelength of the particle is inversely proportional to the mass of the particle and its velocity. Hence, heavier particles have shorter wavelengths. Scan to know more about this topic

Davisson and Germer experiment Scan to know more about this topic

de Broglie wavelengths

 During this experiment the wave nature of electron was verified by observing the diffraction pattern.  In the experimental set up, inter-atomic space of Nickel was taken as single slit. Scattered electron beam was collected by

movable collector and its intensity was plotted.  The experiment was performed by varying the accelerating voltage from 44 V to 68 V. A strong peak of the intensity of scattered electrons appeared for accelerating voltage 54 V at a scattering angle q = 50°.

    This experiment concluded that the electrons exhibit diffraction when they are scattered from crystals whose atoms are spaced appropriately.  Electron wavelength was calculated using diffraction formula of wave optics. Putting experimental values as the 1st dark diffraction angle 50° and the inter-atomic space of nickel d = 0.91Å, obtained l = 0.165 nm.

    Wavelength obtained from theoretical calculation using de 1.227   Broglie formula  λ e = nm  is 0.165 nm by putting V   V = 54V.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. The de Broglie wavelength of an electron having kinetic energy 56 eV is: (1) 0.022 nm (2) 16.4 Å (3) 0.164 nm (4) 0.0164 nm  (CUET 2023, 7th June) 2. The wavelength of matter wave is independent of:

(1) 3. (1) (3)  4.

momentum (2) charge (3) velocity (4) mass According to Einstein’s photoelectric equation: Kmax = hν – φo (2) hν = Kmax – φo Kmax = hν + φo (4) φo = Kmax + hν (where all symbols have their usual meaning) (CUET 2023, 7th June) A graph for variation of photo current with anode potential is given for different intensities of incident radiation:

84 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS



Choose the correct order of anode intensities. (1) I1 > I2 > I3 (2) I1 = I2 = I3 (3) I1 < I2 < I3 (4) I1 < I2 > I3  (CUET 2023, 5th August) 5. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The deBroglie wavelength of particle varies cyclically between two values (l1), (l2) with (l1 > l2) then, the correct statements from the following are: (i) The particle could be moving in a circular orbit with origin as centre. (ii) The particle could be moving in an elliptical orbit with origin as its focus. (iii) When the de-Broglie wavelength is l1, the particle is nearer to origin then when its value is l2. (iv) When the de Broglie wavelength is l2, the particle is nearer to the origin then when its value is l1. (1) (i) and (iv) only (2) (ii) and (iv) only (3) (ii) and (iii) only (4) (iii) and (i) only  (CUET 2022, 5th August) 6. In a typical set-up of photoelectric effect, the original source of blue light is replaced with a source of red light. Which of the following phenomenon is observed? (1) Photoelectric current increases (2) Photoelectric current decreases (3) Stopping potential increases (4) Stopping potential decreases (CUET 2022, 5th August) 7. An electron is accelerated by 100 volt potential difference. Its wavelength will be comparable to that of: (1) Infrared radiator (2) Microwave (3) X-rays (4) Radio waves  (CUET 2022, 5th August) 8. If le lp and la, are the de -Broglie wavelengths of electron, proton and α-particle, respectively. If all are accelerated by same potential, then (1) lp < l < le (2) le < l < lp (3) le < lp < l (4) l < lp < le  (CUET 2022, 6th August) 9. An electron is accelerated to potential V. If mass of electron is m = 9.1 × 10–31 kg, its charge is e = 1.6 × 10–19 C, then de -Broglie wavelength of electron is 125 1.227 nm (1) λ = (2) λ = nm φ V (3) λ =

eh nm 2mV

(4) λ =

3.6 Vm φ

 (CUET 2022, 6th August) 10. Consider the given graph between the stopping potential and frequency of incident light in an experiment of photoelectric effect. The slope of the graph gives: (where symbols have their used meanings)

 11.

(1)  12. (1) (2) (3) (4)  13. (1)

h e

φ (4) eh e (CUET 2022, 8th August) The Kinetic Energy of an electron having de Broglie wavelength triple, then the de Broglie wavelength associated with it becomes: λ (2) λ 3 (3) (4) 3l 3 (CUET 2022, 23th August) Identify the correct statement according to Einstein’s picture of photoelectric effect: Maximum kinetic energy of electrons depends linearly on frequency of incident radiation. Maximum kinetic energy of electrons depends linearly on intensity of incident radiation. The photoelectric current is independent of intensity of incident radiation. Intensity of incident radiation is directly proportional to the frequency of radiation. (CUET 2022, 23rd August) A proton and a triton are accelerated from rest through a potential V. The ratio of their de Broglie wavelength is: 1:1 (2) 1 : 2

(1) h

(2)

(3)

(3) 1: 2 (4) 3 :1  (CUET 2022, 26th August) 14. In the experimental study of photoelectric effect, a graph of stopping potential versus frequency of incident radiation for two metals A and B is plotted and identical slopes for both are obtained with different cut off frequencies (vDB > nOA). So. h (1) Slope for all metals is same = . Here work function of A e > work function of B h (2) Slope for all metals is same = . Here work function of B e > work function of A (3) Slope for all metals is same = h. Here work function of A > work function of B (4) Slope for all metals is same = h. Here work function of B > work function of A  (CUET 2022, 26th August) 15. Photoelectric emission occurs only when the incident light has more than a certain minimum. (1) Intensity (2) Angle of Incidence (3) Speed (4) Frequency  (CUET 2022, 26th August) 16. The work function for a metal surface is 4.14eV. The threshold wavelength for this metal surface is (1) 4125Å (2) 2062.5Å

(3) 3000Å (4) 6000Å

85

DUAL NATURE OF MATTER AND RADIATION

17. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly (1) 1.2 nm (2) 1.2 × 10–3 nm –6  (3) 1.2 × 10 nm (4) 1.2 × 10 nm [NCERT Exemp. Q. 11.2, Page 68]  18. Kinetic energy of electrons emitted in photoelectric effect is (1) directly proportional to the intensity of incident light. (2) inversely proportional to the intensity of incident light. (3) independent of the intensity of incident light. (4) independent of the frequency of light. 19. Threshold wavelength of a photoelectric emission from a material is 600 nm. Which of the following illuminating source will emit photoelectrons? (1) 400 W, infrared lamp (2) 10 W, ultraviolet lamp (3) 100 W, ultraviolet lamp (4) Both (B) & (C) 20. Photoelectrons emitted from a metal have (1) different speeds starting from zero to certain maximum. (2) same kinetic energy. (3) same frequency. (4) Both (2) & (3) 21. At stopping potential, the kinetic energy of emitted photoelectron is (1) minimum. (2) maximum. (3) zero. (4) cannot be predicted 22. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to

(3) H0 (4) H–1/2 [NCERT Exemp. Q. 11.1, Page 68]  23. An electron is moving with an initial velocity v = vo i  and is in a magnetic field B = B j . Then, its de Broglie (1) H  

(2) H1/2

o

(1) (2) (3) (4)  24.

wavelength remains constant. increases with time. decreases with time. increases and decreases periodically. [NCERT Exemp. Q. 11.6, Page 69] An electron (mass m ) with an initial velocity = v v0i ( v0 > 0 )  is in an electric field E = − E i (E = constant > 0). Its de 0

(1) (3) 25. (1) (2) (3) (4) 

0

Broglie wavelength at time t is given by  eE t  λ0 (2) λ 0 1 + 0   eE 0 t   mv0  1 +  m v 0  l0 (4) l0t Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then, no electrons will be emitted as only photons can emit electrons. electrons can be emitted but all with an energy, E0. electrons can be emitted with any energy, with a maximum of E0– f (f is the work function). electron can be emitted with energy, with a maximum of E0. [NCERT Exemp. Q. 11.3, Page 68]

26. Consider the figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of q that (1) will be larger than the earlier value (2)  will be the same as the earlier value (3) will be less than the earlier value (4)  will depend on the target.

 27. Two particles A1 and A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then, (1) their momenta are the same. (2) their energies are the same. (3) energy of A1 is more than the energy of A2. (4) Both (1) and (3)

[NCERT Exemp. Q. 11.4, Page 69] 28. The momentum of a photon is 3.3 × 10–29 kg ms–1. Its frequency is (1) 3 × 103 Hz (2) 6 × 103 Hz (3) 7 × 1012 Hz (4) 1.5 × 1013 Hz

86 Oswaal CUET (UG) Chapterwise Question Bank

[C] COMPETENCY BASED QUESTIONS

29. (1) (2) (3) (4) 30.

The rest mass of photon is Infinite Zero More than zero Equal to that of an electron Photon in motion has mass h hv (1) 0 (2) (3) 2 v c

PHYSICS

(4) hv

[B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies. Reason (R): The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal. 2. Assertion (A): Energy of moving photon varies inversely as the wavelength. Reason (R): Energy of the particle = Mass × (Speed of light)2 3. Assertion (A): If the frequency of an incident photon is twice the threshold frequency then two electrons are emitted. Reason (R): According to Einstein’s equation, Energy of photon = Maximum kinetic energy of photo electron work function. 4. Assertion (A): de Broglie wavelength is significant for microscopic particles. Reason (R): de Broglie wavelength is inversely proportional to the mass of a particle when velocity is kept constant. 5. Assertion (A): If a proton and electron are moving with same velocity, then wavelength of de Broglie wave associated with electron is longer than that associated with proton. Reason (R): The wavelength of de Broglie wave associated with a moving particle is inversely proportional to its mass. 6. Assertion (A): Photoelectric effect demonstrates wave nature of radiation. Reason (R): The number of photoelectrons ejected is proportional to the intensity of incident radiation. 7. Assertion (A): Photosensitivity of a material increases when its work function decreases. Reason (R): Work function = hv0. 8. Assertion (A): In photoelectric emission, all the emitted photoelectrons have the same kinetic energy. Reason (R): In photoelectric emission, incident photons transfer its whole energy. 9. Assertion (A): Electron microscope offers higher resolution than optical microscope. Reason (R): Electron beam as a wave has much smaller wavelength compared to visible light. 10. Assertion (A): According to de Broglie, a wave is associated with each moving particle which is called matter waves. Statement II: Heavier particles have larger wavelengths.

I. Based on following passage answer questions from 1-5. Photocell: A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. It is connected to the external circuit having a high-tension battery B and micro ammeter (mA) as shown in the Figure.



A part of the bulb is left clean for the light to enter it. When light of suitable wavelength falls on the emitter C, photoelectrons are emitted. These photoelectrons are drawn to the collector A. Photocurrent of the order of a few microamperes can be normally obtained from a photo cell. A photocell converts a change in intensity of illumination into a change in photocurrent. 1. Photocell is an application of (1) thermoelectric effect (2) photoelectric effect (3) photoresistive effect (4) None of the above 2. Photosensitive material should be connected to (1) Negative terminal of battery (2) Positive terminal of battery (3) Either positive or negative terminal of battery (4) ground 3. Which of the following statement is true? (1) The photocell is totally painted black. (2) A part of the photocell is left clean. (3) The photo cell is completely transparent. (4) A part of the photocell is made black. 4. The photocurrent generated is in the order of (1) Ampere (2) Milliampere (3) Microampere (4) None of these 5. A photocell converts a change in_______ of incident light into a change in _________. (1) Intensity, photo-voltage (2) Wavelength, photo-voltage (3) Frequency, photo-current (4) Intensity, photo-current II. Based on following passage answer questions from 6-10. A photon is the smallest discrete amount or quantum of electromagnetic radiation. It is the basic unit of all light. According to Einstein, photons have energy equal to their frequency times Planck’s constant. The intensity of the light corresponds to the number of photons. The basic properties of photons are: (i) They have zero mass and rest energy. They only exist as moving particles. (ii) They are elementary particles despite lacking rest mass. (iii) They have no electric charge.

87

DUAL NATURE OF MATTER AND RADIATION

(iv) They are stable. (v) They carry energy and momentum which are dependent on the frequency. (vi) They can have interactions with other particles such as electrons, such as the Compton effect. (vii) They can be destroyed or created by many natural processes, for instance when radiation is absorbed or emitted. (viii) In free space, they travel at the speed of light. 6. Photons have energy equal to their frequency times (1) Rydberg’s constant (2) Planck’s constant (3) Avogadro’s constant (4) Boltzmann constant 7. The intensity of the light corresponds to (1) Number of photons (2) Speed of photons (3) Energy of photons (4) Frequency of photons

8. (1) (3) 9. (1) (2) (3) (4) 10. (1) (2) (3) (4)

Charge of a photon is – e (2) + e 0 (4) None of the above Which of the following statements is wrong? Photons only exist as moving particles. Photons carry energy and momentum. Mass of photon is equal to the mass of electron. Photons travel at the speed of light. Which of the following statements is wrong? Photons can neither be destroyed nor created. Photons can have interactions with other particles. Photons are elementary particles. Photon is the basic unit of all light.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (2)

3. (1)

4. (3)

5. (2)

6. (4)

7. (3)

8. (4)

9. (2)

10. (2)

11. (3)

12. (1)

13. (4)

14. (2)

15. (4)

16. (3)

17. (2)

18. (3)

19. (4)

20. (1)

21. (3)

22. (4)

23. (1)

24. (1)

25. (4)

26. (3)

27. (4)

28. (4)

29. (2)

30. (3)

8. (4)

9. (1)

10. (3)

8. (3)

9. (3)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (4)

4. (1)

5. (1)

6. (2)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (1)

3. (2)

4. (3)

5. (4)

6. (2)

7. (1)

ANSWERS WITH EXPLANATIONS [A] MULTIPLE CHOICE QUESTIONS 1. Option (3) is correct. h Explanation: λ = 2mE Or,

λ=

6.6 × 10−34

2 × 9.1 × 10−31 × 56 × 1.6 × 10−19 \ l = 0.164 nm 2. Option (2) is correct. h h = Explanation: l = mv p So, it depends on mass, velocity and momentum. But, it is independent of charge. 3. Option (1) is correct. Explanation: Einstein theorised that when a photon falls on the surface of a metal, the energy of the photon is transferred to the electron. A part of this energy is used to remove the electron from the metal and the rest is given to the ejected electron as kinetic energy. Kmax = hν – φo Where all symbols have their usual meaning. 4. Option (3) is correct. Explanation: For a fixed collector plate potential, as intensity of incident light increases, emission of photoelectrons increases and the photo current increases. As collector plate potential increases, photo current also increases and ultimately reaches a constant value. This current does not increase further unless the intensity is increased.

5. Option (2) is correct. Explanation: de-Broglie wavelength of a moving particle, h λ= mv Since, l1 > l2 so v1 < v2 and the velocity of the particle becomes greater at a point nearer to the origin on an elliptical orbit. 6. Option (4) is correct. Explanation: The photon of blue light has higher energy as compared to red light. So, the red light emits electrons of reduced kinetic energy than that of blue light and hence the stopping potential also decreases. 7. Option (3) is correct. Explanation: de -Broglie wavelength of electron

le =



le =



1.23 nm V

1.23 100 le = 0.123 nm

88 Oswaal CUET (UG) Chapterwise Question Bank So, the wavelength is comparable to the wavelength of X-rays. Range of wavelength of X-rays is (0.01 – 10) nm. 8. Option (4) is correct. h 2mqV

Explanation: We know that l = With V remaining constant

1 mq

l Since, (mq)e < (mq)p < (mq)a So, l < lp < le 9. Option (2) is correct. Explanation: h l= = p

=

h 2meV 6.62 × 10−34

2 × 9.1 × 10−31 × 1.6 × 10−19 × V

6.62 × 10−9 1.227 = nm m= V 5.4 V 10. Option (2) is correct. Explanation: eV0 = h– f0  h  ϕ  V0 =   v −  0  e  e  h It is an equation of straight line with slope e. 11. Option (3) is correct. Explanation: Initially, or,

l1 =

h (2mk )

Finally,

l2 =

h (6mk )

\

λ1 = λ2





\

l2 =

3K K

λ1 3

12. Option (1) is correct. Explanation: KEmax = hv – φ0 This is an equation of straight line in the form y = mx – C. Hence, the variation is linear. 13. Option (4) is correct. Explanation: As

l=

h 2mqV

m2 λ1 = m1 λ2 Here m1 = mass of proton m2 = mass of triton As mass of triton = 3 (mass of proton) ⇒ m2 = 3m1 ⇒

⇒

λ1 = λ2

3m1 = 3 :1 m1

14. Option (2) is correct. Explanation: By Einstein’s photoelectric equation hn = W + EK

⇒

PHYSICS

hn – W = eV0

hν hν 0 − e e On comparing it with equation of straight line, y = mx + c

⇒

V0 =



m=

h (constant) e

hν 0 e As threshold frequency of metal B is more than metal A, therefore, work function of metal B is more than metal A. 15. Option (4) is correct. Explanation: Threshold frequency is the minimum frequency which is required to initiate photoelectric emission. 16. Option (3) is correct. hc = 4.14eV = Explanation: Work function



c= −

λ0

\

−34

6.6 × 10 × 3 × 108 λ0 = 4.14 × 1.6 × 10−19 −7 = 2.989 × 10 ≈ 3000Å

17. Option (2) is correct. Explanation: Energy of the photon must be equal to the binding energy of proton So, energy of photon = 1 MeV = 106 × 1.6 × 10–19 J hc 6.63 × 10−34 × 3 × 108 6.63 × 3 λ = = = × 10−26 +13 E 1.6 × 10−13 1.60 19.89 = × 10−13 =12.4 × 10−13 =1.24 × 101 × 10−13 1.60 = 1.24 × 10−3 × 10−9 = 1.24 × 10−3 nm 18. Option (3) is correct. Explanation: KE = hv – f So, KE is independent of intensity of incident light. 19. Option (4) is correct. Explanation: The incident wavelength should be less than threshold wavelength for photoelectric emission. IR has a wavelength of more than 600 nm. UV has a wavelength of less than 600 nm. So, photoelectrons emitted when illuminated by UV lamp either 100 W or 10 W. 20. Option (1) is correct. Explanation: When a photon strikes a metal surface, the surface electrons come out with maximum speed and maximum kinetic energy. But if the electron emission takes place from inner side of metal, then some energy of the electron is lost due to collision with other electrons and so their speed becomes less. So, ultimately the electrons come out with different speeds. 21. Option (3) is correct. Explanation: At stopping potential, the kinetic energy of the emitted photoelectron is zero because at this potential the frequency of the photon is equal to the threshold frequency of the electron. So, the photon provide just the enough energy to the electron to come out of the metal surface without any kinetic energy. 22. Option (4) is correct. Explanation: Velocity v, of freely falling body after falling from a height H, will be: v = 2 gH h We know that de Broglie wavelength, λ = p

89

DUAL NATURE OF MATTER AND RADIATION

= λ

h h = mv m 2 gH

h, m, and g are constant h 1 is constant ⇒ λ ∝ ⇒ λ ∝ H −1/ 2 \ m 2g H 23. Option (1) is correct.

 Explanation:= üüüüü v v= oi Magnetic force on moving electron = −e[v i × B j ] = −ev B  k o

o

o

o

j

o

So, the force is perpendicular to v and B both. So, the magnitude of v or mv (p = mv, momentum) will not change. So, the de Broglie wavelength remains same. 24. Option (1) is correct. Explanation: According to de Broglie theory, the wavelength of de Broglie wave is given by

λ=

h h = = p mv

25. Option (4) is correct. Explanation: If a beam of electrons of having energy E0 is incident on a metal surface kept in an evacuated chamber.

h 2mE

As initial velocity of the electron is v0i , the initial de Broglie wavelength of electron, h λ= mv0 Electrostatic force on electron in electric field is,   Fe = eE0i −eE = −e  − E0i  =    F eE0i Acceleration of electron, = a = m m Velocity of the electron after time t, ˆ  eE i    eE  v= v0i +  0  t = v + 0 t i  m   0 m      eE  ⇒ v= v0 1 + 0 i mv 0  

The electrons can be emitted with maximum energy E0 (due to elastic collision) and with any energy less than E0, when part of incident energy of electron is used in liberating the electrons from the surface of metal. 26. Option (3) is correct. Explanation: In Davisson-Germer experiment, the de Broglie wavelength of diffracted beam of electrons 12.27 λd = Å ....(i) V 

V is the applied voltage. If there is a maxima of the diffracted electrons at an angle q, then 2d sinq = l...(ii) From equation (i), as the applied voltage in this experiment increases, the wavelength ld decreases in turn sinq or q decreases by relation (ii). Hence, verifies the option (3). 27. Option (4) is correct. Explanation: h We know the de Broglie wavelength, λ = mv where, mv = p (momentum) of the particle ⇒ But we can express wavelength λ= Hence, p ∝

P1 λ = =1 ⇒ p1 =p2 P2 λ Thus, their moment is same. Then.

Also, = E



h mv

=

1 2 1 2m = mv mv 2 2 m

1 m 2v 2 1 p 2 = 2 m 2 m

As p is constant, E ∝

h h mv0 ⇒ λ = =   eE  eE    m v0 1 + 0 t   1 + 0 t  mv 0    mv0    = ⇒ λ

1 p λ ⇒ 1 =2 λ p2 λ1

But particles have the same de-Broglie wavelength. (l1 = l2 = l).

de Broglie wavelength associated with electron at time t is λ=

λ0 h = , as λ 0 mv0  eE0  t 1 +  mv0 

h h ⇒ p= p λ

∴

1 m

E1 m2 = < 1 ⇒ E1 < E2 E2 m1

28. Option (4) is correct. hv Explanation: p= c Or,

v=

pc h

90 Oswaal CUET (UG) Chapterwise Question Bank 3.3 × 10−29 × 3 × 108 6.6 × 10−34 \ v = 1.5 × 1013 Hz 29. Option (2) is correct. Explanation: The rest mass of a photon is zero. 30. Option (3) is correct. hv Explanation: E= mc 2 Or,

v=

\

m=

hv c2

[B] ASSERTION REASON QUESTIONS

1. Option (1) is correct. Explanation: Photoelectrons produced by monochromatic light have different velocities and hence different energies. Actually all the electrons do not occupy the same level of energy. So, electrons coming out from different levels have different velocities and hence different energies. The electrons coming out from inside the metal surface, face collisions with the other atoms in the metal. So, energies become different. Hence, both the assertion and reason are true and reason is the correct explanation of assertion. 2. Option (2) is correct. hc Explanation: E = hv = l 1 . Hence, the assertion is correct. l According to Einstein, E = mc2s. So, the reason is also correct. But reason is not the correct explanation of assertion. 3. Option (4) is correct. Explanation: One photon can eject one electron only. Hence, the assertion is incorrect. Einstein’s equation related to photoelectric effect is, Energy of photon = Maximum kinetic energy of photo electron - work function. Hence, the reason is correct. 4. Option (1) is correct. h Explanation: de-Broglie wavelength, l = , h and v mv 1 remaining constant,   . m So, as the mass of the particle becomes smaller and smaller the de-Broglie wavelength of the particle becomes more and more significant. 5. Option (1) is correct. h Explanation: l = mv Both proton and electron are moving with same velocity. So, 1 l∝ . So, the reason is correct. Mass of proton > mass of m electron. So, wavelength of electron > wavelength of proton. So, the assertion is also correct and reason is the correct explanation of assertion. 6. Option (2) is correct. Explanation: Photoelectric effect does not demonstrate particle nature of radiation. It demonstrates the wave nature of radiation. So, assertion is false. So, E ∝

PHYSICS

The number of photoelectrons ejected is not proportional to the frequency of radiation but intensity of incident radiation. So, reason is also true. 7. Option (1) is correct. Explanation: Work function = hv0. Where h = Planck’s constant and v0 is the threshold frequency. Hence the reason is correct. Increase of photosensitivity means it is easier to eject photoelectrons from a material, i.e., energy requirement is less. It means that the work function of the material is less. So, photosensitivity of a material increases when its work function decreases. The assertion is also correct but the reason is not the correct explanation of assertion. 8. Option (4) is correct. Explanation: In photoelectric emission, incident photons transfer its whole energy. So, the reason is correct. All the electrons do not come out from the surface. Electrons coming out from inside the metal surface, face collisions with the other atoms in the metal and in the process lose of energy. So, they cannot have same kinetic energy. Hence, the assertion is incorrect. 9. Option (1) is correct. Explanation: Resolution is inversely proportional to the wavelength. Wavelength of visible light is about 5000 times larger compared to the wavelength of electron beam. So, electron microscope offers much higher resolution than an optical microscope. Resolution of optical microscope is less than 200 nm. Resolution of electron microscope may go upto 0.1 nm. 10. Option (3) is correct. Explanation: According to de Broglie hypothesis, moving particles of matter should display wave nature under suitable conditions. This wave is named as matter wave. Hence the assertion is true. de Broglie’s wavelength of particle is given h by l = . mv So,

l m.s

So, heavier particles have shorter wavelengths. Hence, the reason is false.

[C] COMPETENCY BASED QUESTIONS

1. Option (2) is correct. Explanation: Photocell is a technological application of the photoelectric effect. 2. Option (1) is correct. Explanation: Photosensitive material used as emitter should be connected to negative terminal of the battery so that the emitted electrons are repelled by emitter and collected by collector. 3. Option (2) is correct. Explanation: A part of the bulb is left clean for the light to enter in it. 4. Option (3) is correct. Explanation: Photocurrent of the order of a few microampere can be normally obtained from a photo cell. 5. Option (4) is correct. Explanation: A photocell converts a change in the intensity of illumination into a change in photocurrent.

DUAL NATURE OF MATTER AND RADIATION

6. Option (2) is correct. Explanation: Photons have energy equal to their frequency times Planck’s constant. E = hv. 7. Option (1) is correct. Explanation: Intensity of light depends on photons per second per unit area. 8. Option (3) is correct. Explanation: Photon is an uncharged particle.

24

91 9. Option (3) is correct. Explanation: Photons have zero rest mass. 10. Option (1) is correct. Explanation: Photons can be destroyed or created by many natural processes, for instance when radiation is absorbed or emitted.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

8

ATOM AND NUCLEI All positively charged particles are together at one location at centre called nucleus. Size of the nucleus is calculated to be about 10–15 m to 10–14 m. Size of one atom is of the order of 10–10 m. Rutherford’s model of atom:  Rutherford overturned Thomson’s model in 1911 with alpha-particle scattering experiment in which he demonstrated that the atom has a tiny and heavy, positively charged nucleus.  In his model, the atom is made up of a central charge, the nucleus, surrounded by orbiting electrons in circular orbits at the speed of light.

  Revision Notes

Alpha - particle scattering experiment  Experimental set-up: Vacuum

Gold foil

Screen

Source of α-particles

Radioactive element 214 83 Bi was taken as collimated source of alpha particles. Thin foil of highly malleable heavy metal gold was used as target. The detector was made from ZnS.  Alpha-particle trajectory:

Target nucleus

d

Impact parameter: It is the perpendicular distance  between the direction of the given a-particle and the centre of the nucleus. It is represented by ‘b.’  Distance of closest approach: It is the distance between centre of nucleus and the a-particle where it stops and reflects back. It is represented by ‘d.’ This distance gives an approximation of nucleus size. d=

1 2eZe 4πε 0 K

Experimental observations: Most of the a-particles passed  roughly in a straight line (within 1°) without deviation. A very small number of a-particles were deflected. (1 out of 8000)  Conclusions: Most of the space in the atom is mostly empty (only 0.14% scatters more than 1°).

Limitations of Rutherford model:   It could not explain the stability of Scan to know the atom. Rutherford applied classical more about this topic electromagnetism. So, the electrons are attracted to the nucleus. As the electron orbits, it accelerates. Because of the acceleration, the electrons should fall inward toward the nucleus. But, that would Nuclei make any atom highly unstable! But this does not happen since atoms in the natural world are stable.  It could not explain the nature of energy spectrum. Bohr model  Bohr presented his atomic model with the three following postulates:  An electron can revolve in certain stable orbits without emission of radiant energy. These orbits are called stationary states of the atom.  Electron revolves around nucleus only in those orbits for which the angular momentum is the integral multiple of h/2π, where, h is Planck’s constant. Hence, angular momentum, L = nh/2π  An electron may make a transition from one of its specified non-radiating orbit to another of lower energy. When it does so, a photon is radiated having energy equal to energy difference between initial and final state.



First Level



Second Level



Trace the Mind Map

Third Level

s.

s

. Doesn't explain Zeeman's and Stark's effect .

s ck ba

where, n is an integer

.

s

w ra D

Atoms

th

n

n

Energy Level

, it releases

ne r

gy

Hydrogen Spectrum

Bohr Model

En

t ia en Pot

It is also known as principal Quantum Number.

.

spectra

Contribution

Most of – particles passed undeviated or with a small angle –

In 1898, J.J Thomson proposed the first model of atom known as plum-pudding model. In 1911, Rutherford prepared planetary model of atom. In 1913, Niels Bohr prepared a model of Hydrogen atom based on quantum theory.

En

erg

y

Total E nerg y Kin etic

when passed components of different wavelength appear.

ATOM AND NUCLEI

93

lE

on Velocity of electr it in n th Bohr's orb

e.g.,

• •

e.g.,

(A)

s (N)

(Z)

• An electron. • Moderate penetrating power. • Less ionizing power. • Decay of one  particle causes change in atomic no. by 1 and and mass no. by 4.

Isot op

Properties of    rays

Nuclear fusion

gave

Mass-Energy Relation

Nuclei

Contribution

• High frequency e.m. wave. • Highest penetrating power. • Least ionizing power. • No. change in atomic and mass no.

e

es

-particl

es ton Iso

Isobars

(Z) (A)

• Doubly ionized Helium ion. • Least penetrating power. • Highest ionizing power. • Decay of one  particle causes decrease in atomic no. by 2 and and mass no. by 4.

e.g.,

e.g.,

• •

•

1n 1932, James Chadwick discovered neutrons – elementary particles devoid of any electrical charge.

t1/2 = 0.693/ N = N0(1/2)n (n = t/t1/2)

e.g.,

B.E.

B.E.

n

n

0.693

 = 1/  = ty 2

L a w of

ns ity

r a di

ty

a

B.E/A

neutron

n

First Level

200 MeV

Second Level

Third Level

–dN/dt = N N = N0e–t

r

Trace the Mind Map

oac tiv i

us

Nuclear Radi

me r Volu clea u N

Disintegration of heavy elements into comparatively higher elements by emission of   and  radiations.

Radioactivity

Nuclear fission

Binding Energy

Mass Defect

E

showed



e rD



a cl e Nu 

• • Common name

94 Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

95

ATOM AND NUCLEI

 When electron jumps from lower energy level to higher energy level it absorbs energy.  Emission atomic spectra: When Scan to know more about an excited electron comes down from this topic higher energy level to lower energy level, radiation of particular wavelength is emitted. This atomic spectra is known as emission spectra.  Absorption atomic spectra: If Binding energy and fission electrons are excited, they absorb the radiation corresponding to their emission spectra and transit from lower to higher energy level. Black lines appear in the same places of the emission spectra. This type of spectra is known as absorption spectra.  Hydrogen spectrum:  Lyman series: Lines emitted by Scan to know transitions of the electron from an outer orbit more about this topic of quantum number n > 1 to the n = 1 orbit.

Increasing energy of orbits

n=3 n=2 n=1

A photon is emitted with energy E=hv

Speed of electron in nth orbit, vn=

1 e2 1 ⋅ n 4πε 0 ( h / 2π )

Radius of nth orbit, 2

1 1 1  = R  2 − 2 ; n = 2, 3, 4, 5, .... λ 1 n 

 n 2   h  4πε 0 rn =     2  m   2π  e For innermost orbit n = 1; the value of r1 is known as Bohr’s radius a0. h 2ε a0 = 02 = 5.29 × 10−11 m ≈ 0.53Å πme Total energy of an electron in nth orbit, En =

1 1   1 =R  2 − 2  n =3, 4, 5, .... λ n  2

−13.6 eV n2

Negative value shows that electron is bound to nucleus.  Limitation of Bohr’s atomic model:  Bohr’s model is for hydrogenic Scan to know atoms. It does not hold true for a multimore about electron model. this topic  This model fails to explain the effect of magnetic field on the spectra of atoms (Zeeman effect).  This model fails to explain the effect Bohr’s atomic model of electric field on the spectra of atom (Stark effect). Energy levels & Hydrogen spectrum Level n= 5th excited state, n=6 th 4th excited state, n=5 3nd excited state, n=4 2 excited state, n=3 1st excited state,

Energy (eV) lonization

n=2

This is in UV range. Spectral lines of Hydrogen Longest wavelength = 1216 Å atom Shortest wavelength = 912 Å  Balmer series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n > 2 to the n = 2 orbit.

0.00 –0.38 –0.54 –0.85 –1.51 –3.40

Longest wavelength = 6566.4 Å Shortest wavelength = 3648 Å This is in visible range.  Paschen series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n > 3 to the n = 3 orbit. 1 1 1  =R  2 − 2  ; n = 4, 5, 6, .... λ 3 n  Longest wavelength = 18761.14 Å Shortest wavelength = 8208 Å This is in far infrared region.  Brackett series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n > 4 to the n = 4 orbit. 1 1   1 =R  2 − 2  ; n =5, 6, .... n λ 4   Longest wavelength = 40533.33 Å Shortest wavelength = 14592 Å This is in far infrared region.  Pfund series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n > 4 to the n = 4 orbit. 1 1  1 =R  2 − 2  ; n = 6, 7, 8 .... λ n  5

n=1

Ground State –13.60

Longest wavelength = 74618.1 Å Shortest wavelength = 22800 Å This is in far infrared region. Composition and size of nucleus  The atomic nucleus is the small, dense region consisting of protons (positively charged particles) and neutrons (electrically neutral particles) at the centre of an atom.

96 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

 The diameter of the nucleus is in the range of 1.70 fm for Hydrogen to about 11.7 fm for Uranium.  A nucleus of mass number A has a radius R = R0 A1/3 where, R0 = 1.2 × 10–15 m  Nuclear matter density = 2.3 × 1017 kg m–3  The composition of a nucleus is described using the following terms and symbols: Z = Atomic number = Number of protons (equal to the number of electrons) N = Neutron number = Number of neutrons A = Atomic mass number = (Z + N) = Total number of protons and neutrons An atom is represented as ZA X , where X = Symbol of element, A = Mass number, Z = Atomic number. Atomic mass  The atomic mass is the mass of an atom. Atomic mass is often expressed in the unified atomic mass unit (u).  1 u is defined as 1⁄12 of the mass of a carbon-12 atom at rest in its ground state. The protons and neutrons of the nucleus account for nearly all of the total mass of atom, with the electrons and nuclear binding energy making minor contributions. Thus, the numeric value of the atomic mass when expressed in u is nearly the same value as the mass number.

 Have very high ionizing power.  Have very short penetration range.  Have very little biological damaging power when externally exposed to.  Properties of β-rays:  These are fast moving electrons or positrons.  In b– decay, an electron is emitted from nucleus along with antineutrino. [There is no electron in the nucleus. A neutron splits into a proton and an electron.]  After b+-decay, a nucleus transforms into a different nucleus.

Isotopes, Isobars, Isotones  Isotopes : The atoms of an element having the same atomic number (z is same) but different atomic mass numbers (due to the different number of neutrons) are said to be isotopes.

 Example: 106C → 105 B + 10 e +  Deflected by electric and magnetic field.  Velocity is about 90% of the velocity of light.  Affect photographic plates.  Have moderate ionizing power.  Have moderate penetration range.  Have some biological damaging power when externally exposed to.  Properties of γ-rays:  When an excited nucleus makes a transition from lower energy state, γ radiation is emitted.  These are photons of very short wavelength (10-11m to 10-13m).  Carry no charge.  Neither deflected by magnetic field nor by electric field.  Velocity is equal to that of the velocity of light.  Affect photographic plates.  Have very weak ionizing power.  Have very long penetration range.  Have very high biological damaging power when externally exposed to. Radioactive decay law  dN/dt = – λN (λ = disintegration constant) Integrating both sides, N(t) = N0e–lt  Alternative form of radioactive decay: R = R0e–lt (R0 = Radioactive decay rate at t = 0 and R = radioactive decay rate at any subsequent time t) SI unit of radioactive decay is becquerel. 1 becquerel is 1 decay per second. 1 curie = 3.7 × 1010 Bq  Half life time (T1/2): It is the time period in which both N and R reduce to half of initial values. ln 2 T1/2 = = 0.693/λ λ

Examples:

12 6

C,

13 6

C,

14 6

C

Isobars : The atoms of different elements having the same  mass number but different atomic numbers are said to be isobars. 58 58 Examples: 26 Fe , 27 Ni Isotones: The atoms of different elements having different  mass numbers and atomic numbers such that their difference is same are said to be isotones. It means they have same number of neutrons. Examples: 168 O , 157 N , 146 C Radioactivity – alpha, beta, and gamma particles/rays, and their properties Radioactivity: When atoms of some elements become very  heavy, neutrons are unable to bind and some nucleons keep on leaving the nucleus. These elements are known as radioactive elements and the process of spontaneous ejection of nucleons and radiations is known as radioactivity.  There are 3 types of radioactive decay in nature:  α-decay: In this decay, α-particles eject out.  β-decay: In this decay, electrons or positrons eject out.  γ-decay: In this decay, high energy photons are emitted.  Properties of α-rays:  These are doubly ionized helium atoms.  After α-decay, a nucleus transforms into a different nucleus. ZA X → ZA −−42Y + 24 He  Mass number of daughter nucleus reduces by 4 and atomic number reduces by 2.

 Example:

298 92

4 U → 234 90Th + 2 He  Deflected by electric and magnetic field.  Velocity is about 10% of the velocity of light.  Affect photographic plates.

A A 0 Z X → Z +1Y + −1 e + ν  Mass number of daughter nucleus remains same. Atomic number increases by 1.

Example: 1523 P → 1632 S + −10 e + ν  b+ decay, a positron is emitted from nucleus along with neutrino. [In this process, a proton splits into a neutron and a positron.]  After b+-decay, a nucleus transforms into a different nucleus. A A 0 Z X → Z −1Y + 1 e + ν  Mass number of daughter nucleus remains same. Atomic number decreases by 1.

97

ATOM AND NUCLEI

Mean life (τ): It is the time at which both N and R reduce to  e–1 of initial values. τ = 1/λ  Relation between half life time and mean life: ln 2 T1/2 = = τln2 λ  So, from the formula of half life time and radioactive decay, N/N0 = (1/2)n Mass-energy relation  Einstein’s theory of special relativity established the fact that mass is another form of energy. Also, it is possible to convert mass-energy into other forms of energy. According to Einstein, mass-energy equivalence relation is given by, E = mc2 Where E is the equivalent energy of mass m. c is the velocity of light in vacuum.  This enables the understanding of nuclear masses and the interaction of nuclei with each other (fusion, fission, etc.) and evolution of energy. According to law of conservation of energy, the initial energy and final energy of such reactions are same provided the mass-energy is taken into consideration.

 Higher the binding energy per nucleons, more stable is the element.  Most of the atoms where atomic mass number are in the range 30 < A < 200, the binding energy per nucleon is fairly constant and quite high. It is maximum for A = 56 about 8.75 MeV.  For A < 30 and A > 170, binding energy per nucleon is quite low. Nuclear fission and fusion  If a nucleus of lower binding energy is converted into higher binding energy, then energy is released. There are two methods of converting lower binding energy nucleons into higher binding energy nucleons.  Fission: When a heavy nucleus (low binding energy per nucleon) is broken into two lighter nuclei (higher binding energy per nucleon), the process is known as fission. In this process huge amount of energy is released. Example: When a neutron bombards a Uranium target, the Uranium nucleus breaks into two nearly equal size fragments releasing huge amount of energy. 1 0

n + 235 92 U →

236 92

U

→

144 56

Ba + 3689 Kr + 3 01n

1 0

Mass defect  The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect. ΔM = [Zmp + (A – Z )mn] – M Binding energy per nucleon and its variation with mass number  Nuclear binding energy of a nucleus is that quantity of energy which when given to nucleus, its nucleons will become free and will leave the nucleus.  Binding energy, Eb = ΔMc2  Binding energy per nucleon (Eb /A): The average energy required to remove an individual nucleon from nucleus is the binding energy per nucleon. It is the ratio of the binding energy Eb of a nucleus to the number of the nucleons.  Variation of Binding energy per nucleon (Eb /A) with mass number (A):

236 99 1 n + 235 → 133 92 U → 92 U 51 Ba + 41 Nb + 4 0 n Energy released in the fission reaction of nuclei like Uranium is in the order of 200 MeV per fissioning nucleus.  Fusion: When two light nuclei (of low binding energy per nucleon) fuse and form one nucleus of higher binding energy per nucleon, the process is known as fusion. In this process huge amount of energy is released.

Example: 12 H + 12 H → 13 H + 11H + 4.03 MeV 2 H + 12 H → 23 He + n + 3.27 MeV 1 Fusion process gives more energy than fission process. Source of energy of stars is fusion process. Nuclear force:  This high binding energy per nucleon counters the repulsive force between protons and bind both protons and neutrons into the tiny nuclear volume.  The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. But it is a short-range force.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. Match List - I with List - II List - I List -II (Scientist) (Phenomenon) (A) Rutherford (I) Nuclear Model of atom (B) A.H. Becquerel (II) Plum-Pudding Model of atom (C) Niels Bohr (III) Radioactivity

(D) J.J. Thomson (IV) Hydrogen atom model Choose the correct answer from the options given below: (1) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (2) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (3) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)  (CUET 2023, 7th June)

98 Oswaal CUET (UG) Chapterwise Question Bank 2. The correct sequence A, B, C, D, E with reference to atomic spectral series is : 1 1   1 (A) Lyman series = R 2 − 2  λ  A n  1 1   1 (B) Balmer series = R 2 − 2  λ n  B 1 1   1 (C) Pfund series = R 2 − 2  λ C n   1 1   1 (D) Brackett series = R 2 − 2  λ n  D 1 1   1 (E) Paschen series = R 2 − 2  λ n  E Choose the correct answer from the options given below: (1) (A), (B), (C), (D), (E) (2) (B), (A), (E), (C), (D) (3) (A), (B), (E), (D), (C) (4) (E), (D), (C), (B), (A)  (CUET 2022, 7th June) 3. A radioactive isotope has half-life of ‘K’ years. How long will it take, so that activity reduces to 6.25% of its original value? (1) 5 K (2) 4 K (3) 3 K (4) 2 K  (CUET 2022, 7th June) 4. What will be the half-life of a radioactive isotope whose th 1   of initial amount remains unchanged after 2 hours?  16  (1) 60 min (2) 45 min (3) 30 min (4) 15 min  (CUET 2023, 7th June) 5. The variation of the number of undecayed nuclei with time is best represented by (Where N0 -initial number of atoms) (1)



(2)

(3)



(4)

(CUET 2022, 5th August) Bohr model of atom is valid for: Only hydrogen atom Only one electron atom Only one electron atoms and ions All the atoms (CUET 2022, 5th August) Which of the following is correct for nuclear forces? Nuclear forces are much weaker than Coulomb force. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometre. (3) Nuclear forces are charge dependent. (4) Nuclear forces are long range forces.  (CUET 2023, 5th August) 8. The study of hydrogen atom spectrum has mainly five series. Choose the correct sequence of these series in increasing order of their shortest wavelength.  6. (1) (2) (3) (4) 7. (1) (2)

PHYSICS

(i) Lyman Series (ii) Pfund Series (iii) Paschen Series (iv) Brackett Series (v) Balmer Series Choose the correct answer from the options given below: (1) (ii), (i), (iv), (iii), (v) (2) (ii), (iii), (iv), (v), (i) (3) (i), (v), (iii), (iv), (ii) (4) (i), (iii), (v), (ii), (iv)  (CUET 2022, 6th August) 9. Two nuclei have mass number in the ratio 1:27, then the ratio of their radii and densities should be respectively (1) 1:1 and 1:3 (2) 1:9 and 1:1 (3) 1:3 and 1:1 (4) 1:1 and 1:9  (CUET 2022, 6th August) 10. The Bohr model for the spectra of an H-atom A. Will not be applicable to hydrogen in the molecular form. B. Will not be applicable as it is for a He-atom. C. Is valid only at room temperature. D. Predicts continuous as well as discrete spectral lines. Choose the correct answer from the options given below: (1) A, B only (2) A, B, C only (3) A, B, D only (4) B, C, D only  (CUET 2022, 6th August) 11. In a Geiger Marsden experiment, an alpha particle of energy 6 meV hits a nucleus of Z = 72. The distance of closest approach nearly is (1) 35mm (2) 45fm (3) 25fm (4) 30fm  (CUET 2022, 8th August) 12. Arrange the following in increasing order of binding energy per nucleon A. Li B. Fe C. O D. N E. H Choose the correct answer from the options given below: (1) E < A < D < C < B (2) A < B < D < E < C (3) C < B < D < A < E (4) B < D < C < A < E  (CUET 2022, 8th August) 13. Arrange the following lines of Balmer series in the increasing order of their wavelength A. Hδ B. Hb C. H∞ D. Hl E. Ha Choose the correct answer from the options given below: (1) C, A, D, B, E (3) E, B, D, A, C (2) E, B, D, C, A (4) C, A, B, D, E  (CUET 2022, 8th August) 14. For the radioactive material half-life is 10 minutes. If initially there are 800 number of nuclei, the time taken (in minutes) for the disintegration of 600 nuclei is (1) 40 minutes (2) 10 minutes (3) 20 minutes (4) 60 minutes  (CUET 2022, 8th August) 15. The force between a neutron and a proton are (i) Weak (ii) Gravitational (iii) Nuclear

99

ATOM AND NUCLEI

(iv) Colombian (v) Electromagnetic Choose the correct answer from the options given below: (1) (i), (ii) and (iv) only (2) (i), (iii) and (v) only (3) (ii), (iii) and (iv) only (4) (ii) and (iii) only  (CUET 2022, 8th August) 16. The relation between half-life of a radio nuclide denoted by T1/2 and average life of a radio nuclide denoted by τ, is (1) T1/2 = τln2 (2) τ = T1/2 ln2 (3) T1/ 2 =

1 τ

(4) T1/2 = τ

 (CUET 2022, 23rd August) 17. A particular radioactive sample has a half-life of 12 years. The fraction of it that remains undecayed after 48 years, will be (1)

1 4

th

th

(2)

1 8

th

of its initial amount th

1 of its initial amount 32  (CUET 2022, 26th August) 18. Choose the correct statements from the following based on the properties of atomic nucleus. I. Radius of a nucleus of mass no. A depends on its mass (3)

1 16

of its initial amount of its initial amount

(4)

number as R ∝ A3 .

22. Match List-I with List-II List-I (Total energy eV) (A) –3.40

(I) At infinity

(B) –13.6

(II) I excited state

(C) 0

(III) Ground state

(D) –0.85 (IV) III excited state Choose the correct answer from the options given below: (1) (A)- (II), (B)-(III), (C) - (I), (D)-(IV) (2) (A)- (III), (B)-(I), (C)-(II), (D)-(IV) (3) (A)- (IV), (B)-(I), (C)- (III), (D)- (II) (4) (A)- (I), (B)- (II), (C)-(III), (D)-(IV)  (CUET 2022, 26th August) 23. According to Bohr’s postulates, the quantity that shows discrete values is (1) Linear momentum (2) Angular momentum (3) Kinetic Energy (4) Potential Energy  (CUET 2022, 30th August) 24. The expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1), for a large value of n become (1)

me 4 1 × 8ε 02 h3 n 2

(2)

me 4 1 × 4ε02 h3 n

(3)

me 4 1 × 4ε 02 h3 n3

(4)

me 4 1 × 8ε02 h3 2n3

1

II. Radius of nucleus and mass number are related as R ∝ A 3 . III. Density of nuclei is always a constant. IV. Density of nuclei depends on its mass number A. V. Density of nuclei depends on its size, volume. Choose the correct answer from the options given below: (1) I and III only (2) II and III only (3) II and IV only (4) II and V only  (CUET 2022, 26th August) 19. The constancy of the binding energy in the range 30 < A < 170 in binding energy curve, is a consequence of the given fact (1) Nuclear forces are charge independent. (2) Nuclear forces are short ranged. (3) Nuclear forces are the strongest forces. (4) Nuclear forces are attractive in nature.  (CUET 2022, 26th August) 20. For a dynamically stable orbit in a hydrogen atom, according to Rutherford’s atomic model the relation between the orbit radius (r) and the electron velocity (v) is e (1) r = 4πε0 mv 2 (3) r =

e2 4πε 0 mv 2

e2 (2) r = 4πε 0 mv

(4) r =

e 4πε0 m 2v

 (CUET 2023, 26th August) 21. The spectral series to which the emission line belongs, when electron in the hydrogen atom jumps from 4th energy state to 2nd energy state, is (1) Paschen series (2) Lyman series (3) Balmer series (4) Brackett series  (CUET 2022, 26th August)

List-II (States of electron in Hydrogen atom)

 (CUET 2022, 30th August) 25. If the minimum wavelength corresponding to Paschen series of hydrogen spectra is 820 nm then that corresponding to the Balmer series would be (1) 1640 nm (2) 364.4 nm (3) 528.4 nm (4) 298.2 nm  (CUET 2022, 30th August) 26. The half-life of a radioactive material undergoing b decay is 12.5 years. What fraction the material remains undecayed after 25 years? 3 1 (4) 4 8  (CUET 2022, 30th August) 27. Kinetic energy of the emitted a-particle in the a-decay of

(1)

1 4

226 88



(2)

1 2

(3)

Ra will be 226 m ( 88 Ra ) = 226.02540u

m ( 42 He ) = 4.002603u 222 m ( 86 Rn ) = 222.01750u

(1) 6.937MeV (2) 4.85MeV (3) 0.087MeV (4) 2.87MeV  (CUET 2022, 30th August) 28. When two nuclei (A ≤ 10) fuse together to form a heavier nucleus, the (1) binding energy per nucleon increases. (2) binding energy per nucleon decreases. (3) binding energy per nucleon does not change. (4) total binding energy decreases.

100 Oswaal CUET (UG) Chapterwise Question Bank 29. O2 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms (1) is not important because nuclear forces are short-ranged. (2) is as important as electrostatic force for binding the two atoms. (3) cancels the repulsive electrostatic force between the nuclei. (4) is not important because oxygen nucleus have equal number of neutrons and protons.  [NCERT Exemp. Q. 12.5, Page 76] 30. For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, (1) because Bohr model gives incorrect values of angular momentum. (2) because only one of these would have a minimum energy. (3) angular momentum must be in the direction of spin of electron. (4) because electrons go around only in horizontal orbits.  [NCERT Exemp. Q. 12.4, Page 76]

[B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): Volume of the nucleus is proportional to A where A is mass number. Reason (R): Density of nucleus is independent of mass number A. (CUET 2022, 30th August) 2. Assertion (A): Bohr postulated that the electrons in stationary orbits around the nucleus do not radiate. Reason (R): According to classical Physics, all moving electrons radiate. 3. Assertion (A): According to Rutherford, atomic model, the path of an electron is parabolic. Reason (R): Rutherford could not explain the stability of the atom. 4. Assertion (a): In the a-particle scattering experiment, most of the a-particles pass undeviated. Reason (R): Most of the space in the atom is empty. 5. Assertion (A): Bohr model is not applicable for multielectron model. Reason (R): Bohr model cannot account for sub-level (s, p, d, f  ) orbitals and electron spin. 6. Assertion (A): In alpha particle scattering experiment, thin gold foil was used. Reason (R): Nucleus of gold is about 50 times heavier than an a-particle, it is reasonable to assume that it remains stationary throughout the scattering process. 7. Assertion (A): Two atoms of different elements having the same mass number but different atomic numbers are called isobars. Reason (R): Atomic number is the number of protons present and atomic mass is the total number of protons and neutrons present in a nucleus.

PHYSICS

8. Assertion (A): Nuclear force is same between neutronneutron, proton-proton and neutron-proton. Reason (R): Nuclear force is charge independent. 9. Assertion (A): Electrons experience strong nuclear force. Reason (R): Strong nuclear force is charge dependent. 10. Assertion (A): Chlorine has two isotopes having masses 34.98u and 36.98u. The average mass of a chlorine atom is 35.47u. Reason (R): The relative abundances of the isotopes of chlorine are 75.4 and 24.6 per cent, respectively. The average mass is obtained by the weighted average of the masses of the two isotopes.

[C] COMPETENCY BASED QUESTIONS

I. Based on following passage answer questions from 1-5. In a-decay, the mass number of daughter nucleus is four less than that of decaying nucleus (parent nucleus). While atomic number decreases by two. In general, a-decay of parent nucleus

A Z

X results in daughter nucleus A Z



 1. (1) (2) (3) (4)

A− 4 Z −2

Y.

X → ZA−−24 Y + 42 He

From Einstein’s mass energy equivalence relation and energy conservation, it is clear that this spontaneous decay is possible only when mass of decay products is less than the mass of the products. By referring the table of nuclear masses, one can check that the total mass of daughter nucleus and a-particle is indeed less than that of parent nucleus. The disintegration energy or the Q value can be calculated by formula Q = Dmc2 (CUET 2022, 6th August) In a-decay, the mass number of the daughter nucleus is One more than that of parent nucleus One less than that of parent nucleus Two more than that of parent nucleus Four less than that of parent nucleus α

β

β

2. A → B → C → D In the process A → B, a-particle is emitted by nucleus and in process B → C and C → D each, one b-particle is emitted. (1) A, B, C are isobars (2) B, C, D are isobars (3) A, B, C are isobars (4) A, B, C, D all are isotopes 3. a-particle is (1) Nucleus of Hydrogen atom (2) Hydrogen atom (3) Nucleus of Helium atom (4) Helium atom 4. During the a-decay process, the mass of all decay products is (1) Equal to the mass of initial nucleus (2) Less than the mass of initial nucleus (3) Equal to the mass of a-particle (4) Greater than the mass of initial nucleus 5. An a-decay process is given by A A− 4 4 Z X → A − 2 Y + 2 He If mass of nuclei X,Y and He are mx, my and mHe respectively, then Q-value is given by (1) ( mx − m y − mHe ) c 2

(2) ( m y + mHe − mx ) c 2

(3) ( mx + m y − mHe ) c 2

(4) ( mx − m y + mHe ) c 2

101

ATOM AND NUCLEI

II. Based on following passage answer questions from 6-10. When the electron of hydrogen is at the lowest energy level it absorbs light, jumps to a higher energy level. Electron can move only from one energy level to another. It can’t go partway in between levels. In addition, it takes a very discrete amount of energy-no more, no less-to move the electron from one particular level to another. In the visible part of the spectrum, electron of hydrogen atom absorbs light of wavelengths of 410 nm (violet), 434 nm (blue), 486 nm (blue-green) and 656 nm (red) and jumps to the corresponding energy levels. Each of the absorption lines corresponds to a specific electron jump. In the spectrum, lines corresponding to these wavelengths appear black. Electrons can also lose energy and drop down to lower energy levels. When an electron drops down between levels, it emits photons with the same amount of energythe same wavelength-that it would need to absorb in order to move up between those same levels. This is why hydrogen’s emission spectrum is the inverse of its absorption spectrum, with emission lines at 410 nm (violet), 434 nm (blue), 486 nm (blue-green) and 656 nm (red). 6. What is the wavelength of light emitted when an electron drops from the n = 3 level to the n = 2 level within a Bohr model hydrogen atom? (1) 656 nm (2) 656 Å (3) 656 m (4) 656 × 10–6 m 7. Which of the following shows three lines of Lyman emission spectra? (1) n = 4 n=3

8. Which of the following shows three lines of Lyman absorption spectra? (1) n = 4 (2) n = 4 n=3 n=3 n=2

n=1 (3) n = 5

n=2



n = 1 (4) n = 5 n=4

n=4 n=3

n=3

n=2 n = 2 9. The following is the wavelength vs. brightness graph of absorption spectrum of hydrogen. What will be the colour of lines in the spectrum corresponding to 410 nm, 434 nm, 486 nm and 656 nm?

(2) n = 4 n=3

n=2

n=2

n=1 (3) n = 5



n = 1 (4) n = 5 n=4

n=4 n=3

n=3

n=2



n=2

(1) 410 nm (violet), 434 nm (blue), 486 nm (blue-green) and 656 nm (red). (2) 410 nm (white), 434 nm (white), 486 nm (white) and 656 nm (white). (3) 410 nm (Black), 434 nm (Black), 486 nm (Black) and 656 nm (Black). (4) 410 nm (white), 434 nm (black), 486 nm (white) and 656 nm (black). 10. What will be the colour of lines in the emission spectrum corresponding to 410 nm, 434 nm, 486 nm and 656 nm? (1) 410 nm (violet), 434 nm (blue), 486 nm (blue-green) and 656 nm (red). (2) 410 nm (white), 434 nm (white), 486 nm (white) and 656 nm (white). (3) 410 nm (Black), 434 nm (Black), 486 nm (Black) and 656 nm (Black). (4) 410 nm (red), 434 nm (blue), 486 nm (violet) and 656 nm (blue-green).s

102 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (3)

3. (2)

4. (3)

5. (2)

6. (3)

7. (2)

8. (3)

9. (3)

10. (1)

11. (1)

12. (1)

13. (1)

14. (3)

15. (4)

16. (1)

17. (3)

18. (2)

19. (2)

20. (3)

21. (3)

22. (1)

23. (2)

24. (3)

25. (2)

26. (1)

27. (2)

28. (1)

29. (1)

30. (1)

8. (1)

9. (2)

10. (1)

8. (1)

9. (3)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (4)

4. (1)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (2)

3. (3)

4. (2)

5. (1)

6. (1)

7. (2)

ANSWERS WITH EXPLANATIONS [A] MULTIPLE CHOICE QUESTIONS

1. Option (1) is correct. Explanation: J.J. Thomson described an atom as a sphere of positive charge, called the pudding, with electrons embedded in it, like plums in a dessert. This model of the atom was discarded by Ernest Rutherford. He performed the alpha particle scattering experiment and properly described the nucleus of an atom. Niels Bohr removed the discrepancies of Rutherford atomic model and put forward an atomic model of Hydrogen atom. A.H. Becquerel is a French scientist who accidentally discovered radioactivity. 2. Option (3) is correct. Explanation: 1 1  For Lyman series, 1= / λ R 2 − 2  1 n  1   1 For Balmer series, = 1/ λ R 2 − 2  n  2 1 1  For Paschen series, 1= / λ R 2 − 2  3 n  1   1 For Brackett series, = 1/ λ R 2 − 2  n  4 1  1 Pfund series, 1= / λ R 2 − 2  5 n   3. Option (2) is correct. Explanation: Half life = T1/2 = K Now, N = N0(1/2)n ...(i) Where N0 = initial activity 6.25 N (activity at time t) = 6.25% of N 0 = × N 0 = N 0 / 16 100 n = t/T1/2 = t/K Substituting N and n in equation (i) N0/16 = N0(1/2)t/K Or, 1/16 = (1/2)t/K Or, (1/2)4 = (1/2)t/K Or, 4 = t/K ∴ t=4K 4. Option (3) is correct. Explanation: N = N0(1/2)n Or, N0/16 = N0(1/2)n For

Or, 1/16 = (1/2)n Or, (1/2)4 = (1/2)n Or, n=4 Or, t/T1/2 = 4 Or, T1/2 = t/4 ∴ T1/2 = 2/4 hours = 30 minutes 5. Option (2) is correct. Explanation: The variation of the number of undecayed nuclei with time is represented as N = N0e–λt Where N is the number of undecayed nuclei at any time t and N0 is the number of nuclei at t = 0. The relation is exponential. So, the nature of the graph will be No N time

6. Option (3) is correct. Explanation: Bohr’s model is valid only for one electron atoms or ions. Since, Hydrogen has only one electron, the model is valid for Hydrogen only. 7. Option (2) is correct. Explanation: Nuclear force is the strongest force. It is charge independent and very small ranged force. 8. Option (3) is correct. Explanation: Hydrogen Spectral Series: (i) Lyman Series - Transition to first shell from any other shell n > 1. (ii) Pfund Series - Transition to fifth shell from any other shell n > 5. (iii) Paschen Series - Transition to third shell from any other shell n > 3. (iv) Brackett Series - Transition to fourth shell from any other shell n > 4. (v) Balmer Series - Transition to second shell from any other shell n > 2. Hence, sequence should be: (i), (v), (iii), (iv), (ii) 9. Option (3) is correct. Explanation: Radius of Nucleus R ∝ A1/3 Where A is mass number

103

ATOM AND NUCLEI

1 3

1

R1  A1   1  3 = =  =  1: 3 R2  A2   27  Nuclear density does not change with mass number. So, the ratio is ρ1 = 1:1 ρ2 \

10. Option (1) is correct. Explanation: The Bohr model of atom is applicable to atoms only. Bohr model is applicable for one electron structure. Since “He” has 2 electrons. Bohr model is not applicable. 11. Option (1) is correct. Explanation: At the distance of closest approach, 1 Ze × 2e × = K .E 4πε 0 r0 Or,

9 × 109 × 72 × 2 × (1.6 × 10−19 )

2

r0 =6 × 10 × 1.6 × 10−19 r0 =

9 × 109 × 72 × 2 × 1.6 × 10−19 6.0 × 10−3

\ r= 345.6 × 10−7 m ≈ 35µm 0 12. Option (1) is correct. Explanation: The correct sequence of binding energy per nucleon is H < Li < N < O < Fe i.e., E H∞ i.e., C, A, D, B, E 14. Option (3) is correct. Explanation: Initial number of nuclei, N0 = 800 Number of nuclei disintegrated = 600 Number of nuclei left, N = 800 – 600 = 200 n = Number of half-lives  N  1  =   N0   2 

17. Option (3) is correct. Explanation: n

−3

Or,

n

 200   1   =   800   2  \ n=2 So, Time taken for disintegration = 2 × 10 = 20 minutes 15. Option (4) is correct. Explanation: Force between neutron and proton is gravitational and nuclear. There is no Coulombian force since neutron is neutral. 16. Option (1) is correct. Explanation: ln2 T1/ 2 = λ 1 and τ= λ Putting in above equation T1/2 = τln2 Or,

n

1 N = N0   2 N → Number of remaining nuclei N0 → Initial number of nuclei 48 Here, = n = 4 12 n → number of half live As

4

1 ⇒ N= N0   2 N0 ⇒ N= 16 1th i.e., of initial amount remains undecayed. 16 18. Option (2) is correct. Explanation: As volume of nucleus is directly proportional to the mass number A, i.e., 4 3 πR ∝ A 3 ⇒ R3 ∝ A ⇒ R ∝ A1/3 ⇒ R = R0 A1/3 Here, R0 is Empirical constant. 3 Density of nucleus = 4πR03 i.e., density of nucleus is always constant Hence, statement II and III are correct. 19. Option (2) is correct. Explanation: Constancy of Binding energy per nuclear in the range of mass number A from 30 to 170 can be explained on the basis of short ranged nature of nuclear force. 20. Option (3) is correct. Explanation: As electrons are revolving around the nucleus, so, centripetal force on electron is equal to electrostatic force of attraction between electron and proton.

104 Oswaal CUET (UG) Chapterwise Question Bank i.e., ⇒

1 e2 mv 2 = r 4πε 0 r 2 1 e2 r= 4πε0 mv 2

21. (3) is correct. Explanation: Transition between n = 4 to n = 2 belongs to Balmer series. 22. Option (1) is correct. Explanation: As Energy in nth orbit is given by −13.6 E= eV n2 n =1;E = –13.6eV n = 2;E = –3.4eV n = 3;E = –1.51eV n = ∞;E = 0eV 23. Option (2) is correct. nh Explanation: Angular momentum ( L ) = 2π n = principal quantum number = 1, 2, 3, 4, 5...... 24. Option (3) is correct. Explanation: Energy of electron in nth orbit So,

En =

− me 4 z 2 8ε 02 h 2 n 2

= ∆E

me 4 z 2 8ε 02 h 2

1 1   2−  − n ( n 1) 2  

Or,

me 4 z 2  (n − 1) 2 − n 2  ∆E = 2 2  2  8ε0 h  n (n − 1) 2 

Or,

me 4 z 2  2n − 1  ∆E = 2 2  2  8ε0 h  n (n − 1) 2 

Or,

  1   2n  1 −   me z 2n   ∆E = 2 2   8ε0 h  4  1  2   n 1 −     n 

Or,

me 4 z 2 ∆E = 2 2 3 4ε0 h n

4 2

∆E h me 4 z 2 v= 2 3 3 \ 4ε0 h n 25. Option (2) is correct. Explanation:  1 1 1  = R 2 − 2  λ n n 2   1 Or,

v=

n1 3, n2 ≥ 4 For Paschen series,= for \

λ min n1 = 3, n2 = ∞ 1 R = λ1 9

n1 2, n2 ≥ 3 For Balmer series, = for

λ min n1 = 2, n2 = ∞

\

1 R = λ2 4

Now,

λ2 4 = λ1 9

Or,

λ 2=

4 × λ1 9

\

λ2 =

4 × 820 ≈ 364.4 nm 9

PHYSICS

26. Option (1) is correct. Explanation: N N = n0 2 Here, N = number of remaining nuclei N0 = number of initial nuclei n = number of half-life 25 25 = n = = 2 T1/ 2 12.5 \

N 1 1 = = N 0 22 4

27. Option (2) is correct. Explanation: Δm = (226.02540 – 4.002603 – 222.01750)u Δm = 0.005297u Q = Δm × 931.5MeV Or, Q = (0.005297u) × (931.5MeV/u) = 4.9341555MeV) 28. Option (1) is correct. Explanation: When two nuclei (A ≤ 10) fuse together to form a heavier nucleus, binding energy per nucleon increases to achieve better stability. 29. Option (1) is correct. Explanation: As we know that the nuclear force is too much stronger. Only attractive force as compared to electron static repulsive force and nuclear force decreases to zero on increasing distance. So, in case of oxygen molecule, the distance between atoms of oxygen is larger as compared to the distances between nucleons in a nucleus. So that, the force between the nuclei of two oxygen atoms is not important as nuclear forces are shortranged forces. 30. Option (1) is correct. Explanation: According to Bohr’s second postulate of atomic model, angular momentum of revolving electron must be some integral multiple of h/2p. So, the Bohr’s model does not give correct value of angular momentum.

[B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: Radius of nucleus R = R0A1/3 Volume ∝ R3 \ V ∝ R3 So, assertion is true.

Mass of a nucleus = A amu = A × 1.66 × 10−27 kg Volume of nucleus= =

4 3 πR 3 1  4  π  R0 A 3  3  

3

105

ATOM AND NUCLEI



3 4 = π (1.1 × 10−15 ) A m3 3

So, density of nucleus =

=

A × 1.66 × 10−27

3 4 π (1.1 × 10−15 ) A 3

1.66 × 10−27

3 4 π (1.1 × 10−15 ) 3

kg m −3

So, density of nucleus is independent of mass number A. So, reason is also true. And reason also explains the assertion. 2. Option (1) is correct. Explanation: Bohr postulated that electrons in stationary orbits around the nucleus do not radiate. This is the one of Bohr’s postulate. According to this the moving electrons radiate only when they go from one orbit to the next lower orbit. According to classical Physics, all moving electrons radiate. So, both assertion and reason are true but reason does not explain the assertion. 3. Option (4) is correct. Explanation: According to Rutherford, “the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus, with electrons revolving around the nucleus just as planets revolve around the Sun.” So, the assertion is false. According to classical physics, the electron orbiting around the nucleus radiate energy. As a result, the radius of the orbit continuously decreases and the electron falls into the nucleus. So, the stability of an atom is not explained. Hence, the reason is true. 4. Option (1) is correct. Explanation: Most of the a-particles pass roughly in a straight line (within 1°) without deviation. This shows that no force is acting on them. So, the assertion is true. Most of the space in the atom is empty. Only 0.14% of a-particles are scattered more than 1°. So, the reason is also true and reason correctly explains the assertion. 5. Option (1) is correct. Explanation: Bohr model works well for H and He+ having one electron only. But it does not work for multi-electron atoms, since it cannot account for sub-level (s, p, d, f  ) orbitals and electron spin. So, assertion and reason both are true and reason correctly explains the assertion. 6. Option (1) is correct. Explanation: In alpha particle scattering experiment, very thin gold foil was used. So, the assertion is true. Gold (atomic no. 79) is heavier than alpha particle. The reason is also true. But a thin sheet of gold was used because it is vey malleable, therefore, reason does not explain assertion. 7. Option (1) is correct. Explanation: Two atoms of different elements having same mass number but different atomic numbers are called isobars. The assertion is true. Atomic number is the number of protons present and atomic mass number is the total number of protons and neutrons present in a nucleus. The reason is also true. But the reason does not explain the assertion. 8. Option (1) is correct. Explanation: Nuclear force acts between nucleons. It is a powerful attractive force and is charge independent. So, nuclear

force between neutron & neutron, proton & proton and neutron & proton are equal. So, both assertion and reason are true and reason correctly explains the assertion. 9. Option (2) is correct. Explanation: Nuclear force is a powerful attractive force acts as long as the distance between particles is within 10–15m. This force is charge independent. But as distance increases, the effect of nuclear force rapidly falls. Electrons are distributed far away. The distance is beyond the range of the nuclear force. Hence, nuclear force has no effect on electrons. So, the assertion is true but the reason is false. 10. Option (1) is correct. Explanation: The average mass of a chlorine atom is 35.47u So, the assertion is true. The average mass = (75.4 × 34.98 + 24.6 × 36.98)/100 = 35.47u So, the reason is also true and explains the assertion properly.

[C] COMPETENCY BASED QUESTIONS

1. Option (4) is correct. Explanation: a-particle is Helium Nucleus and is represented by 42 He So, the mass number of daughter Nuclei after a-decay becomes A – 4 where A is the mass number of parent nucleus. 2. Option (2) is correct. Explanation: b-decay changes atomic number only while mass number remains unchanged. Isobars have same mass numbers but different atomic numbers. So, the B, C and D have same mass numbers but different atomic numbers. Hence, those are isobars. 3. Option (3) is correct. Explanation: a-particle is Helium Nucleus represented as 42 He . 4. Option (2) is correct. Explanation: ZA X → AA −− 42 Y + 42 He Thus, mass of decay product nucleus is less than the mass of initial nucleus. 5. Option (1) is correct. Explanation: Q value is the difference of mass of parent and daughter nuclei multiplied by square of the velocity of light in vacuum. \ Q value = Dmc2 \ Q = [mx – (my + mHe)]c2 6. Option (1) is correct. Explanation:  1 1 1  = R 2 − 2    λ  nlow nhigh  1  1 1 Or, = 1.097 × 107 ×  2 − 2  λ 2 3  m 656 nm = λ 656 × 10−9 = \ 7. Option (2) is correct. Explanation: Lyman emission spectra: Lines emitted by transitions of the electron from an outer orbit of quantum number n > 1 to the n = 1 orbit. 8. Option (1) is correct. Explanation: Lyman absorption spectra: Lines absorbed by transitions of the electron from n = 1 orbit to an outer orbit of quantum number n > 1.

106 Oswaal CUET (UG) Chapterwise Question Bank 9. Option (3) is correct. Explanation: Electron of Hydrogen atom absorbs light of wavelengths of 410 nm (violet), 434 nm (blue), 486 nm (bluegreen) and 656 nm (red) and jumps to the corresponding energy levels. Each of the absorption lines corresponds to a specific electron jump. In the spectrum, lines corresponding to these wavelengths appear black.

24

PHYSICS

10. Option (1) is correct. Explanation: 410 nm, 434 nm, 486 nm, and 656 nm are the wavelengths of light of colours violet, blue, blue-green and red respectively which are emitted when electron jumps from specific energy level to n = 2.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

9

ELECTRONIC DEVICES

  Revision Notes

Electron energies

Energy bands in solids - conductors, insulators and semiconductors Energy bands: Energy bands consist of large number of closely spaced energy levels that exist in crystalline materials. In solids, there are three important energy bands known as Valence band, Conduction band, forbidden band or forbidden gap.

Energy bands in Semiconductors: Semiconductors, are materials in which, conduction band and valence band are neither overlapped nor have wide gap. In such materials, the energy provided by the heat at room temperature is sufficient to lift the electrons from the valence band to the conduction band. Empty conduction Ec band Eg Ev Valence band

Electron energies

The collection of energy levels of free electrons which move freely around the material is called conduction band. Extra energy required for valence electrons to move to conduction band is known as forbidden energy. The collection of energy levels which are partially or fully filled by electrons are known as valence band. Energy bands in Conductors: In conductors, the conduction and valence bands are overlapped.

Overlapping of conduction and valence bands

Ev Ec

Valence bands Energy bands in Insulators: In insulators, conduction band and valence band are separated by very wide forbidden energy gap. Empty conduction band E

Electron energies

c

Eg Ev Valence band

The gap between conduction band and valence band exceeds by 3 eV as electrons that transfer from valence band to conduction band need more energy. Due to requirement of more energy, insulators do not conduct any electric current.

Semiconductors behave as insulators at 0 K as no electron exists in conduction band. Silicon and Germanium have energy gaps 1.12 eV and 0.75 eV respectively. Intrinsic semiconductors: Pure semiconductors are called intrinsic semiconductors. in electronic circuits as their conductivity is low. For intrinsic semiconductors, the number of free electrons is equal to the number of holes. ne = nh = ni An intrinsic semiconductor (a) at T = 0 K behaves like insulators. (b) At T > 0 K forms thermally generated electron hole pairs and conductivity increases. Extrinsic semiconductors: When a small fixed amount of charged impurity is mixed with intrinsic semiconductors, they become extrinsic. In extrinsic semiconductors: (i) Conductivity increases (ii) Conductivity is controlled by dopant In extrinsic semiconductors, the number of free electrons is not equal to number of holes. There are two types of extrinsic semiconductors: n-type and p-type. n-Type extrinsic semiconductor: On doping an intrinsic semiconductor with pentavalent impurity like Antimony (Sb) or Arsenic (As), extrinsic semiconductor so obtained is known as n-type. n-type semiconductor has large number of free electrons known as majority (charge) carriers and small number of holes known as minority (charge) carriers. ne >> nh Impurity atom in n-type semiconductor is called donor. p-Type extrinsic semiconductor On doping an intrinsic semiconductor with trivalent impurity like Indium (In) or Gallium (Ga), extrinsic semiconductor so obtained is known as p-type.

First Level

Second Level





V

Third Level

For– n–type n e =7×10 15 m –3 for Si doped with P

Trace the Mind Map

(n n) = 5×10 21 m –3

ne

0K.

nn

El

ution

es

ic Dev tron ic c e

trib

Lo gi

n–

p–

.

E B

.

C

.

tting Diode Emi /LE D

cG ate s

t gh Li

p–n

p-n

p–

n–

+

–

E

.

+

B

.

+

C

.

–

. . . . . .

n–

p–

–

 

n Co

n

 



108 Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

n atio Applic

p

109

ELECTRONIC DEVICES

Forward Bias: When the negative terminal of battery is connected to n-side and positive terminal to p-side, the barrier potential and the width of the depletion layer get reduced and only majority carriers can across the junction. Thus the resistance is reduced.

V-I Characteristics of Diode:

Diode as a rectifier: Rectifier is a circuit which converts AC supply into unidirectional DC supply. There are two types of rectifiers: half-wave rectifier and full-wave rectifier. Half-wave rectifier: Primary Vac

The positive terminal of battery repels majority carriers, holes in p-region while negative terminal repels electrons in n-region which pushes them towards the junction. Reverse Bias: When the positive terminal of a battery is connected to n-side and negative terminal to p-side, then barrier potential and width of depletion region increases and only minority charge carriers can across junction.

Secondary

p-type semiconductor has large number of holes known as majority (charge) carriers where number of free electrons is less known as minority (charge) carriers. nh >> ne Impurity atom in p-type semiconductor is known as acceptor. Semiconductor diode – I-V characteristics in forward and reverse bias: Semiconductor diode: When a p-type semiconductor material is suitably joined to n-type semiconductor, the contact surface is called a p-n junction. It is an electrical device that allows current only in one direction. At junction, electrons and holes diffuse and recombine to form the depletion region. It acts as a barrier for majority charge carriers.

RL

V

Vout

The half-wave rectifier with single diode, allows current to flow in one direction. When Vac is in positive cycle, the diode is forward biased and conducts. As a result, voltage becomes available across the load. When Vac is in negative half cycle, the diode is reversed biased and does not conduct. No voltage becomes available across the load. Voltage waveform across load resistor: Vac Vout

Here, negative terminal of battery attracts majority charge carriers, holes, in p-region and positive terminal attracts majority charge carriers, electrons, in n-region which pulls them away from the junction. Thus majority carriers cannot cross the junction. As a result of this, the width of depletion region increases. Resistance increases. But the minority carriers can cross the junction and thus constitutes a feeble current in reverse direction. Beyond certain voltage, diode breaks down and current increases drastically due to avalanche breakdown or zener breakdown.

The output voltage Vdc is not exactly similar to the output of a battery. The output of half-wave rectifier is bumpy as only one half of input AC is rectified. This is called fluctuating d.c. Full-wave rectifier: For rectifying AC input for both half cycles full wave rectifier is used. D1 A

RL

Vout

D2 B

A simple kind of full-wave rectifier uses a centre tap transformer with two diodes.

110 Oswaal CUET (UG) Chapterwise Question Bank In full wave rectifier, for positive half cycle, A is +ve, B is –ve, then only D1 diode conducts, while D2 diode blocks the current. For negative half cycle, A is –ve and B is +ve, only D2 diode conducts while the D1 diode blocks the current. Voltage waveform across load resistor : Vac

Dark current I1 I2 I3 I4

Vout

I-V characteristics of LED, photodiode, solar cell and Zener diode: Scan to know LED (light emitting diode): Light more about emitting diode is a heavily doped diode this topic which emits light when forward biased. Circuit symbol: How does an LED work?

I-V characteristics:

PHYSICS

I4

I3 I2 I1

Solar cell: It a large area p-n junction diode. It produces a potential difference when absorbs photons of sunrays. Circuit symbol:

I-V Characteristics: I V (open circuit voltage) V

Isc

Photodiode: It is a light sensitive semiconductor diode. It produces a potential difference when absorbs photons. It works in reverse bias. Circuit symbol: Anode

Short circuit current

Zener Diode: Zener diode is a heavily doped diode having thin depletion region. It is always operated in reverse biased condition. Circuit symbol: Anode

Cathode

I-V Characteristics: I-V Characteristics: Forward Current Forward Bias Region –Vz Reverse voltage

Reverse Bias

Forward voltage

"Zener" Breakdown Region

Constant Zener Voltage

Reverse Current

Cathode

111

ELECTRONIC DEVICES

– – Resistor RS is used to limit reverse current through diode to a safer value. VS and RS are selected such that the diode operates in breakdown region. Junction transistor: Semiconductor diode, a two terminal device, is made of two semiconductor materials to form a simple p-n junction mainly for rectification purpose. Junction transistor, a three terminal device, is made of three semiconductor materials to form two p-n junctions mainly for amplification purpose. The terminals are: emitter, collector and base. There are two types of junction transistors: (i) n-p-n type A p-type semiconductor material is sandwiched between two n-type semiconductor materials. Emitter Collector N

P

N

Scan to know more about this topic

Base

E

B

The base is common to input and output and is grounded. (ii) Common Emitter or grounded emitter configuration: C B

C

E Circuit symbol (ii) p-n-p-type A n-type semiconductor material is sandwiched between two p-type semiconductor materials. Emitter Collector P

N

Base

P

E

The emitter is common to input and output and is grounded. (iii) Common collector configuration or grounded collector: E B

INPUT

How does a p-n junction diode work?

B

C

OUTPUT

Vs

OUTPUT

Regulated output

C Circuit symbol Transistor action: For normal operation of a transistor, the emitter-base junction is forward biased and the collector-base junction is reversed biased. So, the dynamic resistance of the EB junction is very low whereas the dynamic resistance of the CB junction is very high. In a transistor, a current from a low-resistance input circuit is transferred to a high-resistance output circuit with almost unchanged magnitude. This results in a power gain. In fact, the name “transistor” is coined from “transfer resistor”. Characteristics of a transistor: Transistor circuit configuration: a transistor is a there terminal device but for connecting it to a circuit 4 terminals are required – 2 for input and 2 for output. Hence, one of the three terminals is made common to input and output. Thus three configurations have been developed for operation of a transistor. (i) Common base configuration or grounded base:

OUTPUT

RL

Vz

B

INPUT

Iz Unregulated Supply

E

INPUT

Zener diode as a voltage regulator: Voltage regulation is the measure of ability of a power supply to maintain constant voltage output under variation of either in input voltage or load current. Reverse biased Zener diode maintains constant voltage across load as long as the supply voltage is more than Zener voltage. When input voltage is fixed and the load is varying, the output voltage remains constant as long as the load resistance is maintained above a minimum value. Voltage regulator circuit using Zener diode: IL Rs + +

C

The collector is common to input and output and is grounded.

112 Oswaal CUET (UG) Chapterwise Question Bank Input characteristics in CE configuration:

PHYSICS

IB is plotted against VBE keeping VCE constant at different values. A.C. input resistance is obtained from the graph as  ∆V Rin =  BE  ∆I B

  VCE = 1/slope 

Output characteristics in CE configuration: IC (mA)

(Saturation region)

(Active region)

IB = 0 A O (Cutoff region)

VCE (V)

IC is plotted against VCE keeping IB constant at different values. Transfer characteristics:

The graph has 3 different regions:  Saturation region, In this region, both the junctions are forward biased  Active region In this region, collector junction is reverse biased and emitter junction is forward biased. If transistor is to be used as an amplifier, it must operate in this region.  Cut off region In this region both the junctions are reverse biased. A.C. output resistance is obtained from the graph as  ∆V Rout =  CE  ∆I C

IC is plotted against IB keeping VCE constant. Current amplification factor is obtained from the graph as

  IB = 1/slope 

 ∆I  β =  C  VCE = slope  ∆I B 

Transistor as an amplifier (common emitter configuration):

Cout Cin

~

V1 VBB

RB

IC B IB

C E IE

RC VCC

~

113

ELECTRONIC DEVICES

A small signal a.c. amplifier circuit is designed using n-p-n transistor. Base of the transistor is connected to the positive end of battery VBB through resistor RB so that the input circuit becomes forward biased. Collector is connected to the positive end of battery VCC through resistor RC so that the output circuit becomes reverse biased. In the output circuit, VCE = VCC – ICRL As an a.c. signal is applied at the input, then IB changes. Along with IC also changes. So, VCE also changes. This change is obtained at the output which is a magnified replica of the input a.c. signal. Oscillator: Oscillator is a circuit which converts d.c. energy to a.c. energy at very high frequency. It requires no external signal to generate oscillation.

Truth table of a two input AND gate: Input

Output

0

0

0

0

1

0

1

0

0

1 1 1 Boolean expression of a two input AND gate: Y = A.B OR gate: It produces “LOW” output only when all the inputs are “LOW”. Symbol of a two input OR gate: A Y

B

Truth table of a two input OR gate: Input

Basic parts of an oscillator: Scan to know  An CE configuration amplifier circuit more about  An LC network this topic A small disturbance such as a thermally generated noise is utilized to start an oscillation. This initial voltage starts the feedback process and oscillation. Basic logic The LC network is used as feedback to gates fed a part of the output signal back to the input section of the amplifier. The frequency of oscillation is also controlled by the LC circuit. Logic gates (OR, AND, NOT, NAND and NOR): Logic gates are the primary building Scan to know blocks of any digital system. It is an more about this topic electronic circuit holding one or more than one input and only one output. The correlation between the input and the output is based on a certain logic. Inputs and outputs of logic gates are How does BJT work? defined by two logical values – “HIGH” or “1” and “LOW” or “0”. Logic gates are classified as  Basic logic gates: AND OR and NOT gates  Universal logic gates: NAND and NOR gates (Any digital gate / Boolean expression can be implemented using universal gates) AND gate: It produces “HIGH” output only when all the inputs are “HIGH”. Symbol of a two input AND gate:

0

0

0

0

1

1

1

0

1

1

1

1

Boolean expression of a two input AND gate: Y=A+B NOT Gate: A NOT gate is a single input and single output gate and is also called an inverter. It produces “HIGH” output when input is “LOW” and vice versa. Symbol of a NOT gate: Y

A

Truth table of a NOT gate: Input

Output

0

1

1 0 Boolean expression of a NOT gate: Y= A NAND Gate: NAND gate is AND gate followed by NOT gate. It produces “LOW” output only when all the inputs are “HIGH”. Symbol of a two input NAND gate: A

A

Y

B

Y

B

Truth table of a two input NAND gate: Input

Output

0

0

1

0

1

1

1

0

1

1

1

0

Boolean expression of a NAND gate:

A Y B

Output

Y = A.B

NOR Gate: NOR gate is OR gate followed by NOT gate. It

produces “HIGH” output only when all the inputs are “LOW”.

114 Oswaal CUET (UG) Chapterwise Question Bank Implementation of NOT gate using NAND gate:

Symbol of a two input NOR gate: A

A Y

B

PHYSICS

Y

B

   Truth table of a two input NOR gate: Input

Y

Implementation of NOT gate using NOR gate:

Output

0

0

1

0

1

0

1

0

0

1 1 Boolean expression of a NOR gate:

AND

A

A

NOR

Y

Transistor as a switch:

0

Y = A+ B Implementation of OR gate using NOR gates: A

NOR

NOR

B

Y

Implementation of OR gate using NAND gates:

Implementation of AND gate using NAND gates: A

AND

AND

B

Y

Implementation of AND gate using NOR gates: A

NOR NOR

B

A simple switch using NPN transistor is shown in the figure. When input voltage is more than the threshold voltage of the junction, the transistor moves to the saturation region and creates a short circuit between collector and emitter. Hence, the circuit acts like an ON switch and the device starts functioning. When input voltage is less than the threshold voltage of the junction, the transistor moves to the cut off region and creates an open circuit between collector and emitter. Hence, the circuit acts like an OFF switch and the device stops functioning.

Y

NOR

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. For a photodiode, the I–V characteristics for different illumination intensities I1, I2, I3 and I4 were plotted. Which of the following is correct? mA

Reverse bias I1 I2 I3 I4

volts

A

(1) I1 > I2 > I3 > I4 (2) I4 > I3 > I2 > I1 (3) I2 > I1 > I3 > I4 (4) I4 > I3 > I1 > I2  (CUET 2023, 7th June) 2. Match List - I with List - II

List - I List -II (Electronic Device) (Application) (A) Zener diode (I) Rectifier (B) Capacitor (II) Amplifier (C) Transistor (III) Voltage regulator (D) p-n junction diode (IV) Filter circuit Choose the correct answer from the options given below: (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (3) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)  (CUET 2023, 7th June) 3. If the forward voltage in a p-n junction diode is increased, the width of the depletion region (1) increases (2) decreases (3) fluctuates (4) no change  (CUET 2023, 7th June) 4. Which of the following statements is true for a p-type semiconductor?

115

ELECTRONIC DEVICES

(1) Holes are minority carriers and pentavalent atoms are the dopant (2) Electrons are minority carriers and pentavalent atoms are dopant (3) Holes are majority carriers and trivalent atoms are the dopant (4) Electrons are majority carriers and trivalent atoms are the dopant  (CUET 2022, 5th August) 5. NPN transistor is set up as an amplifier in common emitter mode, output signal voltage obtained across the load resistance of 4 k W is 1 volt. Calculate input signal voltage. Assume current amplification factor = 250 and base resistance = 1 k W. (1) 4 mV (2) 2.5 mV (3) 1 mV (4) Insufficient data  (CUET 2022, 5th august) 6. An intrinsic semiconductor has 5 × 1028 atoms/m3. It is doped by 0.01 ppm concentration of arsenic. If ni = 1.5 × 1016 m3, then number of holes in the n-type semiconductor will be (1) 5 × 1028/m3 (2) 5 × 1011/m3 11 3 (3) 3.0 × 10 /m (4) 4.5 × 1011/m3  (CUET 2022, 5th August) 7. Correct symbol of NPN transistor is Collector (1) Emitter

9. In the circuit diagram D1, D2, D3 and D4 are diodes. Choose the correct answer from the options given below.

(1) (2) (3) (4) 10.

A •

• B (1)

>

Collector

>

Collector >

>

>



Collector >

>

Base

 (CUET 2022, 5th August)

8. The logic circuit given below will act like A• •Y B•

(1) NOT gate (3) OR gate



t2

t3

t4

B

   Input O t1



O



O

t1



O

t1

t2

t3

t1 t2 t3

t4

t4

t3

t2

t2

t4

t3

t4

(1) 5 V (2) 25 V (3) 100 V (4) 50 V  (CUET 2022, 6th August) 12. The I–V characteristics of an operating photodiode are drawn in

Base Emitter



t1

t1 t4 t2 t3 O  (CUET 2022, 6th August) 11. Three amplifiers are connected in series, (cascaded). The first amplifier has voltage gain of 5, second has voltage gain of 10 and third has voltage gain of 20. If the Input signal is 0.1 V, then the final output of AC signal will be

Base

Emitter

(4)

(4)

>

>

(3)

(3)

Base



A Input O

(2)

Emitter

(2)

Y •

ov

>



D1 and D2 are forward biased D1 and D3 are forward biased D2 and D4 are forward biased All diodes are forward biased  (CUET 2022, 6th August) Sketch the output y for the circuit shown below for two inputs.

(2) AND gate (4) XOR gate (CUET 2022, 5th August)

(1) First quadrant (2) Second quadrant (3) Third quadrant (4) Fourth quadrant  (CUET 2022, 6th August) 13. For a Common-Emitter amplifier, the audio signal voltage across the collector resistance of 2 kW is 2 V. Suppose the current amplification factor of the transistor is 100. The base resistance is 1 kW. The input signal voltage will be (1) 0.02 V (2) 0.04 V (3) 0.03 V (4) 0.01 V  (CUET 2022, 23rd August) 14. Choose the correct answer from the following options: A.

116 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

19. A Zener diode having VZ = 12 V and IZ(max) = 40 mA is used in a voltage regular circuit with unregulated power supply giving E = 19 V as shown in the circuit diagram shown below. The minimum value of the series resistor R required when the load resistance of 400 W is connected across the Zener diode, so that the Zener should not burn out, will be IL R I

B.

C.

RL

E IZ

(1) Only diode A is in forward biasing. (2) Just two diodes A and B are in forward biasing. (3) All diodes A, B and C are in forward biasing. (4) None of the diodes A, B and C are in forward biasing.  (CUET 2022, 23rd August) 15. Choose the logic operation for the following circuits: A Y B

(1) OR (2) AND (3) NOR (4) NAND  (CUET 2022, 23rd August) 16. Identify the logic operation carried out by the given circuit. A B

A' Y B'

(1) OR (2) AND (3) NOR (4) NAND  (CUET 2022, 26th August) 17. For transistor action, which of the following statements are correct? (A) The base region must be very thin and lightly doped (B) Base emitter and collector regions should have same size but different doping concentration (C) The collector junction is reverse biased and emitter junction is forward biased (D) Base, emitter and collector regions should have different sizes, but same doping concentration (E) Both the emitter junction as well as the collector junction are forward biased Choose the correct answer from the options given below: (1) (A) and (C) only (2) (A), (B) and (D) only (3) (B), (C) and (D) only (4) (B) and (C) only  (CUET 2022, 26th August) 18. The electron and hole concentration in an extrinsic semiconductor in thermal equilibrium is given by n (1) ne . nh = ni2 (2) e = ni2 nh (3) 

nh = ni2 ne

(4) ne . nh = ni (CUET 2022, 26th August)

(1) 100 W (2) 233 W  20. Match List-I with List-II

(3) 271 W (4) 175 W (CUET 2022, 26th August)

List-I

List-II

(A) Zener diode

(I) Heavily doped operates in forward bias

(B) Photo diode

(II) No external bias required

(C) Solar cell

(III) Heavily doped operates in reverse bias as a voltage regulator

(IV) Fabricated with a transparent window to allow light to fall on it Choose the correct answer from the options given below: (1) (A) - (II), (B) - (III), (C) - (IV), (D) - (I) (2) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (3) (A) - (IV), (B) - (I), (C) - (II), (D) - (III) (4) (A) - (III), (B) - (IV), (C) - (II), (D) - (I)  (CUET 2022, 26th August) 21. Select the correct statement explaining the function of rectifier. (1) It converts DC Signal to AC signal (2) It converts AC Signal to unidirectional pulsating signal (3) It increases the level of DC Signal (4) It decreases the level of AC Signal  (CUET 2022, 30th August) 22. In the circuit shown in figure, if the diode forward voltage drop is 0.3V. the voltage difference between A and B is A 0.2mA (D) Light emitting diode

5k

5k

B (1) 1.3 V (2) 2.3 V (3) 0 (4) 0.7 V (CUET 2022, 30th August) 23. In the circuit shown in figure, the potential barrier for Ge diode is 0.3 V and for Si diode it is 0.7 V. What is the voltage VA? (1) 23.7 V (2) 23 V (3) 24.3 V (4) 24.7 V  (CUET 2021, 23rd September)

117

ELECTRONIC DEVICES

24. In figure, V0 is the potential barrier across a p-n junction, when no battery is connected across the junction (1) (a) and (c) both correspond to forward bias of junction (2) (c) corresponds to forward bias of junction and (a) corresponds to reverse bias of junction (3) (a) corresponds to forward bias and (c) corresponds to reverse bias of junction. (4) (c) and (a) both correspond to reverse bias of junction. (a) (b) (c)

25. Hole is (1) an anti-particle of electron. (2) a vacancy created when an electron leaves a covalent bond. (3) absence of free electrons. (4) an artificially created particle. 26. The conductivity of a semiconductor increases with increase in temperature because (1) number density of free current carriers increases. (2) relaxation time increases. (3) both number density of carriers and relaxation time increase. (4) number density of current carriers increases; relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density. 27. Truth table for the given circuit figure is A D

E

B

(a) (1)

(b) (2)





C

A

B

E

0

0

1

0

1

0

1

0

1

1

1

0

A

B

E

0

0

1

0

1

0

1

0

0

1

1

1

(c) (3)

(d) (4)



A

B

E

0

0

0

0

1

1

1

0

0

1

1

1

A

B

E

0

0

0

0

1

1

1

0

1

1 0 1 28. Semiconductors behave like insulators at ________. (1) 0°C (2) 0 K (3) 273 K (4) None of the these

29. Tetra valent semiconductor is to be doped with ____ valent element to achieve _____ type extrinsic semiconductor. (1) penta, n (2) tri, p (3) penta, p (4) both (1) and (2)

30. Which are the relative sizes of Emitter, base and Collector blocks of a transistor?

(1)

E

(2)

E

(3)

E

(4)

E

B

C

B

C B B

C C

[B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): In light emitting diodes, energy is released in the form of photons when minority carriers recombine with majority carriers near the junction. Reason (R): Light emitting diodes are fabricated using heavily doped p-n junction which emits spontaneous radiation under forward biased condition.  (CUET 2022, 30th August) 2. Assertion (A): The electrical conductivity of a semiconductor increases on doping. Reason (R): Doping always increases the number of electrons in the semiconductor. 

3. Assertion (A): The resistivity of a semiconductor

decreases with temperature. Reason (R): As temperature increases, the electrons gain sufficient energy and jump from valence band to conduction band.

118 Oswaal CUET (UG) Chapterwise Question Bank 4. Assertion (A): For normal operation of a transistor, the

emitter-base junction is forward biased and the collectorbase junction is reversed biased. Reason (R): In a transistor, a current from a low-resistance input circuit is transferred to a high-resistance output circuit with almost unchanged magnitude. This results in a power gain.

5. Assertion (A): The given circuit behaves like an OR gate. A



2. In a full wave rectifier using a 100 secondary turns Centre tapped transformer and two diodes, the tapping is erroneously done over turn no. 30 as shown. The output voltage waveform will be as

30

Y

(1) V0

NOR

Reason (R):Each shorted input NOR gate behaves like a NOT gate. a transistor – Common emitter, common base and common collector. Reason (R): In all the configurations, emitter is common to both input and output. of a power supply to maintain constant voltage output under variation of either in input voltage or load current. Reason (R): Zener diode in forward bias is used as a voltage regulator.

8. Assertion (A): Output characteristics of a transistor in CE configuration is obtained by plotting IB against VBE keeping VCE constant at different values. Reason (R): The output characteristics is similar to the characteristics of a forward biased p-n junction diode. 9. Assertion (A): Semiconductors do not obey Ohm’s law. Reason (R): V- characteristic of semiconductors is linear. 10. Assertion (A): When a p-n junction diode is reverse biased, a feeble reverse current flows known as reverse saturation current. Reason (R): In reverse bias condition, the minority carrier can cross the junction.

[C] COMPETENCY BASED QUESTIONS

I. Based on following passage answer questions from 1-5. Based on the understanding operations of rectifiers answer the questions. Diodes can be used for rectifying an ac voltage (restricting the ac voltage to one direction) with the help of a capacitor or a suitable filter, a dc voltage can be obtained.  (CUET 2022,8th August) X Rectifier

ac

+

C

RL dc

t

(2) V0

t

(3)

7. Assertion (A): Voltage regulation is the measure of ability

RL

100 turn

6. Assertion (A): There are three different configurations of

Vo

70

NOR NOR

B

PHYSICS



V0

t

(4) V0

t

3. Arrange the device in the process of conversion of alternating current to direct current in proper sequence. (A) Filter (B) Rectifier (C) Transformer Choose the correct answer from the options given below: (1) A, B, C (2) C, B, A (3) B, C, A (4) A, C, B 4. In half wave rectification, if the input frequency is 60 Hz, the output frequency is (1) 60 Hz (2) 120 Hz (3) 100 Hz (4) 50 Hz 5. Capacitor is used in the rectifier circuit because (1) It helps to amplify the rectifier output (2) It helps to convert DC into AC. (3) It eliminates ripples/fluctuations and gives almost constant DC voltage output (4) The output voltage obtained by using capacitor as input filter is much higher than the peak voltage of the rectified voltage. I. Based on following passage answer questions from 6-10.

Y

1. Input and output voltage signal for a device X are shown below. Identify X. v  t

(1) Zener diode (3) Half wave rectifier

t

(2) Full wave rectifier (4) Light emitting diode

ND

NA

119

ELECTRONIC DEVICES



The donor concentration is high at n side and acceptor concentration is high at p-side. There is no acceptor concentration at n-side and no donor concentration at p-side. For simplicity, let us assume that there is no hole concentration in n-side and no electron concentration in p-side. After diffusion and drift of electrons and holes, depletion region is formed around the junction where there is no free charge. There are only immobile ions: +ve ions in n-side and –ve ions in p-side.

(2)

(3)

6. Which of the following figure properly depicts the acceptor (1)

concentration?

(4)

(2)

8. Which following figure properly depicts the carrier (1) (3)

(2)

(4) (3)

7. Which of the following figure properly depicts the donor (1)

concentration?

(4)

concentration?

120 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

9. Which of the following figure properly depicts the immobile charge concentration in depletion region? (1)

X

(2)

X

(3)

X

(4)

X

10. (1) (2) (3) (4)

The depletion region Will increase when the p-n junction is reverse biased Will decrease when the p-n junction is forward biased Will never change for any type of bias Both (1) and (2)

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (3)

3. (2)

4. (3)

5. (3)

6. (4)

7. (1)

8. (3)

9. (2)

10. (3)

11. (3)

12. (3)

13. (4)

14. (3)

15. (3)

16. (1)

17. (1)

18. (1)

19. (1)

20. (4)

21. (2)

22. (2)

23. (1)

24. (2)

25. (2)

26. (4)

27. (3)

28. (2)

29. (4)

30. (2)

8. (2)

9. (3)

10. (1)

8. (1)

9. (1)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (3)

3. (1)

4. (1)

5. (4)

6. (3)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (3)

3. (2)

4. (1)

5. (3)

6. (1)

7. (2)

121

ELECTRONIC DEVICES

ANSWERS WITH EXPLANATIONS [A] MULTIPLE CHOICE QUESTIONS

1. Option (1) is correct. Explanation: As the intensity increases, the current increases. 2. Option (3) is correct. Explanation: Zener diode is mainly used as a voltage regulator. Rectifier is a major application of p-n junction diode. Output of rectifier is pulsating d.c. Hence, it requires filtration which is done by using capacitors. Amplifier is a major application of bi-junction transistors. 3. Option (2) is correct. Explanation: Under forward biasing the applied potential difference causes a field which acts opposite to the potential barrier. This results in reducing the potential barrier , and hence the width of depletion layer decreases. 4. Option (3) is correct. Explanation: Holes and electrons are the majority and minority carriers in p-type semiconductor respectively. Electrons and holes are majority and minority carriers in n-type semiconductor respectively. When pentavalent impurity is doped, n-type semiconductor is formed. When trivalent element is doped p-type semiconductor is formed. 5. Option (3) is correct V 1 Explanation: Ic = c = = 0.25 mA Rc 4 × 103

β=

IC IB

I c 0.25 × 10−3 Or, I = = 10−6 A B= β 250 Input signal voltage = Vi = IBRB = 10–6 × 103 = 1 mV 6. Option (4) is correct. Explanation: 0.01 atom of Arsenic is doped out of 106 atom of intrinsic semiconductor. In 5 × 1028 atom of intrinsic semiconductor number of Arsenic 5 × 1028 = 5 × 1020 106 0.01 1 Arsenic atom donates 1 excess electron. So, number of excess electrons = 5 × 1020 = ne From law of mass action, n n = n i 2 atoms doped =

e h

n 2 (1.5 × 1016 ) 2 nh = i = ne 5 × 1020 = 0.45 × 1012 = 4.5 × 1011/m3 So, number of holes = 4.5 × 1011/m3 7. Option (1) is correct Explanation: There should be arrow only on emitter and it should be away from the base. 8. Option (3) is correct. A' A• •Y B' B•

Explanation: When A = 0 , B = 0, A’ = 1, B’ = 1 , So, Y = 1 When A = 0, B = 1, A’ = 1, B’ = 0 , So, Y = 1 When A = 1, B = 0, A’ = 0, B’ = 1 , So, Y = 1 When A = 1, B = 1, A’ = 0, B’ = 0 , So, Y = 0 It is the characteristic of an OR gate. 9. Option (2) is correct. Explanation: From the given circuit diagram Since, VA > VB D1 is forward biased Since, VA > VC D2 is reverse biased Since, VC > VD D3 is forward biased Since, VB > VD D4 is reverse biased So, only D1 and D3 are forward biased. 10. Option (3) is correct. Explanation: A • Y •

ov • B

Y = B.1 = B So, the output is HIGH when B is HIGH and LOW when B is LOW. 11. Option (3) is correct. Explanation: Net gain of the cascaded amplifier system, Anet = A1 × A2 × A3 Or, Anet = 5 × 10 × 20 = 1000

Now,

Anet =

Output voltage Input voltage

Output voltage 0.1 So, Output voltage = 1000 × 0.1 = 100V 12. Option (3) is correct. Explanation: I–V Characteristics of photodiodes are drawn in 3rd quadrant since photodiodes are operated in reverse biased condition. 13. Option (4) is correct Explanation: RC = 2 kW V0 = 2V RB = 1 kW b = 100 Vin = IBRB …(i)

Or,

1000 =

Vin = Equating (i) and (ii)

I C RB  β

IB =

IC β

Putting IC =

VC RC



VC 2 = = 10µA RCβ 2000 × 100



IB =

So, Input voltage = IBRB = 10 × 10–6 × 103 = 10 mV = 0.01 V

…(ii)

122 Oswaal CUET (UG) Chapterwise Question Bank 14. Option (3) is correct Explanation: In all cases cathode is at lower potential than that of anode. So, the diode is forward biased in all the three cases. 15. Option (3) is correct. Explanation: When A = 0, B = 0, then inputs of the last NAND gate is 1, 1. So, Y = 0. When A = 0, B = 1, then inputs of the last NAND gate is 1, 0. So, Y = 1. When A = 1, B = 0, then inputs of the last NAND gate is 0, 1. So, Y = 1. When A = 1, B = 1, then inputs of the last NAND gate is 0, 0. So, Y = 1. This is the characteristics of an OR gate. 16. Option (1) is correct. A

A+B=A+B=OR Gate Y

A+B

B 17. Option (1) is correct Explanation: For Transistor: Base region must be thin and lightly doped as iB = 5% of iE Collector Junction should be reverse biased and Emitter Junction should be forward biased. 18. Option (1) is correct Explanation: ni2 = ne . nh ni  intrinsic charge density ne  electron density nh  hole density 19. Option (1) is correct Explanation:

R I

IL I2

IZ

E Here ⇒ As  ⇒

RL

IZ = 40 mA VZ = 12 V VZ 12 = 300Ω RZ = = I Z 40 × 10−3 VZ = VL (As they are connected in parallel) VL = ILRL

21. Option (2) is correct Explanation: Rectified converts AC signal to unidirectional pulsating DC signal. 22. Option (2) is correct Explanation: From Kirchhoff ’s voltage law VA – VB = 0.2 × 10–3 × 5 × 103 + 0.3 + 0.2 × 10–3 × 5 × 103 \ VA – VB = 1 + 0.3 + 1 = 2.3 V 23. Option (1) is correct Explanation: Since, potential barrier of Ge diode is 0.3 V, it will be the easier path for current to flow. Ge diode is forward biased. So, for ideal diode it may be considered as a short circuit path. In this path potential drop will be 0.3 V. So, VA will be 24 – 0.3 = 23.7 V 24. Option (2) is correct. Explanation: When p-n junction is in forward bias, it compresses or decreases the depletion layer so as the potential barrier. In reverse bias potential barrier increases. 25. Option (2) is correct Explanation : Atoms of semiconductor are binded by covalent bonds between the atoms of same or different types. Due to thermal agitation when an electron leaves its position and become free, then it leaves a vacancy, this vacancy in the bond (covalent) is called hole. 26. Option (4) is correct Explanation: In semiconductor, the density of charge carriers (electron, holes) is very small, so its resistance is high. When temperature increases the density of charge carriers increases which increases the conductivity. As temperature of semiconductor increases, the speed of free electrons increases which decreases the relaxation time. As the density of charge carriers is small, so there is a small effect on decrease of relaxation time. 27. Option (3) is correct. Explanation: From the given circuit diagram, we can see that this is AND Gate and their output will be multiple of its input. So, the output of C = AB, and D = AB. The Resultant output of E is equal to the AND multiplication of C and D, Output as inputs of E. We know that NOT gate is also known as inverter that means if we apply ‘0’ as an input then output will be ‘1’ and vice versa. Resultant Truth Table A

B

A

C

D

E

0

0

1

0

0

0

0

1

1

0

1

1

I = 40 + 30 = 70 mA

1

0

0

0

0

0

19 − 12 7 × 103 = −3 70 × 10 70 = 100 W 20. Option (4) is correct Explanation: Zener Diode  Acts as voltage Regulator Photo Diode  Allows light to fall on it. A photodiode is a PN-Junction diode that consumes light energy to produce an electric current. Solar cell  Converts sunlight directly into Electricity without external bias. LED  forward biased device used to produce light.

1

1

0

1

0

1

VL 12 = = 30 mA RL 400 I = IZ + IL

⇒

Now,

IL =

R=

PHYSICS

28. Option (2) is correct. Explanation: At 0 K temperature, all electrons of semiconductor are immovable from their shell as they do not have sufficient energy. So, no free electron is available as charge carrier. This makes the semiconductors to behave like insulators. 29. Option (4) is correct. Explanation: When a semiconductor having 4 valence electrons is doped with an element having 3 valence electrons, then an extra hole is generated and the semiconductor becomes p-type extrinsic semiconductor.

123

ELECTRONIC DEVICES

When a semiconductor having 4 valence electrons is doped with an element having 5 valence electrons, then an extra electron is available and the semiconductor becomes n-type extrinsic semiconductor 30. Option (2) is correct. Explanation: The size of the emitter is larger than the base and slightly smaller than the collector. Size of base is the smallest and that of the collector is the largest.

[B] ASSERTION REASON QUESTIONS

1. Option (1) is correct. Explanation: It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. When the diode is forward biased, electrons are sent from n  p (where they are minority carriers) and holes are sent from p  n (where they are minority carriers). At the junction boundary, the concentration oF minority carriers increases as compared to the equilibrium concentration. Thus, at the junction boundary on either side of the junction, excess minority carriers recombine with majority carriers near the junction. On recombination, the energy is released in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. 2. Option (3) is correct. Explanation: On doping, number of free carriers increases, hence the electrical conductivity of a semiconductor increases on doping. So, the assertion is correct. Doping may increase the number of electrons or holes in the semiconductor depending on the type of doping. If the dopant is donor type, then number of electrons increases and if the dopant is acceptor type, then the number of holes increases. So, the reason is incorrect. 3. Option (1) is correct. Explanation: At 0 K, there is no electron in conduction band of an intrinsic semiconductor. Intrinsic semiconductor at this temperature behaves like an insulator. As temperature increases, the electrons gain sufficient energy and jump from valence band to conduction band. So, assertion is correct. Free charge carriers being available, the resistivity decreases. As temperature increases, more and more free charge carriers become available and the resistivity decreases more and more. So, reason is also correct. 4. Option (1) is correct. Explanation: For normal operation of a transistor, the emitterbase junction is forward biased and the collector-base junction is reversed biased. So, the dynamic resistance of the EB junction is very low whereas the dynamic resistance of the CB junction is very high. So, the assertion is correct. In a transistor, a current from a low-resistance input circuit is transferred to a high-resistance output circuit with almost unchanged magnitude. This results in a power gain. In fact, the name “transistor” is coined from “transfer resistor”. Hence, the reason is also correct. 5. Option (4) is correct. Explanation: In the given circuit, When A = 0, B = 0, Y = 0 When A = 0, B = 1, Y = 0 When A = 1, B = 0, Y = 0 When A = 1, B = 1, Y = 1 It is the characteristics of an AND gate. So, the given circuit behaves like an AND gate. Hence, the assertion is incorrect.

When two inputs of a NOR gate are shorted then and connected to logic 1, then both the inputs are logic 1 and the output is logic 0. When two inputs of a NOR gate are shorted then and connected to logic 0, then both the inputs are logic 0 and the output is logic 1. This is the characteristics of a NOT gate. Hence, the reason is correct. 6. Option (3) is correct. Explanation: There are three different configurations of a transistor – Common emitter, common base and common collector. So, the assertion is correct. In common emitter configuration emitter is common to both input and output. In common base configuration base is common to both input and output. In common collector configuration collector is common to both input and output. Hence the reason is incorrect. 7. Option (3) is correct. Explanation: Voltage regulation is the measure of ability of a power supply to maintain constant voltage output under variation of either in input voltage or load current. So, assertion is correct. Reverse biased Zener diode maintains constant voltage across load as long as the supply voltage is more than Zener voltage. So, in a voltage regulator, Zener diode in reverse bias is used as a voltage regulator. So, reason is incorrect. 8. Option (2) is correct. Explanation: Input characteristics of a transistor in CE configuration is obtained by plotting IB against VBE keeping VCE constant at different values. So, the assertion is incorrect. The input characteristics is similar to the characteristics of a forward biased p-n junction diode. So, the reason is also incorrect. 9. Option (3) is correct. Explanation: Semiconductors do not obey Ohm’s law. So, the assertion is correct. V-I characteristic of semiconductor is nonlinear. Hence, the reason is incorrect. 10. Option (1) is correct. Explanation: When a p-n junction is reverse biased, then the majority charge carriers cannot cross the junction. So, no forward current flows. But in reverse direction, a feeble current flows which is known as reverse saturation current. So, the assertion is correct. In p-side there are few electrons as minority charge carriers and in n-side, there are few holes as minority charge carriers. In reverse bias condition, the holes at n-side feel a pull exerted by the negative polarity of the voltage source connected to the p-side. Similarly, the electrons at p-side feel a pull exerted by the positive polarity of the voltage source connected to the n-side. So, these minority carrier now can cross the junction and give rise to a feeble current in the opposite direction. Hence, the reason is also correct.

[C] COMPETENCY BASED QUESTIONS

1. Option (3) is correct. Explanation: Half wave rectifier rectifies one one-half cycle. 2. Option (3) is correct. Explanation: As tapping in transformer is not in centre, the peak values of the output corresponding to positive and negative half cycles will be unequal. Since, the ratios of number of turns is 70:30, therefore output peaks corresponding to positive and negative half cycles will be in the ratio 70:30.

124 Oswaal CUET (UG) Chapterwise Question Bank 3. Option (2) is correct. Explanation: The input a.c. is to be first stepped down to the required amplitude using a transformer. Now this a.c. is to be rectified by using diodes. Since, after rectification, the output is pulsating d.c., it is to be smoothened using a filter. 4. Option (1) is correct. Explanation: Only one half cycle of each full cycle is rectified in half wave rectifier. So, output frequency is equal to the input frequency. 5. Option (3) is correct. Explanation: Output of a rectifier is called pulsating d.c. It is not a pure d.c. It contains ripples. Ripples are removed by using capacitor as a filter. 6. Option (1) is correct. Explanation: The acceptor concentration is high at p-side. There is no acceptor concentration at n-side.

24

PHYSICS

7. Option (2) is correct. Explanation: The donor concentration is high at n side. There is no donor concentration at p-side. 8. Option (1) is correct. Explanation: In p-side there are only holes, there is no electron concentration. In n-side there are only electrons, there is no hole concentration. At depletion region, there is no charge carrier. 9. Option (1) is correct. Explanation: After recombination of free electrons and holes, around the depletion region, there will be positively charged ions in the n-side and negatively charged ions in p-side. 10. Option (4) is correct. Explanation: In forward bias, applied voltage does not support potential barrier. As a result, the depletion layer width decreases. In reverse bias, applied voltage supports potential barrier. As a result, depletion layer widens.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

10

COMMUNICATION SYSTEMS

  Revision Notes

Elements of a communication system: Communication is the method of transmitting and receiving information/data. Block diagram of a generalized communication system: Communication system Information source

Message Input transducer signal

Transmitter

Transmitted signal

Channel

Message Output Received Receiver transducer signal signal

User of Information

Noise

Information source: Information source generates signal in  various forms which are to be transmitted.  Input transducer: Converts the non-electrical information into electrical message signal.  Transmitter: Converts the message signal into a form, suitable for transmission over a medium. It also provides interface to match with the transmission medium.  Channel: Channel is the medium through which the signal reaches from source to destination. Channel may be of two types: Guided and Unguided. Guided medium are different types of cables used for communication – telephone cable, twisted pair cable, coaxial cable, optical fibre cable, waveguide. Unguided medium is space.  Noise: Noise is the undesirable electric signals which creep into the communication system and interfere with the desired signal. Source of noise may be external (due to electrical sparks, discharges etc.) or internal (from transmitter, receiver and channel).  Receiver: Receiver receives the transmitted signal along with the noise and recovers the original message signal.  Output transducer: Converts the electrical message signal into the original form. Bandwidth of signals (speech, TV and digital data): In a communication system, the message signal can be voice, music, picture or computer data. Each of these signals has different ranges of frequencies. The type of communication system needed for a given signal depends on the band of frequencies which is considered essential for the communication process.  For speech signals, frequency range 300 Hz to 3100 Hz is considered adequate. Therefore, speech signal requires a

bandwidth of 2800 Hz (300Hz – 3100 Hz) for commercial telephonic communication.  To transmit music, an approximate bandwidth of 20 kHz is required because of the high frequencies produced by the musical instruments.  Video signals for transmission of pictures require about 4.2 MHz of bandwidth. A TV signal contains both voice and picture and is usually allocated 6 MHz of bandwidth for transmission.  Digital data requires infinite bandwidth theoretically. For practical purposes, as the bandwidth is limited, the received waves become a distorted version of the transmitted one. If the bandwidth is large enough the information is not lost. Bandwidth of transmission medium: The commonly used transmission media are wire, free space and fiber optic cable.  Coaxial cable is a widely used wire medium, which offers a bandwidth of approximately 750 MHz.  Communication through free space using radio waves takes place over a very wide range of frequencies: from a few hundreds of kHz to a few GHz. This range of frequencies is further subdivided and allocated for various services. Service

Frequency band

Comments

Standard AM broadcast

540-1600 kHz

FM broadcast

88-108 MHz

Television

54-72 MHz

VHF

76-88 MHz

TV

174-216 MHz

UHF

420-890 MHz

TV

and

S k

Bandwidth

.

+

+

VC



Square Law Device V2

Television

c

Mobile to base station Base station to Mobile

V1

t

Vam

R

ave

v es

ew

Spa c

Sky w a

•

V2

• C

Low pass filter

RL Message signal

Second Level

Trace the Mind Map

is

.

Third Level

when superimposed

H H

,

,

k

a

z

• Propagation of ratio waves through atmosphere following the surface of the earth. • Attenuation is high for ground wave propagation and increase with increase in frequency. • Large antenna is required for ground wave. Length of antenna is directly proportional to the wavelength of the eletromagnetic wave. • Suitable for low and medium frequency i.e., from few hundred kHz to 2 MHz.

In transmitted signal, only amplitude vary while frequency remains same.

The process of transmitting the

the

First Level

54–72 MHz VHF 76–88 MHz TV 176–216 MHz UHF 420–890 MHz TV

Downlink

k

Bandwidth

Uplink

m

Vam



Vm

Square law device •



Band Pass Filter

Mo n io lat u d 

126 Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

127

COMMUNICATION SYSTEMS

Cellular Mobile Radio

896-901 MHz

Mobile to base station

840-935 MHz

Base station to mobile

Satellite 5.925-6.425 GHz Up link Communication 3.7-4.2 GHz Down link  Optical communication using fibers is performed in the frequency range of 1 THz to 1000 THz (microwaves to ultraviolet). An optical fiber can offer a transmission bandwidth in excess of 100 GHz. Scan to know Propagation of electromagnetic waves more about this topic in the atmosphere, sky, and space wave propagation: Earth’s atmosphere plays a vital role in propagation of electromagnetic wave. There are three ways of such propagation: Propagation of electromagnetic  Ground wave propagation: waves (i) Propagation of radio waves through atmosphere following the surface of the earth are called ground wave propagation. (ii) Ground wave propagation is suitable for low and medium frequency, i.e., from few hundred KHz to 2 MHz. (iii) Ground wave propagates along the curvature of the earth. (iv) Attenuation is high for ground wave propagation and increases with increase in frequency. (v) Large antenna is required for ground wave. Length of antenna is directly proportional to the wavelength of the electromagnetic wave.  Sky wave propagation: (i) Propagation of radio waves through atmosphere and reflection from ionosphere is known as sky wave propagation. (ii) Radio waves of frequency between 3 MHz to 30MHz are suitable for sky wave propagation. (iii) For reflection from ionosphere there is a critical frequency in sky wave propagation. Radio waves having frequencies below the critical frequency are reflected by the ionosphere. Radio waves having frequencies above the critical frequency penetrate the ionosphere and do not return to earth. The values of critical frequencies are found to be 4MHz, 5MHz and 6MHz and 8MHz for D, E, F1 and F2 layers of ionosphere respectively.  Space wave propagation: (i) Propagation of radio waves in a straight line from transmitting antenna to the receiving antenna is known as space wave propagation. Space waves are used for line-ofsight (LOS) communication as well as satellite communication. (ii) Radio waves of frequency above 40MHz are suitable for space wave propagation. At these frequencies, the antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. Due to curvature of earth, the signal may get blocked after certain distance. To increase the range the placement of antennas should be at sufficiently high points. The maximum line-of-sight distance dM between the two antennas having heights hT and hR above the earth is given by dM =

2 RhT +

RhR (R is the radius of earth).

Need for modulation:  Size of antenna for low frequency transmission is minimum λ   × wavelength  which is impractical. 4 

 Low frequency signal gets easily Scan to know attenuated and hence cannot be transmitted more about over a long distance. this topic  Transmission of two or more low frequency signals cannot be transmitted simultaneously.  For low frequency signal transmission, Need for interference of two or more signals is Modulation unavoidable.  For low frequency signal transmission, noise creeps into the signal.  At the receiving end, quality of low frequency signal deteriorates. For all these reasons it is preferable to put low frequency signal on a high Scan to know frequency signal and this process is known more about this topic a modulation. Production and detection of an amplitude-modulated wave: Amplitude modulation: The amplitude Amplitude of a high frequency carrier wave is modulation modulated according to the amplitude of the message signal. m Vm

Modulating signal

to

t

c Vc

t

Carrier wave Vc Amplitude modulated wave

0

Vm

Vmax t

Vmin

In amplitude modulation the amplitude of the carrier is varied in accordance with the information signal. Let c(t) = Ac sin ωct represent carrier wave and m(t) = Am sin ωmt represent the message or the modulating signal where ωm = 2πfm is the angular frequency of the message signal. The modulated signal cm(t) can be written as cm(t) = (Ac + Am sin ωm t)sinωct

 A  = Ac 1 + m sin ωmt sinωct  .....(i) A c   From Eq. (i), we can write, cm(t) = Acsin ωct + μAcsin ωmtsinωct.....(ii) Here μ = Am/Ac is the modulation index; in practice, μ is kept ≤ 1 to avoid distortion. 1 Using the trigonometric relation sin AsinB = [cos (A – B) 2 – cos(A + B)], we can write cm(t) of Eq. (ii) as cm (= t ) Acsinωct +

µAc µA cos ( ωc − ωm ) t − c cos ( ωc + ωm ) t 2 2

Here ωc – ωm and ωc + ωm are respectively called the lower side and upper side frequencies. The modulated signal now consists of the carrier wave of frequency ωc plus two sinusoidal waves each with a frequency slightly different from, known as side bands.

128 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

Frequency spectrum:  Carrier

VC mVC 2

LSB i=e' (fc–fm)

0

USB

(fc+fm) Frequency

Production of amplitude modulated wave: Vm +

V1

+

Square V 2 Law Device

Band Vam Pass Filter

VC m(t) Am sin m t (Modulating Signal)

+

x(t)

Square Law y(t) Device

c(t) AC sin C t (carrier)

Bandpass Filter Centred At C

AM Wave

Bx(t)+ 2 Cx(t)

Fig: Block diagram of a simple modulator for obtaining an AM signal Here the modulating signal Amsinωmt is added to the power amplifier which provides the necessary power and then carrier signal Acsinωct to produce the signal x ( t ) . This signal the modulated signal is fed to an antenna of appropriate size for radiation. = x ( t ) Amsinωmt + Acsinωct is passed through a square law device which is a non-linear device which produces an output Detection of amplitude modulated wave: = y ( t ) Bx ( t ) + Cx 2 ( t ) ....(i) R Square law where B and C are constants. Thus, • • device y= ( t ) BAm sin ωmt + BAc sin ωc

+C  Am2 sin 2ωmt + Ac2sin 2ωct + 2 Am Ac sin ωmtsinωct  ...(ii)  BAm sin ωmt + BAc sin ωct = +

CAm2 CA2 CA2 + Ac2 − m cos2ωmt − c cos2ωct 2 2 2

Vam

V2

C

RL Message signal

• Low pass

filter +CA A cos ( ωc − ωm ) t − CAm Ac cos ( ωc + ωm ) t .....(iii) m c where sin 2 A= (1 − cos2 A ) / 2 In Eq.(iii), there are sinusoids of frequencies ωm , 2ωm , ωc , 2ωc , ωc − ωm and ωc + ωm . As shown in Fig. above the signal is passed through a band pass filter which rejects dc and the sinusoids of frequencies ωm , 2ωm and ωc and retains the frequencies ωc , ωc − ωm and ωc + ωm . It is to be mentioned that the modulated signal cannot be Fig: Block diagram of a detector for AM signal. transmitted as such. The modulator is to be followed by a The modulated signal is passed through a square law device (a rectifier) to produce the output shown below.

This envelope of signal (b) is the message signal. In order to retrieve the message signal, the signal is passed through an envelope detector (which may consist of a simple RC low pass filter circuit).

129

COMMUNICATION SYSTEMS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. Two antennas A and B have maximum line of sight range RAB = 45.5 km. If the height of antenna A is 32 m then the height of antenna B is: (Given radius of earth is = 6400 km) (1) 25 m (2) 100 m (3) 50 m (4) 75 m  (CUET 2023, 7th June) 2. A carrier wave of peak voltage 24 V is used to transmit a message signal. What should be the peak Voltage of the modulating signal in order to have a modulation index of 70%? (1) 16.8 V (2) 2.9 V (3) 18 V (4) 16.8 × 102 V  (CUET 2023, 7th June) 3. Arrange the following elements of a generalised communication system in the sequential order: (A) Channel (B) Information source (C) Transmitter (D) Receiver (E) User of information Choose the correct answer from the options given below: (1) (B), (D), (A), (C), (E) (2) (E), (D), (C), (B), (A) (3) (A), (B), (C), (D), (E) (4) (B), (C), (A), (D), (E)  (CUET 2023, 7th June) 4. A 50 m long antenna is mounted on a 200 m tall building. The complex can become transmission tower for waves with wave length (1) ∼ 400 m (2) ∼ 200 m (3) ∼ 20m (4) ∼ 4 m  (CUET 2022, 8th August) 5. A block diagram of a typical receiver in a communication system is as shown in figure. The various components in correct sequence is:

A. B. C. D. E. (1) (2) (3) (4)  6.

Detector Amplifier 1 IF stage Amplifier 2 Receiving Antenna Choose the correct answer from the options given below: (i)-E, (ii)-B, (iii)-D, (iv)-C, (v)-A (i)-E, (ii)-B, (iii)-C, (iv)-A, (v)-D (i)-E, (ii)-D, (iii)-A, (iv)-B, (v)-C (i)-E, (ii)-A , (iii)-B, (iv)-C, (v)-D (CUET 2022, 23rd August) In the detection of amplitude modulated wave, the carrier frequency is usually changed to a long frequency by (1) Amplifier (2) Detector (3) IF stage (4) Receiving Antenna  (CUET 2022, 23rd August) 7. Some wireless communications are given as following. Select the correct arrangement of them into their increasing order of frequencies from low to high band range.

(i) Satellite communication (ii) Sky wave communication (iii) A.M. Broad Cast (iv) F.M. Broad Cast Choose the correct answer from the options given below: (1) (iii) → (i) → (iv) → (ii) (2) (iii) → (iv) → (i) → (ii) (3) (iii) → (ii) → (i) → (iv) (4) (iii) → (ii) → (iv) → (i)  (CUET 2022, 30th August) 8. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of modulating signal in order to have a modulation index of 75%? (1) 3 V (2) 6 V (3) 9 V (4) 12 V (CUET 2022, 30th August)  9. Mode of communication used for T.V. transmission (1) Space wave communication (2) Sky wave propagation (3) Ground wave propagation (4) Both (1) and (2) 10. Modulation is a process of superimposition of (1) Low frequency signal on high frequency signal (2) High frequency signal on low frequency signal (3) Low frequency signal on low frequency signal (4) None of the above 11. The maximum peak to peak voltage of an amplitude modulated wave is 30mV and the minimum peak to peak voltage is 10mV. The modulation index is (1) 10% (2) 50% (3) 5% (4) 100% 12. Which of the following waves may be transmitted through sky wave propagation? (1) Wave of frequency 15kHz (2) Wave of frequency 15MHz (3) Wave of having 15Hz (4) All of the above 13. An electromagnetic wave of frequency 7 MHz will, penetrate (1) D layer of ionosphere (2) D and E layers of ionosphere (3) D, E and F1 layers of ionosphere (4) D, E, F1 and F2 layers of ionosphere 14. The principle of communication through optical fibre is (1) Reflection (2) Refraction (3) Total internal reflection (4) Total internal refraction 15. Which of the following carrier signals can be used for amplitude modulation of an audio signal 1KHz? (1) Wave of frequency 1 KHz (2) Wave of frequency 0.5 KHz (3) Wave of frequency 1 MHz (4) None of the above 16. Over modulated amplitude modulated waves have modulation index (1) 0 (2) > 1 (3) < 1 (4) 1 17. Number of side bands of an amplitude modulated wave is (1) 1 (2) 2 (3) 3 (4) 0

130 Oswaal CUET (UG) Chapterwise Question Bank 18. The last stage of production of AM waves using square law device is a (1) Band pass filter (2) Band reject filter (3) Low pass filter (4) High pass filter 19. The last stage of AM wave detector using square law device is a (1) Band pass filter (2) Band reject filter (3) Low pass filter (4) High pass filter 20. Three waves A, B and C of frequencies 1,600 kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication? (1) A is transmitted via space wave while B and C are transmitted via sky wave. (2) A is transmitted via ground wave, B via sky wave and C via space wave. (3) B and C are transmitted via ground wave while A is transmitted via sky wave. (4) B is transmitted via ground wave while A and C are transmitted via space wave.  [NCERT Exemp. Q. 15.1, Page 98] 21. A 100 m long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with λ (1) ∼ 400 m. (2) ∼ 25 m. (3) ∼ 150 m. (4) ∼ 2,400 m.  [NCERT Exemp. Q. 15.2, Page 98] 22. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be (1) 1.003 MHz and 0.997 MHz. (2) 3,001 kHz and 2,997 kHz. (3) 1,003 kHz and 1,000 kHz. (4) 1 MHz and 0.997 MHz.  [NCERT Exemp. Q. 15.4, Page 99] 23. A message signal of frequency wm is superposed on a carrier wave of frequency wc to get an amplitude modulated wave (AM). The frequency of the AM wave will be (1) wm (2) wc (3) (wm + wc)/2 (4) (wm – wc)/2  [NCERT Exemp. Q. 15.5, Page 99] 24. I-V characteristics of four devices are shown in figure.

(i)



(ii)

(iii)



(iv)

Identify devices that can be used for modulation: (1) (i) and (iii). (2) only (iii). (3) (ii) and some regions of (iv). (4) All the devices can be used.  [NCERT Exemp. Q. 15.6, Page 99] 25. Frequency spectrum of amplitude modulated wave is represented by

PHYSICS

(1)



(2)



(3)



(4)

26. A basic communication system consists of (A) transmitter, (B) information source, (C) user of information, (D) channel and (E) receiver. Choose the correct sequence in which these are arranged in a basic communication system: (1) ABCDE. (2) BADEC. (3) BDACE. (4) BEADC.  [NCERT Exemp. Q. 15.8, Page 100] 27. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true? (1) The bandwidth required for amplitude modulation is 1.5MHz. (2) The bandwidth required for amplitude modulation is 6 kHz. (3) The bandwidth required for amplitude modulation is 3 MHz. (4) The bandwidth requirement is infinite. 28. If a TV telecast has to cover 200km, the height of the antenna should be (1) 6250m (2) 3.125m (3) 15.625m (4) 3125m 29. Transducer converts (1) signals of various physical forms to electrical signals, and vice versa. (2) signals of various physical forms to electrical signals only. (3) signals of electrical signals to various physical forms only. (4) None of the above.

131

COMMUNICATION SYSTEMS

30. Which figure shows overmodulated AM wave?

Fig. (a)

Fig. (b)

(1) Fig. (a) (3) Fig. (c)

Fig. (c) (2) Fig. (b) (4) Fig. (a) and (c)

[B] ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements – Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true.1. 1. Assertion (A): A radio wave having frequency greater than 30MHz is transmitted through sky wave propagation. Reason (R): Ionosphere reflects electromagnetic waves having frequencies greater than certain critical frequency. 2. Assertion (A): For speech signals, frequency range 300 Hz to 3100 Hz is considered adequate. Reason (R): The bandwidth of the speech signal is 3400 Hz. 3. Assertion (A): A very large antenna of impractical size is required to transmit an electromagnetic wave of frequency 20KHz. Reason (R): The antenna should have a size comparable to the wavelength of the signal. 4. Assertion (A): The square law device used in an amplitude modulated wave detector basically behaves like a half wave rectifier. Reason (R): The envelope detector used in an amplitude modulated wave detector is basically a low pass RC filter. 5. Assertion (A): A communication satellite is placed above the ionosphere. Reason (R): Radio wave reflection from satellite is sky wave propagation. 6. Assertion (A): The amplitude modulated signal consists of the carrier wave of frequency wc and four sinusoidal waves each with a frequency slightly different from wc , known as side bands. Reason (R): The amplitudes of side bands are equal and less than the amplitude of the carrier wave. 7. Assertion (A): High frequency waves are suitable for ground wave propagation. Reason (R): Ground wave propagates along the curvature of the earth.

8. Assertion (A): When the height of the tower used for LOS communication is increased by 21%, the maximum distance coverage decreases by 10%. Reason (R): Maximum distance coverage through LOS communication decreases as the height of tower increases. 9. Assertion (A): In undermodulated amplitude modulation, then the carrier amplitude will neither fall to zero, nor will it rise to twice its unmodulated level. Reason (R): Amplitude modulation with modulation index 100% is called perfect modulation. 10. Assertion (A): Medium used for ground wave propagation, sky wave propagation, space wave propagation air. Reason (R): Unguided media transport electromagnetic waves without using a physical conductor.

[C] COMPETENCY BASED QUESTIONS

I. Based on following passage answer questions from 1-5. A village is situated at a distance 50km from a TV transmission station. The TV transmission antenna is 32m high whereas the receiving antenna in the village is 30m high. Villagers are unable to watch the TV programmes. 1. What should be the minimum height of the receiving antenna to receive the signal in the village? (1) 70m (2) 40m (3) 32m (4) 56m 2. What is the type of wave propagation taking place in this TV signal transmission? (1) Ground wave propagation (2) Sky wave propagation (3) Space wave propagation (4) None of these 3. What type of wave propagation is suitable if a flying aircraft desire to receive the TV signal? (1) Ground wave propagation (2) Sky wave propagation (3) Space wave propagation (4) None of these 4. What may be the alternate wave propagation to receive the TV signal in the village? (1) Ground wave propagation (2) Sky wave propagation (3) Optical fibre cable communication (4) Both (2) and (3) 5. Why is it not possible to receive the signal by ground wave propagation technique? (1) Ground wave propagation is suitable for low and medium frequency only. (2) Attenuation is high for ground wave propagation and increases with increase in frequency. (3) Large antenna, whose dimension is proportional to the wavelength of the wave, is required for ground wave. (4) All of the above. II. Based on following passage answer questions from 6-10. Modulation index is a measure of extent of modulation done on a carrier signal. It describes how much the modulated variable of the carrier signal varies around its unmodulated level. It is defined differently in each modulation scheme. In Amplitude modulation, it is defined as the ratio of the amplitude of modulating signal to that of the carrier signal. m = AC / Am It is also defined as m = (Amax – Amin) / (Amax + Amin) Its value is kept less than 1 to avoid overmodulation which leads to distortions in the modulated signal and makes it very hard to demodulate and extract the modulating signal.

132 Oswaal CUET (UG) Chapterwise Question Bank

PHYSICS

6. A carrier wave frequency 1.5 MHz and amplitude 50 V

is modulated by a sinusoidal wave of frequency 10 kHz producing 50% amplitude modulation. The amplitude of the AM wave is (1) Amax = 50V, Amin = 25V (2) Amax = 50V, Amin = 50V (3) Amax = 75V, Amin = 25V (4) Amax = 25V, Amin = 25V 7. A carrier wave frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz producing 50% amplitude modulation. The frequency spectrum of the AM wave is

(1)

(3)

(4)

8. A carrier wave frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz. The bandwidth is (1) 20kHz (2) 20MHz (3) 1kHz (4) 0.02kHz

(2)

9. A carrier wave frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz. Which of the following diagrams correctly displays 60% amplitude modulation?

10. When the amplitude modulation as shown below occur, the modulation index is

(1) = 100% (3) < 100%

(2) > 100% (4) = 0%

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (1)

3. (4)

4. (2)

5. (2)

6. (3)

7. (4)

8. (3)

9. (4)

10. (1)

11. (2)

12. (2)

13. (3)

14. (3)

15. (3)

16. (2)

17. (2)

18. (1)

19. (3)

20. (2)

21. (1)

22. (1)

23. (2)

24. (3)

25. (2)

26. (2)

27. (2)

28. (4)

29. (1)

30. (2)

8. (2)

9. (1)

10. (1)

8. (1)

9. (1)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (2)

2. (2)

3. (1)

4. (1)

5. (3)

6. (4)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (3)

3. (3)

4. (4)

5. (4)

6. (3)

7. (1)

133

COMMUNICATION SYSTEMS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Option (3) is correct. Explanation: d M = Or, Or,

dM =

2 RhT +

2 RhR

2 × 6400 × ( 32 / 1000 ) +

45.5 = 20.2 +

2 × 6400 × hR

12800 × hR

Or, 25.3 = 113 hR ∴ hR = 0.0484 km = 48.4 m 2. Option (1) is correct. Explanation: Or, 0.70 = Am / 24 ∴ Am = 0.7 x 24 = 16.8 V

3. Option (4) is correct. Explanation:

4. Option (2) is correct Explanation: Wavelength transmission depends on the length of the antenna. If l is the wavelength and l is the length of the antenna, then the relation between them is l = 4l \ l = 4 × 5 = 200 m 5. Option (2) is correct Explanation:

6. Option (3) is correct Explanation: In IF stage the original information bearing signal is extracted from carrier wave. 7. Option (4) is correct Explanation: The Amplitude Modulated (AM radio) uses frequency range is 535 to 1605 kHz. Lowest frequency band. Sky wave propagation, uses frequency range 3 MHz to 30 MHz FM radio broadcast uses the frequency range 88 MHz to 108 kHz. For satellite communication uses the frequency range 5.925 to 6.425 GHz for uplink and 3.7 to 4.2GHz for downlink. Highest frequency band. 8. Option (3) is correct Explanation: V Modulation index = m Vc 75 Vm = 100 12 \ Vm = 9 V 9. Option (4) is correct. Explanation: T.V. transmission is either line of sight signal propagation or by reflection of signal from satellite. 10. Option (1) is correct. Explanation: Modulation is a process of superimposition of low frequency signal, known as modulating signal, on high frequency signal known as carrier signal. 11. Option (2) is correct. Explanation: Vmax = 30/2 = 15 mV Vmin = 10/2 = 5mV m = (Vmax – Vmin ) / (Vmax + Vmin) Or, m = (15 – 5)/(15 + 5) ∴ m = 0.5 = 50%

Or,

12. Option (2) is correct. Explanation: Waves having frequency range 3 – 30 MHz are suitable for sky wave propagation. 13. Option (3) is correct. Explanation: The values of critical frequency are found to be 4 MHz, 5 MHz and 6 and 8 MHz for D, E, F1 and F2 layers of ionosphere respectively. Hence, the wave penetrates all the layers of ionosphere except F2. 14. Option (3) is correct. Explanation: Optical fibre works on the principle of total internal reflection. 15. Option (3) is correct. Explanation: Frequency of carrier signal should be higher than the frequency of modulating signal. 16. Option (2) is correct Explanation: Amplitude modulated waves having modulation index greater than 1 are over modulated and distorted. 17. Option (2) is correct. Explanation: The equation of an AM wave is mVc mVc cos ( ωc − ωm ) t − cos ( ωc + ωm ) t . V= Vcsinωct + am 2 2 It shows that there are two side bands:

mVc cos ( ωc − ωm ) t and 2

mVc cos ( ωc + ωm ) t . 2 18. Option (1) is correct. Explanation: Vm

+

Σ

+

VC

V1

Square Law Device

V2

Band Pass Filter

Vam

134 Oswaal CUET (UG) Chapterwise Question Bank 19. Option (3) is correct. Explanation: R Square law device •

• C

V2

Vam

RL Message signal

• Low pass filter

20. Option (2) is correct. Explanation: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation: (i) Ground wave propagation (Frequency range: Few hundred KHz to 2MHz) (ii) Sky wave propagation (Frequency range: 3–30 MHz) (iii) Space wave propagation (Frequency range: Above 40 MHz) So, A is transmitted via ground wave, B via sky wave and C via sky wave. 21. Option (1) is correct. Explanation: Length of antenna = L = 100 m Wavelength of the wave which can be transmitted = λ = 4L = 400m 22. Option (1) is correct. Explanation: Side band frequencies of amplitude modulated wave are (fc + fm) and (fc – fm). (fc + fm) is the frequency of upper side band (USB). (fc – fm) is the frequency of lower side band (LSB).

PHYSICS

According to the problem, frequency of the carrier signal is fc = 1 MHz and frequency of speech signal = 3 kHz = 3 × 10−3 MHz = 0.003 MHz So, the frequencies of side bands are (1 + 0.003) = 1.003 MHz and (1 − 0.003) = 0.997MHz. 23. Option (2) is correct. Explanation: The process of changing the amplitude of carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM). In AM, frequency of the carrier wave remains unchanged or we can say that the frequency of modulated wave is equal to the frequency of carrier wave. Now, according to the problem, frequency of carrier wave is fc . Thus, the amplitude modulated wave also has frequency wc. 24. Option (3) is correct. Explanation: A square law modulator is the device which can produce modulated waves by the application of the message signal and the carrier wave. Square law modulator is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices. And by (ii) corresponds to square law device which shows non-linear relations. Some part of (iv) also follows square law. 25. Option (2) is correct. Explanation: The equation of amplitude modulated wave is: mVc mVc cos ( ωc − ωm ) t − cos ( ωc + ωm ) t 2 2 It has the carrier wave, LSB and USB. Frequency of LSB is less than carrier wave frequency and frequency of USB is more than the carrier wave frequency. Hence, LSB will be situated at the left side and USB will be situated at the right side of the carrier. V= Vcsinωct + am

26. Option (2) is correct Explanation: A basic communication system consists of an information source, a transmitter, a link (channel) and a receiver or a communication system is the set-up used in the transmission and reception of information from one place to another. The whole system consists of several elements in a sequence. It can be represented as the diagram given below:

27. Option (2) is correct. Explanation: Bandwidth requirement modulation is 2fm = 2 × 3KHz = 6KHz. 28. Option (4) is correct. Explanation: d = 2Rh Or,

h = d2/2R

Or,

h=

for

amplitude

( 200 ×10 )

3 2

2 × 6400 × 103 ∴ h = 3125 m 29. Option (1) is correct Explanation: At the transmitting end, signal from information source may not be in electrical form. So, transducer converts it into electrical form. At the receiving end, it is required to convert the electrical signal into various physical forms. Transducer does this conversion.

30. Option (2) is correct. Explanation: An over-modulated AM wave is a signal that has been modulated so significantly that the carrier wave clips or distorts. This can happen when: (i) The modulation index is greater than 1 (ii) The modulating signal voltage is much greater than the carrier voltage (iii) The instantaneous level of the modulating signal exceeds the value necessary to produce 100% modulation of the carrier

[B] ASSERTION REASON QUESTIONS

1. Option (2) is correct. Explanation: The frequency range for sky wave propagation is 3 – 30 MHz. Hence, A radio wave having frequency greater than 30MHz cannot be transmitted through sky wave propagation. The assertion is incorrect.

135

COMMUNICATION SYSTEMS

Ionosphere reflects electromagnetic waves having frequencies lower than certain critical frequency. Electromagnetic wave having frequency greater than the critical frequency penetrates ionosphere. Hence, the reason is also incorrect. 2. Option (2) is correct. Explanation: For speech signals, frequency range 300 Hz to 3100 Hz is considered adequate. The assertion is correct. The bandwidth of the speech signal is 3100 Hz – 300Hz = 2800 Hz. So, the reason is incorrect. 3. Option (1) is correct. Explanation: The antenna should have a size comparable to the wavelength of the signal. It should be at least of size of one fourth of the wavelength. So, the reason is correct. For electromagnetic wave of frequency 20KHz is 15km. So, the size of antenna should be at least 15km/4 = 3.75 km which is impractical. So, the assertion is also correct. 4. Option (1) is correct. Explanation: The output of the square law device used in an amplitude modulated wave detector is as shown:

So, it basically behaves like a half wave rectifier. Assertion is correct. The envelope detector used in an amplitude modulated wave detector is basically a low pass RC filter as shown: R envelop detector Square law device • Vam

• V2

C

RL Message signal

• Low pass filter

So, reason is also correct. 5. Option (3) is correct. Explanation: Communication satellite is placed approximately at a height 36000 km from the surface of earth whereas ionosphere is stretched upto 650km approximately. So, assertion is correct. Radio wave reflection from satellite is space wave propagation since it has to penetrate ionosphere. So, reason is incorrect 6. Option (4) is correct. Explanation: An amplitude modulated wave is represented as: mVc mVc Vam = Vcsin wct + cos(wc – wm)t – cos(wc + wm)t 2 2 So, it has the carrier wave of frequency wc and two side bands of frequency slightly different from wc . Hence, the assertion is incorrect. The amplitude of each side bands is mVc/2. Hence, the amplitudes are equal and less than the amplitude of the carrier wave. Hence, the reason is correct. 7. Option (3) is correct. Explanation: Ground wave propagates along the curvature of the earth. Hence, the reason is correct. Ground wave propagation is suitable for low and medium frequency i.e., from few hundred KHz to 2 MHz. As the

frequency increases, attenuation increases. Hence, high frequency waves are not suitable for ground wave propagation. So, the assertion is incorrect. 8. Option (2) is correct. Explanation: Maximum distance coverage = d = 2Rh So, d∝ h So, maximum distance coverage through LOS communication increases as the height of tower increases. So, the reason is incorrect. Since, d∝ h So,

d2/d1 =

d2/d1 =

h2 h1 1.21h1 h1

d2/d1 = 1.21 = 1.1 So, maximum distance coverage increases by 10%. Hence, the assertion is also incorrect. 9. Option (1) is correct. Explanation: Modulation index, in an amplitude modulation, indicates the maximum and minimum variation of carrier amplitude during modulation. When modulation index is less than 100% then the carrier amplitude neither falls to zero, nor rises to twice its unmodulated level. Hence, it is called undermodulated. So, the assertion is correct. When modulation index is 100%, then the carrier amplitude falls to zero, and rises to twice its unmodulated level during modulation. When modulation index is more than 100%, then the modulated wave gets distorted and it is called overmodulated. So, amplitude modulation with modulation index 100% is the perfect modulation since there is maximum variation of carrier amplitude without any distortion. Hence, reason is also correct. 10. Option (1) is correct. Explanation: Ground wave propagation and sky wave propagation takes place in troposphere. Space wave propagation involves ionosphere. There is physical conductor carrying the signal. Hence, the air medium is called unguided medium. Hence, the assertion and reason both are correct.

[C] COMPETENCY BASED QUESTIONS

1. Option (1) is correct. Explanation: The maximum distance coverage = 50 km Maximum distance coverage = 2 RhT + 2 RhR 2 × 6400 × 1000 × 32 + 2 × 6400 × 1000 × hR Or, 50000 = 20238.6 + 2 × 6400 × 1000 × hR

Or, 50000 =

Or, 885740929.96 = 2 × 6400 × 1000 × hR ∴ hR = 69.2 m ≈ 70 m 2. Option (3) is correct Explanation: Propagation of radio waves in a straight line from transmitting antenna to the receiving antenna is known as space wave propagation. Space waves are used for line-ofsight (LOS) communication as well as satellite communication. 3. Option (3) is correct Explanation: Propagation of radio waves in a straight line from transmitting antenna to the receiving antenna is known as space wave propagation. Space waves are used for line-ofsight (LOS) communication as well as satellite communication.

136 Oswaal CUET (UG) Chapterwise Question Bank 4. Option (4) is correct. Explanation: Ground wave propagation will not be suitable since high frequency wave will get absorbed by earth crust easily. Sky wave propagation, i.e., by reflecting the signal from ionosphere is possible. Signal may be propagated through high bandwidth optical fibre cable. 5. Option (4) is correct. 6. Option (3) is correct. Explanation: Amplitude of unmodulated carrier wave is 50V. Modulation index = 50% So, the variation of amplitude unmodulated wave is 50V ± 50% of 50V So, Amax = 50V + 25V = 75V, Amin = 50V – 25V = 25V 7. Option (1) is correct. Explanation: In the frequency spectrum, there are the carrier wave and two side bands.

PHYSICS

fLSB = fc – fm = 1.5 × 106 – 104 = 1.49 MHz fUSB = fc + fm = 1.5 × 106 + 104 = 1.51 MHz Amplitudes of the side bands are equal and less than the amplitude of the carrier wave. 8. Option (1) is correct. Explanation: Band width = 2 × fm = 2 × 10KHz = 20KHz 9. Option (1) is correct. Explanation: Amplitude of unmodulated carrier wave is 50V. Modulation index = 60% So, the variation of amplitude unmodulated wave is 50V ± 60% of 50V So, Amax = 50V + 30V = 80V, Amin = 50V – 30V = 20V 10. Option (2) is correct. Explanation: When the modulation index is greater than 100%, then the amplitude modulated wave is distorted.

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