71 JEE Main Physics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers 4th Edition 9788194761617, 8194761611

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Contents

Topic 2 : Dimensions of Physical Quantities





Topic 2 : Non-uniform Motion



Topic 3 : Relative Velocity



P-26 – P-35



Topic 1 : Vectors



Topic 2 : Motion in a Plane with Constant Acceleration



Topic 3 : Projectile Motion



Laws of Motion

P-36 – P-53





Topic 4 : Relative Velocity in Two Dimensions & Uniform Circular Motion



Topic 1 : Ist, IInd & IIIrd Laws of Motion



Topic 2 : Motion of Connected Bodies, Pulley & Equilibrium of Forces



Topic 3 : Friction



Work, Energy and Power

P-54 – P-75





Topic 4 : Circular Motion, Banking of Road



Topic 1 : Work



Topic 2 : Energy



Topic 3 : Power



System of Particles and Rotational Motion





Topic 4 : Collisions

6.

Motion in a Plane





Topic 4 : Motion Under Gravity

5.



P-14 – P-25

Topic 1 : Distance, Displacement & Uniform Motion

4.



Motion in a Straight Line





Topic 3 : Errors in Measurements

3.



P-1 – P-13

Topic 1 : Unit of Physical Quantities

2.



Physical World, Units and Measurements





1.

Topic 1 : Centre of Mass, Centre of Gravity & Principle of Moments Topic 2 : Angular Displacement, Velocity and Acceleration Topic 3 : Torque, Couple and Angular Momentum Topic 4 : Moment of Inertia and Rotational K.E. Topic 5 : Rolling Motion

P-76 – P-112

P-113 – P-130



Gravitation



7.

Topic 1 : Kepler’s Laws of Planetary Motion



Topic 2 : Newton’s Universal Law of Gravitation



Topic 3 : Acceleration due to Gravity



Topic 4 : Gravitational Field and Potential Energy



Topic 5 : Motion of Satellites, Escape Speed and Orbital Velocity 

Mechanical Properties of Solids



8.

Topic 1 : Hooke’s Law & Young’s Modulus



Topic 2 : Bulk and Rigidity Modulus and Work Done in Stretching a Wire 

Mechanical Properties of Fluids



9.

Topic 2 : Fluid Flow, Reynold’s Number and Bernoulli’s Principle



Topic 3 : Viscosity and Terminal Velocity



Topic 4 : Surface Tension, Surface Energy and Capillarity 



P-155 – P-168



Topic 1 : Termometer & Termal Expansion



Topic 2 : Calorimetry and Heat Transfer



Topic 3 : Newton’s Law of Cooling 



11. Termodynamics

P-169 – P-185



Topic 1 : First Law of Termodynamics



Topic 2 : Specifc Heat Capacity and Termodynamical Processes



Topic 3 : Carnot Engine, Refrigerators and Second Law of Termodynamics 



12. Kinetic Teory

P-186 – P-198



Topic 1 : Kinetic Teory of an Ideal Gas and Gas Laws



Topic 2 : Speed of Gas, Pressure and Kinetic Energy



Topic 3 : Degree of Freedom, Specifc Heat Capacity, and Mean Free Path 



13. Oscillations

P-199 – P-218



Topic 1 : Displacement, Phase, Velocity and Acceleration in S.H.M.



Topic 2 : Energy in Simple Harmonic Motion



Topic 3 : Time Period, Frequency, Simple Pendulum and Spring Pendulum







Topic 4 : Damped, Forced Oscillations and Resonance

14. Waves

P-139 – P-154

Topic 1 : Pressure, Density, Pascal’s Law and Archimedes’ Principle

10. Termal Properties of Matter



P-131 – P-138

Topic 1 : Basic of Mechanical Waves, Progressive and Stationary Waves Topic 2 : Vibration of String and Organ Pipe

P-219 – P-234



Topic 3 : Beats, Interference and Superposition of Waves



15. Electric Charges and Fields

P-235 – P-253





Topic 4 : Musical Sound and Doppler’s Efect



Topic 1 : Electric Charges and Coulomb’s Law



Topic 2 : Electric Field and Electric Field Lines



16. Electrostatic Potential and Capacitance

Topic 2 : Electric Potential Energy and Work Done in Carrying a Charge



P-281 – P-311





Topic 3 : Capacitors, Grouping of Capacitor and Energy Stored in a Capacitor



Topic 1 : Electric Current, Drif of Electrons, Ohm’s Law, Resistance and Resistivity



Topic 2 : Combination of Resistances



Topic 3 : Kirchhof ’s Laws, Cells, Termo e.m.f. Electrolysis



Topic 4 : Heating Efect of Current



P-312 – P-339



Topic 1 : Motion of Charged Particle in Magnetic Field



Topic 2 : Magnetic Field Lines, Biot-Savart’s Law and Ampere’s Circuital Law



Topic 3 : Force and Torque on Current Carrying Conductor



P-340 – P-347





Topic 4 : Galvanometer and its Conversion into Ammeter and Voltmeter

19. Magnetism and Matter

Topic 1 : Magnetism, Gauss’s Law, Magnetic Moment, Properties of Magnet



Topic 2 : Te Earth Magnetism, Magnetic Materials and their Properties







Topic 3 : Magnetic Equipment

20. Electromagnetic Induction

P-348 – P-360



Topic 1 : Magnetic Flux, Faraday’s and Lenz’s Law







Topic 2 : Motional and Static EMI and Application of EMI

21. Alternating Current





Topic 5 : Wheatstone Bridge and Diferent Measuring Instruments

18. Moving Charges and Magnetism



P-254 – P-280

Topic 1 : Electrostatic Potential and Equipotential Surfaces

17. Current Electricity







Topic 3 : Electric Dipole, Electric Flux and Gauss’s Law

Topic 1 : Alternating Current, Voltage and Power Topic 2 : AC Circuit, LCR Circuit, Quality and Power Factor Topic 3 : Transformers and LC Oscillations

P-361 – P-376

P-377 – P-388





22. Electromagnetic Waves

Topic 1 : Electromagnetic Waves, Conduction and Displacement Current



23. Ray Optics and Optical Instruments

P-389 – P-414





Topic 2 : Electromagnetic Spectrum



Topic 1 : Plane Mirror, Spherical Mirror and Refection of Light



Topic 2 : Refraction of Light at Plane Surface and Total Internal Refection



Topic 3 : Refraction at Curved Surface Lenses and Power of Lens



Topic 4 : Prism and Dispersion of Light



24. Wave Optics

P-415 – P-432





Topic 5 : Optical Instruments



Topic 1 : Wavefront, Interference of Light, Coherent and Incoherent Sources



Topic 2 : Young’s Double Slit Experiment



25. Dual Nature of Radiation and Matter

P-433 – P-448





Topic 3 : Difraction, Polarisation of Light and Resolving Power



Topic 1 : Matter Waves, Cathode and Positive Rays



Topic 2 : Photon, Photoelectric Efect X-rays and Davisson-Germer Experiment P-449 – P-460





26. Atoms

Topic 1 : Atomic Structure and Rutherford’s Nuclear Model



Topic 2 : Bohr’s Model and the Spectra of the Hydrogen Atom P-461 – P-472





27. Nuclei

Topic 1 : Composition and Size of the Nuclei



Topic 2 : Mass-Energy Equivalence and Nuclear Reactions



Topic 3 : Radioactivity 



28. Semiconductor Electronics : Materials, Devices and Simple Circuits

Topic 1 : Solids, Semiconductors and P-N Junction Diode



Topic 2 : Junction Transistor







Topic 3 : Digital Electronics and Logic Gates

29. Communication Systems

P-473 – P-493

P-494 – P-500

Topic 1 : Communication Systems MT-1 – MT-8

Mock Test 2 with Solutions

MT-9 – MT-16





Mock Test 1 with Solutions

Opening/ Closing Rank for TOP NITs & List of NITs in India Opening/ Closing Rank for Top NITS

College Name NIT Trichy NIT Rourkela NIT Surathkal NIT Warangal NIT Calicut NIT Kurukshetra NIT Durgapur MNIT Allahabad NIT Silchar MNIT Jaipur

OR/CR ORF CR OR CR OR CR OR CR OR CR OR

CSE 2060 5317 2253 9420 960 3181 978 2341 2201 10222 2268

ECE 5325 8011 8571 12009 3378 5608 2919 2919 8023 14769 8320

ME 4154 12970 11662 20304 6315 11788 4340 10209 10629 20480 11195

EE/EEE 5708 10353 4084 19168 3456 6801 5270 8152 9703 18966 9454

CR

6170

12067

18115

16273

OR CR OR CR OR CR OR

5611 12095 1449 4051 8699 23882 1148

12509 16098 3600 7128 17899 40841 3881

14511 22753 5884 11145 21851 49215 9277

13595 19325 5879 8790 32579 56958 4119

CR

3831

7868

11426

9179

List of NITs in India

After IITs, NITs form the second layer of topmost engineering Institutes in India Rank Of (Amongst NITs)

Name

State

NIRF Score

NIRF Ranking

1

NIT Trichy

Tamil Nadu

61.62

10

2

NIT Rourkela

Odisha

57.75

16

3

NIT Karnataka

Karnataka

55.25

21

4

NIT Warangal

Telangana

53.21

26

5

NIT Calicut

Kerala

52.69

28

6

V-NIT

Maharashtra

51.27

31

7

NIT Kurukshetra

Haryana

47.58

41

7

MN-NIT

Uttar Pradesh

47.49

42

8

NIT Durgapur

West Bengal

46.47

46

9

NIT Silchar

Assam

45.61

51

10

M-NIT

Rajasthan

45.20

53

11

SV-NIT

Gujarat

41.88

58

12

NIT Hamirpur

Himachal Pradesh

41.48

60

13

MA-NIT

Madhya Pradesh

40.98

62

14

NITIE

Maharashtra

40.48

66

15

NIT Meghalaya

Meghalaya

40.32

67

16

NIT Agartala

Tripura

39.53

70

17

NIT Raipur

Chattisgarh

39.09

74

18

NIT Goa

Goa

37.06

87

FAQs - Frequently Asked Questions at the end of each paper of the examination, given to them at the start of the paper.

ANSWER:

QUESTION: During examination can I change my answers?

1.

Institute of Science (IISc), Bangalore

2.

Indian Institute of Petroleum and Energy (IIPE), Visakhapatnam

ANSWER: Candidate will have the option to change previously saved answer of any question, anytime during the entire duration of the test.





QUESTION: Which Colleges other than IITs accept JEE Advanced scores?



3. Indian Institutes of Science Education and Research (IISER), Bhopal

4. Indian Institutes of Science Education and Research (IISER), Mohali

QUESTION: How can I change a previously saved answer during the CBT of JEE (Advanced)-2018?



5. Indian Institutes of Science Education and Research (IISER), Kolkata

6. Indian Institutes of Science Education and Research (IISER), Pune

7. Indian Institutes of Science Education and Research (IISER), Thiruvananthapuram

QUESTION: Will I be given a printout/hard copy of the questions papers along with my responses to questions in Paper-I and Paper-II after the completion of the respective papers?



8. Indian Institute of Space Science and Technology (IIST), Thiruvananthapuram

ANSWER: To change the answer of a question that has already been answered and saved, first select the corresponding question from the Question Palette, then click on “Clear Response” to clear the previously entered answer and subsequently follow the procedure for answering that type of question.

ANSWER: No. QUESTION: How will I be getting a copy of the questions papers and my responses to questions in Paper-I and Paper-II?

QUESTION: If I am absent in one of the papers (Paper 1, Paper 2), will my result be declared?

ANSWER: The responses of all the candidates who have appeared for both Paper 1 and Paper 2, recorded during the exam, along with the questions of each paper, will be electronically mailed to their registered email ids, by Friday, May 25, 2018, 10:00 IST.



9. Rajiv Gandhi Institute of Petroleum Technology (RGIPT), Rae Bareli

ANSWER: NO. You will be considered absent in JEE (Advanced)-2018 and the result will not be prepared/declared. It is compulsory to appear in both the papers for result preparation. QUESTION: Do I have to choose my question paper language at the time of JEE (Advanced)-2018 registration? ANSWER: NO. There is no need to indicate question paper language at the time of JEE (Advanced)-2018 registration. Candidates will have the option to choose their preferred language (English or Hindi), as the default language for viewing the questions, at the start of the Computer Based Test (CBT) examination of JEE (Advanced)-2018. QUESTION: Can I change the language (from English to Hindi and vice versa) of viewing the questions during the CBT of JEE (Advanced)-2018? ANSWER: Questions will be displayed on the screen of the Candidate in the chosen default language (English or Hindi). Further, the candidate can also switch/toggle between English or Hindi languages, as the viewing language of any question, anytime during the entire period of the examination. The candidate will also be having the option of changing default question viewing language anytime during the examination. QUESTION: Will I be given rough sheets for my calculations during the CBT of JEE (Advanced)-2018? ANSWER: Yes, you will be given “Scribble Pad” (containing blank sheets, for rough work) at the start of every paper of JEE (Advanced)-2018. You can do all your calculations inside this “Scribble Pad”. Candidates MUST submit their signed Scribble Pads

QUESTION: Suppose two candidates have same JEE (Advanced)-2018 aggregate marks. Will the two candidates be given the same rank? ANSWER: If the aggregate marks scored by two or more candidates are the same, then the following tie-break policy will be used for awarding ranks: Step 1: Candidates having higher positive marks will be awarded higher rank. If the tie breaking criterion at Step 1 fails to break the tie, then the following criterion at Step 2 will be followed. Step 2: Higher rank will be assigned to the candidate who has obtained higher marks in Mathematics. If this does not break the tie, higher rank will be assigned to the candidate who has obtained higher marks in Physics. If there is a tie even after this, candidates will be assigned the same rank. QUESTION: I have read in newspapers that for the academic year 2018-2019, supernumerary seats for female candidates would be there in IITs. Does this mean that the non-females will get reduced number of seats in IITs in 2018? ANSWER: A decision has been taken at the level of the IIT Council to, inter alia, improve the gender balance in the undergraduate programs at the IITs from the current (approximately) 8% to 14% in 2018-19 by creating supernumerary seats specifically for female candidates, without any reduction in the number of seats that was made available to non-female candidates in the previous academic year (i.e. academic year 2017-2018).

1

P-1

Physical World, Units and Measurements

Physical World, Units and Measurements 6.

TOPIC 1 Unit of Physical Quantities 1.

2.

The density of a material in SI unit is 128 kg m–3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is: [10 Jan. 2019 I] (a) 40 (b) 16 (c) 640 (d) 410 A metal sample carrying a current along X-axis with density Jx is subjected to a magnetic field Bz (along z-axis). The electric field Ey developed along Y-axis is directly proportional to Jx as well as Bz. The constant of proportionality has SI unit [Online April 25, 2013] 3 2 As m m2 m (a) (b) (c) (d) As As m3 A

Dimensions of Physical TOPIC 2 Quantities 3.

4.

5.

1 E 1 , y= and z = are B CR m0 e0 defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and e 0 , m 0 , - free space permittivity and permeability respectively. Then : [Sep. 05, 2020 (II)] (a) x, y and z have the same dimension. (b) Only x and z have the same dimension. (c) Only x and y have the same dimension. (d) Only y and z have the same dimension. Dimensional formula for thermal conductivity is (here K denotes the temperature : [Sep. 04, 2020 (I)] –2 –2 (a) MLT K (b) MLT K–2 –3 (c) MLT K (d) MLT–3 K–1 A quantity x is given by (IFv2/WL4) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of : [Sep. 04, 2020 (II)] (a) planck’s constant (b) force constant (c) energy density (d) coefficient of viscosity The quantities x =

7.

8.

Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is : [Sep. 03, 2020 (II)] (a) ML2T–2 (b) ML0T–3 (c) M2L0T–1 (d) MLT–2 If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be : [Sep. 02, 2020 (I)] 2 –1 2 –3 (a) FA V (b) FA V (c) FA2V–2 (d) FA–1V0 If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is : [Sep. 02, 2020 (II)] (a) [P2AT –2] (b) [PA–1T–2] (c) [PA1/ 2T-1 ]

9.

(d) [P1/ 2AT-1]

Which of the following combinations has the dimension of electrical resistance (Î0 is the permittivity of vacuum and m0 is the permeability of vacuum)? [12 April 2019 I] (a)

m0 e0

(b)

m0 e0

(c)

e0 m0

e0 (d) m 0

10. In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units ? [10 April 2019 II] –3 –2 8 4 –1 –2 4 2 (a) [M L T A ] (b) [M L T A ] (c) [M–2 L0 T–4 A–2] (d) [M–2 L–2 T6 A3] 11. In SI units, the dimensions of

Î0 is: [8 April 2019 I] m0

(a) A–1TML3 (b) AT2 M–1L–1 (c) AT–3ML3/2 (d) A2T3 M–1L–2 12. Let l, r, c and v represent inductance, resistance, capacitance and voltage, respectively. The dimension of l in SI units will be : [12 Jan. 2019 II] rcv – 2 –1 (a) [LA ] (b) [A ] (c) [LTA] (d) [LT2]

P-2

Physics

13. The force of interaction between two atoms is given by æ x2 ö F = ab exp ç ç akT ÷÷ ; where x is the distance, k is the è ø Boltzmann constant and T is temperature and a and b are two constants. The dimensions of b is: [11 Jan. 2019 I] (a) M0L2T–4 (b) M2LT–4 (c) MLT–2 (d) M2 L2 T–2

14. If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be :

[11 Jan. 2019 II]

(a)

V–2A2F–2

V–2A2F2

(b)

(c)

V–4 A–2 F

(d) V–4A2F

hc5 where c is speed of G light, G universal gravitational constant and h is the Planck’s constant. Dimension of f is that of : [9 Jan. 2019 I] (a) area (b) energy (c) momentum (d) volume 16. Expression for time in terms of G (universal gravitational constant), h (Planck's constant) and c (speed of light) is proportional to: [9 Jan. 2019 II]

15. A quantity f is given by f =

(a) (c)

hc5 G

(b)

Gh

(d)

5

c3 Gh

Gh

c c3 17. The dimensions of stopping potential V0 in photoelectric effect in units of Planck’s constant ‘h’, speed of light ‘c’ and Gravitational constant ‘G’ and ampere A is: [8 Jan. 2019 I] (a) hl/3G2/3cl/3 A –1 (b) h2/3c5/3G1/3A –1 (c) h–2/3 e–1/3 G4/3 A–1 (d) h2G3/2C1/3 A–1 2 B , where B is magnetic field and m0 18. The dimensions of 2m 0 is the magnetic permeability of vacuum, is: [8 Jan. 2019 II] (a) MLT–2 (b) ML2T–1 (c) ML2T–2 (d) ML–1T–2 19. The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length? [Online April 15, 2018] 1 ö2

1 æ Gh 2 3 (a) G2hc (b) ç 3 ÷ (c) 2 h 2 c (d) Gh c G èc ø 20. Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be : [Online April 8, 2017]

(a) [ M ]=[ T–1 C–2 h ] (b) [ M ]=[ T–1 C2 h ] –1 –2 –1 (c) [ M ]=[ T C h ] (d) [ M ]=[ T C–2 h ] 21. A, B, C and D are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation AD = C ln (BD) holds true. Then which of the combination is not a meaningful quantity ? [Online April 10, 2016] C AD 2 (b) A2 –B2C2 BD C A (A - C) -C (c) (d) B D 22. In the following 'I' refers to current and other symbols have their usual meaning, Choose the option that corresponds to the dimensions of electrical conductivity : [Online April 9, 2016] (a) M–1 L–3T 3 I (b) M–1 L–3 T3 I2 (c) M–1L3T3 I (d) ML–3 T–3 I2 23. If electronic charge e, electron mass m, speed of light in vacuum c and Planck’s constant h are taken as fundamental quantities, the permeability of vacuum m0 can be expressed in units of : [Online April 11, 2015]

(a)

æ hc ö (b) ç 2 ÷ è me ø

æ h ö (a) ç 2 ÷ è me ø

æ mc 2 ö (d) çç 2 ÷÷ è he ø 24. If the capacitance of a nanocapacitor is measured in terms of a unit ‘u’ made by combining the electric charge ‘e’, Bohr radius ‘a0’, Planck’s constant ‘h’ and speed of light ‘c’ then: [Online April 10, 2015]

æ h ö (c) ç 2 ÷ è ce ø

(a) u =

e2 h a0

(b) u =

(c) u =

e2c ha 0

(d) u =

hc 2

e a0 e2 a 0 hc

25. From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is: [Online April 12, 2014] (a)

(b)

(c)

(d)

ch 2peo2

e2 2pe o Gme2

(me = mass of electron)

m o eo G

c2 he 2

2p m o eo h G ce2

P-3

Physical World, Units and Measurements

26. In terms of resistance R and time T, the dimensions of ratio m of the permeability m and permittivity e is: e [Online April 11, 2014] (a) [RT–2] (b) [R2T–1] (c) [R2] (d) [R2T2] 27. Let [ Î0 ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then: [2013] –1 –3 2 –1 –3 4 (a) Î0 = [M L T A] (b) Î0 = [M L T A2] (c) Î0 = [M1 L2 T1 A2] (d) Î0 = [M1 L2 T1 A] 28. If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t = r 2b s c d a / 2 . It is observed that the d . The value of b s [Online April 23, 2013]

time period is directly proportional to should therefore be : (a)

3 4

(b)

3

2 3 (d) 3 2 29. The dimensions of angular momentum, latent heat and capacitance are, respectively. [Online April 22, 2013]

(c)

(a) ML2 T1A 2 , L2 T -2 , M -1L-2 T 2 (b) ML2 T -2 , L2 T 2 , M -1L-2 T 4 A 2 (c) ML2 T -1, L2 T -2 , ML2 TA 2 (d) ML2 T -1 , L2 T -2 , M -1L-2 T 4 A 2 30. Given that K = energy, V = velocity, T = time. If they are chosen as the fundamental units, then what is dimensional formula for surface tension? [Online May 7, 2012] (a) [KV–2T –2 ] (b) [K2 V2T–2 ] (c) [K2V–2 T–2 ] (d) [KV2 T2 ] 31. The dimensions of magnetic field in M, L, T and C (coulomb) is given as [2008] (a) [MLT–1 C–1] (b) [MT2 C–2] (c) [MT–1 C–1] (d) [MT–2 C–1] 32. Which of the following units denotes the dimension ML2

, where Q denotes the electric charge? [2006] Q2 2 (a) Wb/m (b) Henry (H) (c) H/m2 (d) Weber (Wb) 33. Out of the following pair, which one does NOT have identical dimensions ? [2005] (a) Impulse and momentum (b) Angular momentum and planck’s constant (c) Work and torque (d) Moment of inertia and moment of a force 34. Which one of the following represents the correct dimensions of the coefficient of viscosity? [2004]

-1 -1 (a) éë ML T ùû

-1 (b) éë MLT ùû

-1 -2 (c) éë ML T ùû

-2 -2 (d) éë ML T ùû

35. Dimensions of meaning, are

1 , where symbols have their usual mo eo [2003]

(b) [L-2 T 2 ]

(a) [L-1T]

(c) [L2 T -2 ] (d) [LT -1 ] 36. The physical quantities not having same dimensions are (a) torque and work [2003] (b) momentum and planck’s constant (c) stress and young’s modulus (d) speed and (m o e o ) -1 / 2 37. Identify the pair whose dimensions are equal [2002] (a) torque and work (b) stress and energy (c) force and stress (d) force and work

TOPIC 3 Errors in Measurements 38. A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively : [Sep. 06, 2020 (I)] (a) Negative, 2 mm (b) Positive, 10 mm (c) Positive, 0.1 mm (d) Positive, 0.1 mm 39. The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in æ x ö the density of the sphere is çè ÷ %. If the relative errors 100 ø in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is ______. [NA Sep. 06, 2020 (I)] 40. A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm, The average of these four reading is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as : [Sep. 06, 2020 (II)] (a) (5.5375 ± 0.0739) mm (b) (5.5375 ± 0.0740) mm (c) (5.538 ± 0.074) mm (d) (5.54 ± 0.07) mm 41. A physical quantity z depends on four observables a, b, c and d, as z =

2 2 3 a b

. The percentages of error in the meacd 3 surement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is : [Sep. 05, 2020 (I)]

P-4

42.

43.

44.

45.

46.

47.

48.

Physics

(a) 12.25% (b) 16.5% (c) 13.5% (d) 14.5% Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as : [Sep. 03, 2020 (I)] (a) 2.121 cm (b) 2.124 cm (c) 2.125 cm (d) 2.123 cm The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale betwen 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division) [Sep. 02, 2020 (I)] (a) 3.2 cm (b) 3.21 cm (c) 3.07 cm (d) 2.99 cm If the screw on a screw-gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is: [9 Jan. 2020 I] (a) 0.001 cm (b) 0.02 mm (c) 0.01 cm (d) 0.001 mm For the four sets of three measured physical quantities as given below. Which of the following options is correct? [9 Jan. 2020 II] (A) A1 = 24.36, B1 = 0.0724, C1 = 256.2 (B) A2 = 24.44, B2 = 16.082, C2 = 240.2 (C) A3 = 25.2, B3 = 19.2812, C3 = 236.183 (D) A4 = 25, B4 = 236.191, C4 = 19.5 (a) A4 + B4 + C4 < A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 (b) A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4 (c) A4 + B4 + C4 < A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 (d) A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 < A4 + B4 + C4 A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is: [8 Jan. 2020 II] (a) 5.40% (b) 3.40% (c) 4.40% (d) 2.40% In the density measurement of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01) m, respectively. The error in the measurement of density is: [9 April 2019 I] 3 (a) 0.01 kg/m (b) 0.10 kg/m3 3 (c) 0.013 kg/m (d) 0.07 kg/m3 The area of a square is 5.29 cm 2. The area of 7 such squares taking into account the significant figures is: [9 April 2019 II]

(a) 37cm2 49.

50.

51.

52.

53.

54.

(b) 37.030 cm2

(c) 37.03 cm2 (d) 37.0 cm2 In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to : [8 April 2019 II] (a) 0.7% (b) 0.2% (c) 3.5% (d) 6.8% The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 µm diameter of a wire is: [12 Jan. 2019 I] (a) 50 (b) 200 (c) 100 (d) 500 The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures? [10 Jan. 2019 II] 3 (a) 4264 ± 81 cm (b) 4264.4 ± 81.0 cm3 3 (c) 4260 ± 80 cm (d) 4300 ± 80 cm3 The pitch and the number of divisions, on the circular scale for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 division below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of the sheet is: [9 Jan. 2019 II] (a) 5.755 mm (b) 5.950 mm (c) 5.725 mm (d) 5.740 mm The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is: [2018] (a) 2.5% (b) 3.5% (c) 4.5% (d) 6% The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity A =

P 3Q 2 . RS

The maximum percentage error in the value of A will be [Online April 16, 2018] (a) 8.5% (b) 6.0% (c) 7.5% (d) 6.5% 55. The relative uncertainty in the period of a satellite orbiting around the earth is 10–2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is [Online April 16, 2018] (a) 3×10–2 (b) 10–2 (c) 2 × 10–2 (d) 6 × 10–2

P-5

Physical World, Units and Measurements

56. The relative error in the determination of the surface area of a sphere is a. Then the relative error in the determination of its volume is [Online April 15, 2018] 2 2 3 a (b) a a (c) (d) a 3 3 2 In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is: [Online April 15, 2018] (a) 0.0430 cm (b) 0.3150 cm (c) 0.4300 cm (d) 0.2150 cm The following observations were taken for determining surface tensiton T of water by capillary method : Diameter of capilary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m Using g = 9.80 m/s2 and the simplified relation rhg T= ´ 10 3 N/m, the possible error in surface tension 2 is closest to : [2017] (a) 2. 4 % (b) 10 % (c) 0.15% (d) 1.5% A physical quantity P is described by the relation P = a1/2 b2 c3 d –4 If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be : [Online April 9, 2017] (a) 8% (b) 12% (c) 32% (d) 25% A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line? [2016] (a) 0.70 mm (b) 0.50 mm (c) 0.75 mm (d) 0.80 mm A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s, and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be: [2016] (a) 92 ± 1.8 s (b) 92 ± 3s (c) 92 ± 1.5 s (d) 92 ± 5.0 s

(a)

57.

58.

59.

60.

61.

62. The period of oscillation of a simple pendulum is L . Measured value of L is 20.0 cm known to 1 mm g accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is : [2015] (a) 1% (b) 5% (c) 2% (d) 3%

T = 2p

63. Diameter of a steel ball is measured using a Vernier callipers which has divisions of 0.1 cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as: [Online April 10, 2015] S.No. MS(cm) VS divisions 1.

0.5

8

2. 3.

0.5 0.5

4 6

If the zero error is – 0.03 cm, then mean corrected diameter is: (a) 0.52 cm (b) 0.59 cm (c) 0.56 cm (d) 0.53 cm 64. The current voltage relation of a diode is given by I = ( e1000V T - 1) mA, where the applied voltage V is in

volts and the temperature T is in degree kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA? [2014] (a) 0.2 mA (b) 0.02 mA (c) 0.5 mA (d) 0.05 mA 65. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? [2014] (a) A meter scale. (b) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm. (c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm. (d) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm. 66. Match List - I (Event) with List-II (Order of the time interval for happening of the event) and select the correct option from the options given below the lists: [Online April 19, 2014] List - I (1) Rotation period of earth

List - II (i) 10 5 s

(2) Revolution (ii) 10 7 s period of earth (3) Period of light (iii) 10 –15 s wave (4) Period of (iv) 10 –3 s sound wave (a) (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv)

(b) (1)-(ii), (2)-(i), (3)-(iv), (4)-(iii) (c) (1)-(i), (2)-(ii), (3)-(iv), (4)-(iii) (d) (1)-(ii), (2)-(i), (3)-(iii), (4)-(iv)

P-6

Physics

67. In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potential wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of volmeter will be: [Online April 12, 2014] (a) – 0. 15 volt (b) 0.15 volt (c) 0.5 volt (d) – 0.05 volt 68. An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be: [Online April 9, 2014] (a) 1.7% (b) 2.7% (c) 4.4% (d) 2.27% 69. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is [2012] (a) 6% (b) zero (c) 1% (d) 3% 70. A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the Vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data is [2012] (a) 58.59 degree (b) 58.77 degree (c) 58.65 degree (d) 59 degree 71. N divisions on the main scale of a vernier calliper coincide with (N + 1) divisions of the vernier scale. If each division of main scale is ‘a’ units, then the least count of the instrument is [Online May 19, 2012] (a) a (c)

N ´a N +1

(b)

a N

(d)

a N +1

72. A student measured the diameter of a wire using a screw gauge with the least count 0.001 cm and listed the measurements. The measured value should be recorded as [Online May 12, 2012] (a) 5.3200 cm (b) 5.3 cm (c) 5.32 cm (d) 5.320 cm 73. A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is [2011] (a) 0.052 cm (b) 0.026 cm (c) 0.005 cm (d) 0.52 cm 74. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are [2010] (a) 5, 1, 2 (b) 5, 1, 5 (c) 5, 5, 2 (d) 4, 4, 2 75. In an experiment the angles are required to be measured using an instrument, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half- a degree (= 0.5°), then the least count of the instrument is: [2009] (a) half minute (b) one degree (c) half degree (d) one minute 76. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The magnitude of its momentum is recorded as [2008] (a) 17.6 kg ms–1 (b) 17.565 kg ms–1 (c) 17.56 kg ms–1 (d) 17.57 kg ms–1 77. Two full turns of the circular scale of a screw gauge cover a distance of 1mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is [2008] (a) 3.32 mm (b) 3.73 mm (c) 3.67 mm (d) 3.38 mm

P-7

Physical World, Units and Measurements

1.

(a) Density of material in SI unit, =

=

2.

128 ( 50 g )( 20)

( 25cm ) ( 4) 3

3

128 ( 20) = 40 units 64

=

=

6.

(b) According to question

E y µ J x BZ \ Constant of proportionality K=

Ey

=

BZ J x

C m3 = J x As

I E ] = C (speed of light) and J = Area B (a) We know that

1

Speed of light, c =

m 0 e0

=x

7.

8.

dQ dT = kA dt dx

\x =

[ML2 T -3 ]

IFv 2 WL4

(d) Young's modulus, Y =

F A

stress strain

Dl = FA –1V0 l0

(c) Energy, E µ AaT b Pc E = kAaT b P c

...(i)

M 1L2T -2 = M c L2a + cT b - c by comparison c=1 2a + c = 2 b – c = –2 c = 1, a = 1/2, b = –1

æ dQ ö çè ÷ dt ø Þk= æ dT ö Aç ÷ è dx ø

5.

= M1L0 T -3 .

TL2

Dimension of momentum, P = M 1L1T -1 Dimension of area, A = L2 Dimension of time, T = T 1 Putting these value in equation (i), we get

l l = = Speed Rc t Thus, x, y, z will have the same dimension of speed.

= [MLT -3K -1 ] [L2 ][KL-1 ] (c) Dimension of Force F = M1L1T–2 Dimension of velocity V = L1T–1 Dimension of work = M1L2T–2 Dimension of length = L Moment of inertia = ML2

= M1L-1T -2 = Energy density

where k is a dimensionless constant and a, b and c are the exponents.

\z =

[k ] =

M1L2 T -2

or,

E =y B Time constant, t = Rc = t

(d) From formula,

L3

Energy Time Area Dimension of Energy, E = ML2T–2 Dimension of Time = T Dimension of Area = L2 \ Dimension of Solar constant

ÞY =

Also, c =

4.

M1L-2 T -2

(b) Solar constant =

=

[As

3.

(M1L2 T -2 )(L4 )

128kg

m3 Density of material in new system =

(M1L2 )(M1L1T -2 )(L1T -2 )2

\ E = A1/ 2T -1 P1 9.

(a)

m0 = e0

m 20 = m0 c e0 m 0

m0c ® MLT–2A–2 × LT–1 ML2T–3A–2 Dimensions of resistance 10. (a) X = 5YZ2 X ÞY µ 2 Z

æ 1 ö = c÷ ç è m0e0 ø

...(i)

P-8

Physics

X = Capacitance =

Q Q2 [ A2T 2 ] = = V W [ ML2T -2 ]

X = [M–1L–2T4A2] F IL Z = [MT–2A–1]

[Q F = ILB]

15. (b) Dimension of [h] = [ML2T–1]

[ M -1 L-2T 4 A2 ]

[G] = [M–1L3T–2] Hence dimension of (Using (i))

é e0 ù é e0 ù e 02 ú = e C[LT 11. (d) ê m ú = m e = ê T–1]×[e0] 0 êë m 0 e 0 úû 0 0 ëê 0 úû é ù 1 = Cú êQ m0 e0 êë úû 2 q Q F= 4pe 0 r 2

Þ [e 0 ] =

[ AT ]2 -2

2

= [ A2 M -1 L-3T 4 ]

[ MLT ] ´ [ L ] é e0 ù -1 2 -1 -3 4 \ ê ú = [ LT ] ´ [ A M L T ] m ëê 0 úû

= [ M -1L-2T 3 A2 ] 12. (b) As we know, élù êë r úû = [ T ] and [ cv] = [ AT ] é l ù é T ù \ê ú=ê = é A –1 ù ë rcv û ë AT úû ë û 13. (b) Force of interaction between two atoms, æ -x2 ö ç ÷ ç akT ÷ ø abeè

Since exponential terms are dimensionless é x2 ù \ ê akT ú = M0L0T0 êë úû Þ

so [Y] = [V–4FA2] = [V–4A2F] [C] = [LT–1]

[ MT -2 A-1 ]2 Y = [M–3L–2T8A4]

F=

solving above equations we get: a = – 4, b = 1, c = 2

Z =B=

Y=

b = 1, a + b + c = – 1, –a –2b –2c = – 2

L2 = M0 L0T 0 [a ]ML2T-2

Þ [a] = M–1T2 [F] = [a] [b] MLT–2 = M–1T2[b] Þ [b] = M2LT–4 14. (d) Let [Y] = [V]a [F]b [A]c [ML–1T–2] = [LT–1]a [MLT–2]b [LT –2]c [ML–1T–2] = [MbLa+b+c T–a–2b–2c] Comparing power both side of similar terms we get,

é hC 5 ù é ML2T -1 ù × é L5T -5 ù û ë û ê ú=ë é M -1 L3T -2 ù ê G ú ë û ë û = [ML2T–2] = energy 16. (c) Let t µ Gx hy Cz Dimensions of G = [M–1L3T–2], h = [ML2T–1] and C = [LT–1] [T] = [M–1L3T–2]x[ML2T–1]y[LT–1]z [M0L0T1] = [M–x+y L3x+2y+z T–2x–y–z] By comparing the powers of M, L, T both the sides –x+ y=0 Þx=y 3x + 2y + z = 0 Þ 5x + z = 0 ..... (i) –2x – y –z = 1 Þ 3x + z = –1 ..... (ii) Solving eqns. (i) and (ii),

Gh 1 5 x=y= ,z=- \ tµ 2 2 C5 17. (None)

Stopping potential (V0 ) µ h x I yG Z C r Here,h = Planck’s constant = éë ML2T -1 ùû I = current = [A] G = Gravitational constant = [M–1L3T–2] and c = speed of light = [LT–1] V0 = potential= [ML2T–3A–1] \ [ML2T–3A–1]=[ML2T–1]x [A]y[M–1L3T–2]z[LT–1]r Mx – z; L2x+3z+r; T–x–2z–r; Ay Comparing dimension of M, L, T, A, we get y = –1, x = 0, z = – 1 , r = 5 \ V0 µ h0 I –1G –1C 5 18. (d) The quantity field. Þ

B2 is the energy density of magnetic 2m0

é B 2 ù Energy Force ´ displacement = ê ú= (displacement)3 êë 2m0 úû Volume é ML2T –2 ù –1 –2 =ê ú = ML T 3 êë L úû

P-9

Physical World, Units and Measurements

19. (b) Plank length is a unit of length, lp = 1.616229 × 10–35 m

lp =

hG

c3 20. (a) Let mass, related as M µ Tx Cy hz M1L0 T0 = (T')x (L1T–1)y (M1L2T–1)z M1L0 T0 = Mz Ly + 2z + Tx– y–z z= 1 y + 2z = 0 x – y– z = 0 y = –2 x+ 2–1=0 x = –1 –1 –2 1 M = [T C h ] 21. (d) Dimension of A ¹ dimension of (C) Hence A – C is not possible. 22. (b) We know that resistivity RA θ< l 1 l Conductivity = resistivity < RA lI < ( Q V = RI) VA [L][I] W W < QV< < é [ML2 T,2 ù q it ê ú ´[L2 ] ê [I][T] ú êë úû < [M,1L,3T 3 ][I2 ] < [M,1L,3T3I2 ]

23. (c) Let µ0 related with e, m, c and h as follows. m0 = keambcchd

[MLT–2A–2] = [AT]a [M]b [LT–1]c [ML2T–1]d = [Mb + d Lc + 2d Ta – c – d Aa] On comparing both sides we get a = – 2 ...(i) b+ d = 1 ...(ii) c + 2d = 1 ...(iii) a – c – d = –2 ...(iv) By equation (i), (ii), (iii) & (iv) we get, a = – 2, b = 0, c = – 1, d = 1 é h ù \ [m 0 ] = ê 2 ú ë ce û 24. (d) Let unit ‘u’ related with e, a0, h and c as follows. [u] = [e]a [a0]b [h]c [C]d Using dimensional method, [M–1L–2T+4A+2] = [A1T1]a[L]b[ML2T–1]c[LT–1]d [M–1L–2T+4A+2] = [Mc Lb+2c+d Ta–c–d Aa] a = 2, b = 1, c = – 1, d = – 1 e 2 a0 hc 25. (b) The dimensional formulae of

\

u=

0 0 1 1ù

e = éM L T A ë û

e0 = é M -1L3T 4 A 2 ù ë û G = é M -1L3T -2 ù and me = é M1L0T 0 ù ë û ë û Now,

=

e2 2 pe0 Gm e2

é M0 L0 T1A1 ù ë û

2

2 p é M -1L-3T 4 A 2 ù é M -1L3T -2 ù é M1L0 T 0 ù ë ûë ûë û

2

éT 2A 2 ù ë û = 2p éM -1-1+ 2 L-3+ 3 T 4 - 2 A 2 ù ë û

=

éT 2 A 2 ù 1 ë û = 0 0 2 2ù 2p é 2p M L T A ë û

1 e2 is dimensionless thus the combination 2p 2 pe 0 Gm e2 would have the same value in different systems of units. 26. (c) Dimensions of m = [MLT–2A–2] Dimensions of Î = [M–1L–3T4A2] Dimensions of R = [ML2T–3A–2]

Q

[MLT -2 A -2 ] Dimensions of m = Dimensions of Î [M -1L-3T 4 A 2 ] = [M2L4T–6A–4 ] = [R2] 1 q1q 2 27. (b) As we know, F = 4 pe0 R 2

\

Þ e0 =

q1q2

4pFR 2

Hence, e0 =

C2

[AT]2 = N.m2 [MLT -2 ][L2 ] = [M–1 L–3 T4 A2]

28. (c) 29. (d) Angular momentum = m × v × r = ML2 T–1 Q ML2 T -2 = = L2T–2 m M Charge = M -1L-2 T 4 A 2 Capacitance C = P.d.

Latent heat L =

30. (a) Surface tension, T = T2

F F l T2 = . . l l l T2 -2

=V ) l2 Therefore, surface tension = [KV–2T–2] 31. (c) Magnitude of Lorentz formula F = qvB sin q

(As, F.l = K (energy);

B=

F MLT -2 = = [ MT -1C -1] qv C ´ LT -1

P-10

Physics

32. (b) Mutual inductance =

f BA = I I

[ MT -1Q -1L2 ]

= ML2Q -2 [QT -1 ] 33. (d) Moment of Inertia, I = MR2 [I] = [ML2] r r r Moment of force, t = r ´ F r t = [ L][ MLT -2 ] = [ML2T -2 ] 34. (a) According to, Stokes law, F F = 6phrv Þ h = 6pr v [Henry] =

h=

[ MLT –2 ] –1

Þ h = [ ML-1T -1 ]

[ L][ LT ] 35. (c) As we know, the velocity of light in free space is given by

c=

1 1 \ = e 2 = Z12T 2 m 0 e0 mo eo

1 2 2 mo eo = C [m/s]

= [LT–1]2 = [M0L2T–2] 36. (b) Momentum, = mv = [MLT–1] Planck’s constant, 2 –2 E = [ ML T ] = [ML2T–1] [T –1 ] v r r 37. (a) Work W = F × s = Fs cos q r r Q A × B = AB cos q

h=

= [ MLT -2 ][ L] = [ML2T -2 ] ; r r r Torque, t = r ´ F Þ t = rF sin q r r Q A ´ B = AB sin q

= [ L ] [MLT -2 ] = [ ML2T -2 ] 38. (b) Given : No. of division on circular scale of screw gauge = 50 Pitch = 0.5 mm Least count of screw gauge

=

Pitch No. of division on circular scale

0.5 mm = 1 ´ 105 m = 10 mm 50 And nature of zero error is positive. 39. (1050) =

Density, r =

6 M M Þ r = MD -3 = 3 p V 4 æ Dö pç ÷ 3 è 2ø

æ Dr ö Dm æ DD ö \%ç ÷ = + 3ç = 6 + 3 ´ 1.5 = 10.5% è D ÷ø è rø m æ Dr ö 1050 æ x ö %ç ÷ = %=ç % è 100 ÷ø è r ø 100

\ x = 1050.00 40. (d) Average diameter, dav = 5.5375 mm Deviation of data, Dd = 0.07395 mm As the measured data are upto two digits after decimal, therefore answer should be in two digits after decimal.

\ d = (5.54 ± 0.07) mm 41. (d) Given : Z =

a 2b 2/ 3 cd 3

Percentage error in Z, =

DZ 2 Da 2 Db 1 Dc 3Dd = + + + Z a 3 b 2 c d

2 1 ´ 1.5 + ´ 4 + 3 ´ 2.5 = 14.5%. 3 2 42. (a) Thickness = M.S. Reading + Circular Scale Reading (L.C.) = 2´ 2+

Here LC =

Pitch 0.1 = = 0.002 cm per Circular scale division 50

division So, correct measurement is measurement of integral multiple of L.C. 43. (c) L.C. of vernier callipers = 1 MSD – 1 VSD

9ö æ = ç1 - ÷ ´ 1 = 0.1 mm = 0.01 cm 10 è ø Here 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. Zero error = 7 × 0.1 = 0.7 mm = 0.07 cm. Length of the cylinder = measured value – zero error = (3.1 + 4 × 0.01) – 0.07 = 3.07 cm. 44. (d) When screw on a screw-gauge is given six rotations, it moves by 3mm on the main scale 3 = 0.5mm 6 Pitch 0.5 mm = \ Least count L.C. = CSD 50 1 mm = 0.01 mm = 0.001cm = 100 45. (None) D1 = A1 + B1 + C1 = 24.36 + 0.0724 + 256.2 = 280.6 D2 = A2 + B2 + C2 = 24.44 + 16.082 + 240.2 = 280.7 D3 = A3 + B3 + C3 = 25.2 + 19.2812 + 236.183= 280.7

\

Pitch =

P-11

Physical World, Units and Measurements

D4 = A4 + B4 + C4 = 25 + 236.191 + 19.5 = 281 None of the option matches. 46. (c) Given, Length of simple pendulum, l = 25.0 cm Time of 40 oscillation, T = 50s Time period of pendulum

T = 2p

l g

4p2 l 4p 2l Þg= 2 g T Dg Dl 2DT = + Þ Fractional error in g = g l T Þ T2 =

Dg æ 0.1 ö æ 1ö =ç + 2 ç ÷ = 0.044 ÷ è 50 ø g è 25.0 ø Dg \ Percentage error in g = ´ 100 = 4.4% g 47. (Bonus) d = M = M = Ml -3 V l3 Þ

0.10 æ 0.01ö Dd DM Dl 3 = + 3ç = +3 è 0.10 ÷ø = 0.31 kg/m , 10.00 d M l 48. (d) A = 7 × 5.29 = 37.03 cm2 The result should have three significant figures, so A = 37.0 cm2 49. (d) We have

T = 2p

l 2 l or g = 4p 2 g T

Dg DR DT ´ 100 = ´ 100 + 2 ´ 100 g Q T 0.1 æ 1ö ´ 100 + 2 ç ÷ ´ 100 è 30 ø 55 = 0.18 + 6.67 = 6.8% 50. (b) Least count of main scale of screw gauge = 1 mm Least count of screw gauge =

=

Pitch Number of division on circular scale

10-3 N Þ N = 200 51. (c) 52. (c) Least count of screw gauge, 5 ´ 10 -6 =

Pitch LC = No. of division = 0.5 × 10–3 = 0.5 × 10–2 mm + ve error = 3 × 0.5 × 10–2 mm = 1.5 × 10–2 mm = 0.015 mm Reading = MSR + CSR – (+ve error) = 5.5 mm + (48 × 0.5 × 10–2) – 0.015 = 5.5 + 0.24 – 0.015 = 5.725 mm

53. (c) = 1.5 % + 3 (1%) = 4.5% 54. (d) Maximum percentage error in A = 3(% error in P) + 2(% error in Q) 1 + (% error in R) + 1(% error in S) 2 1 = 3 ´ 0.5 + 2 ´ 1 + ´ 3 + 1 ´ 1.5 2 = 1.5 + 2 + 1.5 + 1.5 = 6.5% 55. (c) From Kepler's law, time period of a satellite, r3 4p 2 3 T2 = r Gm GM Relative uncertainty in the mass of the earth T = 2p

DM DT =2 = 2 ´ 10-2 M T

(Q 4p & G constant and

Dr negligible) r Ds Dr 56. (c) Relative error in Surface area, = 2 ´ = a and s r Dv Dr relative error in volume, = 3´ v r \ Relative error in volume w.r.t. relative error in area, Dv 3 = a v 2 Value of 1 part on main scale 57. (d) Least count = Number of parts on vernier scale 0.25 cm = 5 × 10–4 cm = 5×100 Reading = 4 × 0.05 cm + 30 × 5 × 10–4 cm = (0.2 + 0.0150) cm = 0.2150 cm (Thickness of wire)

relative uncertainty in radius

rhg ´ 103 2 Relative error in surface tension,

58. (d) Surface tension, T =

DT Dr Dh = + + 0 (Q g, 2 & 103 are constant) T r h Percentage error DT æ 10 –2 ´ 0.01 10 –2 ´ 0.01ö = ç + ÷ 100 T è 1.25 ´ 10 –2 1.45 ´ 10 –2 ø = (0.8 + 0.689) = (1.489) = 1.489% @ 1.5% 59. (c) Given, P = a1/2 b2 c2 d–4, Maximum relative error, DP 1 Da Db Dc Dd = +2 +3 +4 P 2 a b c d 1 = ´ 2 + 2 ´ 1 + 3 ´ 3 + 4 ´ 5 = 32% 2 0.5 60. (d) L.C. = = 0.01 mm 50 Zero error = 5 × 0.01 = 0.05 mm (Negative) Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm 100 ´

P-12

Physics

| DT1 | + | DT2 | + | DT3 | + | DT4 | 4 2 +1+ 3 + 0 = = 1.5 4 As the resolution of measuring clock is 1.5 therefore the mean time should be 92 ± 1.5 L 62. (d) As, g = 4p2 T2 Dg DL DT So, ´100 = ´100 + 2 ´100 g L T

61. (c) DT =

0.1 1 ´100 + 2 ´ ´100 = 2.72 ; 3% = 20 90 0.1 63. (b) Least count = = 0.01 cm 10 d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm 0.61 + 0.57 + 0.59 Mean diameter = 3

68. (b) According to the question. t = (90 ± 1) or,

l = (20 ± 0.1) or,

t = 2p

69.

When, I = 5mA, e1000 V /T = 6mA 1000 T Error = ± 0.01 (By exponential function) )´

1000 ´ (0.01) = 0.2 mA 300 65. (b) Measured length of rod = 3.50 cm For Vernier Scale with 1 Main Scale Division = 1 mm 9 Main Scale Division = 10 Vernier Scale Division, Least count = 1 MSD –1 VSD = 0.1 mm 66. (a) Rotation period of earth is about 24 hrs ; 105 s Revolution period of earth is about 365 days ; 107 s Speed of light wave C = 3 × 108 m/s Wavelength of visible light of spectrum l = 4000 – 7800 Å

70.

= (6 mA) ´

1 C = f l æç and T = ö÷ è fø Therefore period of light wave is 10–15 s (approx) 67. (d) In a voltmeter V µl V = kl Now, it is given E = 1.1 volt for l1 = 440 cm and V = 0.5 volt for l2 = 220 cm Let the error in reading of voltmeter be DV then, 1.1 = 400 K and (0.5 – DV) = 220 K. Þ

1.1 0.5 - DV = 440 220

\

DV = -0.05 volt

l 4p 2l Þ g= 2 g t

1 ö æ 0.1 Dg Dt ö æ Dl + 2 ´ ÷ = 0.027 = ± ç +2 ÷ = ç 90 ø è l è 20 g t ø Dg % = 2.7% \ g (a) According to ohm’s law, V = IR V R= I Absolute error ´102 \ Percentage error = Measurement DV DI ´100 = ´ 100 = 3% where, V I DR DV DI ´ 100 = ´ 102 + ´ 102 then, R V I = 3% + 3% = 6% (c) Q Reading of Vernier = Main scale reading + Vernier scale reading × least count. Main scale reading = 58.5 Vernier scale reading = 09 division least count of Vernier = 0.5°/30 0.5° Thus, R = 58.5° + 9 × 30 R = 58.65° (d) No. of divisions on main scale = N No. of divisions on vernier scale = N + 1 size of main scale division = a Let size of vernier scale division be b then we have aN aN = b (N + 1) Þ b = N +1 aN Least count is a – b = a – N +1 a é N +1 - N ù = aê ú = N +1 ë N +1 û (d) The least count (L.C.) of a screw guage is the smallest length which can be measured accurately with it. or,

I = (e1000 V /T - 1) mA (given)

V /T

Dl 0.1 = l 20

Dg %=? g As we know,

= 0.59 cm 64. (a) The current voltage relation of diode is

Also, dI = (e1000

Dt 1 = t 90

71.

72.

1 cm 1000 Hence measured value should be recorded upto 3 decimal places i.e., 5.320 cm

As least count is 0.001 cm =

P-13

Physical World, Units and Measurements

73. (a) Least count, L.C. =

1 mm 100

Diameter of wire = MSR + CSR × L.C. Q 1 mm = 0.1 cm = 0+

1 × 52 = 0.52 mm = 0.052 cm 100

74. (a) Number of significant figures in 23.023 = 5 Number of significant figures in 0.0003 = 1 Number of significant figures in 2.1 × 10–3 = 2 So, the radiation belongs to X-rays part of the spectrum. 75. (d) 30 Divisions of V.S. coincide with 29 divisions of M.S. 29 \ 1 V.S.D = MSD 30 L.C. = 1 MSD – 1VSD 29 = 1 MSD MSD 30

1

MSD 30 1 ´ 0.5° = 1 minute. = 30 76. (a) Momentum, p = m × v

=

Given, mass of a body = 3.513 kg speed of body = (3.513) × (5.00) = 17.565 kg m/s = 17.6 (Rounding off to get three significant figures) 77. (d) Least count of screw gauge = 0.01 mm Q

0.5 mm 50

Reading = [M.S.R. + C.S.R. × L.C.] – (zero error) = [3 + 35 × 0.01] – (–0.03) = 3.38 mm

14

2

Physics

Motion in a Straight Line Distance, Displacement & TOPIC 1 Uniform Motion 1.

TOPIC 2 Non-uniform Motion

A particle is moving with speed v = b x along positive x-axis. Calculate the speed of the particle at time t = t (assume that the particle is at origin at t = 0). [12 Apr. 2019 II]

6.

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is: [05 Sep. 2020 (II)]

b2 t b2 t b2 t (b) (c) b2 t (d) 2 4 2 All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. [2018]

v (m/s) 4 2 0 –2

(a) 2.

distance

velocity

(a)

3.

4.

5.

B S

D

t (in s)

1 2 3 4 5 6 C

37 49 m (b) 12 m (c) 11 m (d) m 3 4 The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be __________. [NA 4 Sep. 2020 (II)]

(a) position

(b)

time

7.

velocity

position

(c)

A

time

(d)

10 8 u –1 6 (ms ) 4 2

time

A car covers the first half of the distance between two places at 40 km/h and other half at 60 km/h. The average speed of the car is [Online May 7, 2012] (a) 40 km/h (b) 45 km/h (c) 48 km/h (d) 60 km/h The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is [2007] (a) v0 + g /2 + f (b) v0 + 2g + 3f (c) v0 + g /2 + f/3 (d) v0 + g + f A particle located at x = 0 at time t = 0, starts moving along with the positive x-direction with a velocity 'v' that varies as v = a x . The displacement of the particle varies with time as [2006] 2 1/2 3 (a) t (b) t (c) t (d) t

1 2 3 4 5 time (s)

8.

9.

The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is ______. [NA 9 Jan 2020 I] A bullet of mass 20g has an initial speed of 1 ms–1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5×10–2 N, the speed of the bullet after emerging from the other side of the wall is close to : [10 Apr. 2019 II] (a) 0.1 ms–1 (b) 0.7 ms–1 (c) 0.3 ms–1 (d) 0.4 ms–1

P-15

Motion in a Straight Line

(a) a +

b2 4c

(b) a +

b2 3c

b2 b2 (d) a + c 2c 11. A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figures that correctly represents the motion qualitatively (a = acceleration, v = velocity, x = displacement, t = time) [8 Apr. 2019 II]

(c) a +

14. An automobile, travelling at 40 km/h, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding) [Online April 15, 2018] (a) 75 m (b) 160 m (c) 100 m (d) 150 m 15. The velocity-time graphs of a car and a scooter are shown in the figure. (i) the difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively [Online April 15, 2018] (a) 337.5m and 25s

(b) 225.5m and 10s (c) 112.5m and 22.5s

A Car B

45 Velocity (ms –1) ®

10. The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where, a, b and c are constants. When the particle attains zero acceleration, then its velocity will be: [9 Apr. 2019 II]

30

(B)

(C)

(D)

G

15 O 0

(A)

F Scooter

E

5

C D 10 15 20 25 Time in (s) ®

(d) 11.2.5m and 15s 16. A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a point P in a field at a distance d from highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum? [Online April 15, 2018] P d

(a) (B), (C) (b) (A) (c) (A), (B), (C) (d) (A), (B), (D) 12. A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s? [10 Jan. 2019 II] v (m/s) 3

Q

M

d d (c) (d) d 2 2 17. Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity ? [Online April 8, 2017]

(a)

d 3

R

(b)

(a)

2

(b) Velocity

Velocity

1 0

13.

1 2 3 4 5 6 7 8 9 10 (a) 10 m (b) 6 m (c) 3 m (d) 9 m In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then 'v' is equal to: [9 Jan. 2019 II] 2a1 a 2 (a) (b) t 2a1 a 2 t a1 + a 2 a1 + a 2 (c) a1 a 2 t (d) t 2

Time

(c)

Time

(d) Velocity

Velocity

Distance

Distance

18. The distance travelled by a body moving along a line in time t is proportional to t3. The acceleration-time (a, t) graph for the motion of the body will be [Online May 12, 2012]

P-16

Physics (x1 – x2)

(x1 – x2)

a

a

(a)

(b)

(c)

t

t

a (d)

t

The graph of an object’s motion (along the x-axis) is shown in the figure. The instantaneous velocity of the object at points A and B are vA and vB respectively. Then [Online May 7, 2012]

23.

x(m) 15

10 B 5 Dt = 8

0

20.

24.

A Dx = 4 m 10

25.

20 t (s)

(a) vA = vB = 0.5 m/s (b) vA = 0.5 m/s < vB (c) vA = 0.5 m/s > vB (d) vA = vB = 2 m/s An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by

26.

dv = -2.5 v where v is the instantaneous speed. The time dt

21.

taken by the object, to come to rest, would be: [2011] (a) 2 s (b) 4 s (c) 8 s (d) 1 s A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x1(t) after time ‘t’; and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time ‘t’? [2008] (x1 – x2)

(x1 – x2)

(a)

O

t

(b)

(d)

O

t

f to come to rest. If the 2 total distance traversed is 15 S , then [2005] 1 (a) S = ft 2 (b) S = f t 6 1 2 1 2 ft (c) S = ft (d) S = 4 72 A particle is moving eastwards with a velocity of 5 ms–1. In 10 seconds the velocity changes to 5 ms–1 northwards. The average acceleration in this time is [2005] 1 -2 (a) ms towards north 2 1 (b) ms - 2 towards north - east 2 1 (c) ms - 2 towards north - west 2 (d) zero The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is [2005] (a) 2bv3 (b) –2abv 2 (c) 2av2 (d) –2av 3 An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be [2004] (a) 60 m (b) 40 m (c) 20 m (d) 80 m A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is [2003] (a) 12 m (b) 18 m (c) 24 m (d) 6 m If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? [2002] (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm. Speeds of two identical cars are u and 4u at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is [2002] (a) 1 : 1 (b) 1 : 4 (c) 1 : 8 (d) 1 : 16

and then decelerates at the rate

t 19.

t

22. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t

a

(c)

O

O

27.

28.

TOPIC 3 Relative Velocity t

29. Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8

P-17

Motion in a Straight Line

km/hour. Speed (in ms–1) of this person as observed from train B will be close to : (take the distance between the tracks as negligible) [2 Sep. 2020 (I)] (a) 29.5 ms–1 (b) 28.5 ms–1 (c) 31.5 ms–1q (d) 30.5 ms–1 30. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/h. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in same direction, and (ii) in the opposite directions is: [12 Jan. 2019 II] 11 5 3 25 (a) (b) (c) (d) 5 2 2 11 31. A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60º with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is: [12 Jan. 2019 II]

32.

2v v 3 v (a) (b) (c) v (d) 3 2 2 A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of : [Online April 9, 2017]

(a) 33.

34.

110 s

(b)

(c)

u+v 2

uv

(b)

1 2 u + v2 2

(d)

æ u 2 + v2 ö ç ÷ 2 ø è

TOPIC 4 Motion Under Gravity 35.

2 æhö ç ÷ 3 ègø

h g

(b) t = 1.8

æhö (c) t = 3.4 ç ÷ ègø

(d) t =

2h 3g

36. A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h . The velocity versus height of the ball during its motion 2 may be represented graphically by : (graph are drawn schematically and on not to scale) [4 Sep. 2020 (I)]

v

v h/2

(a)

h/2 h

h(v) (b)

v

v (c)

h(v)

h

h h/2

h(v) (d)

h h/2

h(v)

120 s

(c) 10 2 s (d) 15 s A person climbs up a stalled escalator in 60 s. If standing on the same but escalator running with constant velocity he takes 40 s. How much time is taken by the person to walk up the moving escalator? [Online April 12, 2014] (a) 37 s (b) 27 s (c) 24 s (d) 45 s A goods train accelerating uniformly on a straight railway track, approaches an electric pole standing on the side of track. Its engine passes the pole with velocity u and the guard’s room passes with velocity v. The middle wagon of the train passes the pole with a velocity. [Online May 19, 2012] (a)

(a) t =

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the accelertion due to gravity] : [5 Sep. 2020 (I)]

37. A ball is dropped from the top of a 100 m high tower on a 1 planet. In the last s before hitting the ground, it covers a 2 distance of 19 m. Acceleration due to gravity (in ms–2) near the surface on that planet is _______. [NA 8 Jan. 2020 II] 38. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? [2017]

(a)

(c)

(b)

(d)

39. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

P-18

Physics

(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/ s2) [2015]

y h

(The figures are schematic and not drawn to scale) (a) 240

(b)

(y2 – y1) m

240

(c)

(c)

240

41.

240

12

t(s)

t(s)

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is: [2014] 2 2 2 2 (a) 2gH = n u (b) gH = (n – 2) u d (c) 2gH = nu2 (n – 2) (d) gH = (n – 2)u2 Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be : [2009] v

(a)

y h

t

O –v1

t

v +v1

(b)

O –v1

y

t1

2t1

4t1

t

t

O

t

t

42.

(y2 – y1) m

12

+v1

y h

t(s)

(d) 12

t

2t1

v v1

8

(y2 – y1 ) m

t® 8

40.

t(s)

12

t1

t

(y2 – y1) m

(d) 8

O

h

t

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ? [2005] (a) 182 m (b) 91 m (c) 111m (d) 293m 43. A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position T of the ball at second [2004] 3 8h (a) meters from the ground 9 7h meters from the ground (b) 9 (c)

h meters from the ground 9

17 h meters from the ground 18 44. From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If vA and vB are their respective velocities on reaching the ground, then [2002] (a) vB > vA (b) vA = vB (c) vA > vB (d) their velocities depend on their masses.

(d)

P-19

Motion in a Straight Line

1.

x

(b) Given, v = b x dx = b x1/2 or dt x

or

òx

-1/2

0

t

x

Þ 2 x = at Þ x =

dx = ò bdt

6.

0

(a)

1/2

x b2t 2 or = 6t or x = 1/ 2 4 Differentiating w. r. t. time, we get dx b2 ´ 2t = dt 4

(t = t)

a2 2 t 4

4 A B v(m/s) 2 O S D 0 1 2 3 4 5 6 –2 C

t

(in s)

1 13 = 3 3 1 5 SD = 2 - = 3 3 Distance covered by the body = area of v-t graph = ar (OABS) + ar (SCD) OS = 4 +

2

b t 2 (b) Graphs in option (c) position-time and option (a) velocity-position are corresponding to velocity-time graph option (d) and its distance-time graph is as given below. Hence distance-time graph option (b) is incorrect. distance

or v = 2.

t

é2 x ù ò x = a ò dt ; ê 1 ú = a[t ]t0 0 0 ë û0 dx

1 æ 13 ö 1 5 32 5 37 + = m ç + 1÷ ´ 4 + ´ ´ 2 = 2è 3 ø 2 3 3 3 3 u (20) A 8 =

7.

time 3.

(c)

Average speed =

Total distance travelled x = Total time taken T

x = 48 km/h x x + 2 ´ 40 2 ´ 60 dx (c) We know that, v = dt Þ dx = v dt

B t 5 Distance travelled = Area of speed-time graph O

=

4.

x

t

0

0

8.

5.

(3) Distance X varies with time t as x2 = at2 + 2bt + c dx = 2at + 2b dt dx dx (at + b) Þx = at + b Þ = dt dt x Þ 2x

Integrating, ò dx = ò v dt t

1 ´ 5 ´ 8 = 20 m 2

=

t

é gt 2 ft 3 ù or x = ò (v0 + gt + ft ) dt = êv0 t + + ú 2 3 úû ëê 0 0 gt 2 ft 3 or, x = v0 t + + 2 3 g f At t = 1, x = v0 + + . 2 3 (a) v = a x , dx dx =a x Þ Þ = a dt dt x Integrating both sides, 2

2

Þx

d 2 x æ dx ö +ç ÷ =a dt 2 è dt ø 2

æ dx ö æ at + b ö a -ç ÷ a -ç ÷ 2 d x è dt ø = è x ø Þ = x x dt 2

=

Þ

ax 2 - ( at + b ) x3 a µ x–3

2

=

ac - b 2

x3 Hence, n = 3

2

P-20

9.

Physics

(b) From the third equation of motion

But, a =

a1 – a 2 Putting this value of t0 in equation (i)

F m 2

=

é 2.5 ´ 10-2 ù 20 Þ v 2 = (1)2 - (2) ê ú -3 ëê 20 ´10 ûú 100 Þ v2 = 1 –

Þv=

1

m/s = 0.7m/s 2 10. (b) x = at + bt2 – ct3 dx d = (at + bt 2 + ct 3 ) dt dt = a + 2bt – 3ct2

11.

dv d = (a + 2bt - 3ct 2 ) dt dt

æb ö \ t =ç ÷ è 3c ø

2ö 2 æ æb ö æ b ö = a+b a + 2 b 3 c ç ÷ and v = ç ÷ ç ÷ 3c ø è è 3c ø è 3c ø (d) For constant acceleration, there is straight line ®

parallel to t-axis on a - t . ®

®

Inclined straight line on v - t , and parabola on x - t . 12. (d) Position of the particle, S = area under graph (time t = 0 to 5s) 1 = ´ 2 ´ 2 + 2 ´ 2 + 3´1= 9m 2 13. (c) Let time taken by A to reach finishing point is t0 \ Time taken by B to reach finishing point = t0 + t x

vA = a1t0 vB = a2(t0 + t)

vA – vB = v Þ v = a1 t0 – a2 (t0 + t) = (a1 – a2)t0–a2t ...(i) 1 1 x B = x A = a1 t 02 = a 2 (t 0 + t) 2 2 2 Þ a1 t 0 = a 2 ( t 0 + t )

(

)

a2 t – a 2t =

a1a 2 t + a 2 t – a 2 t

v –u 1 2 and S = ut + at t 2 1 (45) Distance travelled by car in 15 sec = (15)2 2 15 675 = m 2 Distance travelled by scooter in 15 seconds = 30 × 15 = 450 (Q distance = speed × time) Difference between distance travelled by car and scooter in 15 sec, 450 – 337.5 = 112.5 m Let car catches scooter in time t;

675 + 45(t –15) = 30t 2 337.5 + 45t – 675 = 30t

Þ 15t = 337.5

Þ t = 22.5 sec 16. (a) Let the car turn of the highway at a distance 'x' from the point M. So, RM = x And if speed of car in field is v, then time taken by the car to cover the distance QR = QM – x on the highway, QM - x .....(i) 2v Time taken to travel the distance 'RP ' in the field t1 =

d 2 + x2 ..... (ii) v Total time elapsed to move the car from Q to P t2 =

u=0

Þ

a1 + a 2

– a 2t

15. (c) Using equation, a =

Velocity, v =

or 0 = 2b – 3c × 2t

(

a1 – a 2

or, v = a1a 2 t 14. (b) According to question, u1 = 40 km/h, v1 = 0 and s1 = 40 m using v2 – u2 = 2as; 02 – 402 = 2a × 40 ...(i) Again, 02 – 802 = 2as ...(ii) From eqn. (i) and (ii) Stopping distance, s = 160 m

1 2

Acceleration,

a2t

v =( al – a2 )

æFö \v = u - 2ç ÷ S èmø 2

a 2t

Þ to =

v2 – u2 = 2aS

)

a1 – a 2 t 0 = a 2 t

QM - x d 2 + x2 + v 2v dt =0 For 't ' to be minimum dx ù 1é 1 x ê- + ú =0 v ëê 2 d 2 + x 2 ûú t = t1 + t2 =

d

d = or x = 3 22 - 1

Q

P d R

M

P-21

Motion in a Straight Line

17. (c) According to question, object is moving with constant negative acceleration i.e., a = – constant (C) vdv = -C dx vdv = – Cdx

v2 v2 k = - Cx + k x=+ 2 2C C Hence, graph (3) represents correctly. 18. (b) Distance along a line i.e., displacement (s) = t3 (Q s µ t 3 given) By double differentiation of displacement, we get acceleration. 3

2

ds dt dv d 3t = = 3t 2 and a = = = 6t dt dt dt dt a = 6t or a µ t Hence graph (b) is correct. V=

19. (a)

Instantaneous velocity v =

Dx Dt

Dx A 4 m = = 0.5 m/s Dt A 8s Dx 8m and vB = B = = 0.5 m/s Dt B 16s i.e., vA = vB = 0.5 m/s

From graph, vA =

dv 20. (a) Given, = -2.5 v dt dv Þ = – 2.5 dt v 0

-½

This equation is of parabola. For t < For t =

v ; the slope is negative a v ; the slope is zero a

v ; the slope is positive a These characteristics are represented by graph (b). 22. (d) Let car starts from A from rest and moves up to point B with acceleration f.

For t >

1 2 ft1 2 Distance, BC = (ft1)t

Distance, AB = S =

t

dv = -2.5ò dt 0

A f B t1

Þ

0

é v +½ ù t = -2.5 [ t ]0 Þ ê (½) ú êë úû 6.25

C f /2 D 2t 1

t

............. (i)

f t1t = 12 S

1 2 f t1 = S 2

Þ – 2(6.25)½ = – 2.5t Þ – 2 × 2.5 = –2.5t

............ (ii)

Dividing (i) by (ii), we get t1 =

Þ t = 25 21. (b) For the body starting from rest, distance travelled (x1) is given by 1 x1 = 0 + at2 2 1 2 Þ x1 = at 2

Þ S=

2

23. (c)

v2 N D v = v 2 + (-v 1 )

W

t

t 6

1 ætö f t2 fç ÷ = 2 è 6ø 72

x1 – x2

v/a

( ft1 )2 u2 = = ft12 = 2S 2a 2( f / 2)

Distance, CD =

15 S Total distance, AD = AB + BC + CD = 15S AD = S + BC + 2S Þ S + f t1t + 2 S = 15 S

Integrating,

ò6.25 v

For the body moving with constant speed x2 = vt 1 \ x1 - x2 = at 2 - vt 2 at t = 0, x1 – x2 = 0

90° - v1

uur Initial velocity, v1 = 5iˆ,

v1

S

E

P-22

Physics

uur Final velocity, v2 = 5 ˆj, uur ur ur Change in velocity D v = (v 2 - v 1 )

=

5 2 + 52 + 0 = 5 2m/s [As | v1 | = | v2 | = 5 m/s] uur Dv Avg. acceleration = t

250 ´ 250 » = –16 ms–2. 324 ´ 2 ´ 6 Case-2 : Initial velocity, u = 100 km/hr

a=–

5 2 1 = m / s2 10 2 5 tan q = = -1 -5 which means q is in the second quadrant. (towards north-west) 24. (d) Given, t = ax2 + bx; Diff. with respect to time (t) d d dx dx (t ) = a ( x 2 ) + b = a.2 x + b.v. dt dt dt dt Þ 1 = 2axv + bv = v(2ax + b)(v = velocity) 1 2ax + b = . v Again differentiating, we get

500 ´ 500 = 24m 324 ´ 32 27. (a) In first case

s=

u1 = u ; v1 =

u , s = 3 cm, a1 = ? 2 1

...(i)

2

æ dx ö = v÷ çèQ ø dt

–u2 8 In second case: Assuming the same retardation

Þ

a=

u2 = u /2 ; v2 = 0 ; s2 = ?; a2 = v22 - u 22 = 2a2 ´ s2

…(i) 5 18

100 m/s 3 and (0)2 – u¢2 = –2ad¢ or u¢2 = 2ad¢ …(ii) (ii) divided by (i) gives, d' 4 = Þ d ' = 4 ´ 20 = 80m d 26. (c) Fir first case : Initial velocity, 5 u = 50 ´ m / s, 18 v = 0,s = 6m, a = a

Using, v 2 - u 2 = 2as

2

5ö æ Þ - ç 100 ´ ÷ = 2 × (–16) × 5 18 ø è

æ uö 2 çè ÷ø - u = 2 × a × 3 2

In second case speed, u¢ = 120 ´ =

2

5ö æ Þ 02 - ç100 ´ ÷ = 2as 18 ø è

Using, v12 - u12 = 2a1s1

dx 1 dv +0= - 2 dt v dt

dv = –2av3 dt 25. (d) In first case speed, 5 50 u = 60 ´ m/s = m/s 18 3 d = 20m, Let retardation be a then (0)2 – u2 = –2ad or u2 = 2ad

5 m/sec 18 v = 0, s = s, a = a As v2 – u2 = 2as

= 100 ´

=

Þ a=

2

5ö æ Þ - ç 50 ´ ÷ = 2 ´ a ´ 6 è 18 ø

v12 + v22 + 2v1v2 cos 90

=

2a

2

5ö æ Þ 02 - ç 50 ´ ÷ = 2 ´ a ´ 6 è 18 ø

-u 2 8

...(ii)

æ –u2 ö u2 = 2ç ÷ ´ s2 4 è 8 ø Þ s2 = 1 cm 28. (d) For first car u1 = u, v1 = 0, a1 = – a, s1 = s1

\ 0-

As v12 - u12 = 2a1s1 Þ –u2 = –2as1 Þ u2 = 2as1

u2 2a For second car u2 = 4u, v1 = 0, a2 = – a, s2 = s2

Þ s1 =

\

v22 - u22 = 2a2 s2

Þ

–(4u)2 = 2(–a)s2

...(i)

P-23

Motion in a Straight Line

Þ 16 u2 = 2as2 8u 2 Þ s2 = a Dividing (i) and (ii),

29.

...(ii)

s1 u 2 a 1 = × 2 = s2 2a 8u 16 (a) According to question, train A and B are running on parallel tracks in the opposite direction.

1.8 km/h

36 km/h A

B

v

R (Observer) Distance, PQ = vp × t (Distance = speed × time) Distance, QR = V.t

PQ QR

120 = 24 second. 5 34 (d) Let 'S' be the distance between two ends 'a' be the constant acceleration As we know v2 – u2 = 2aS

vc2 = u 2 + aS vc2 = u 2 +

v2 - u 2 2

u 2 + v2 2 35. (c) For upward motion of helicopter,

vc =

v2 = u 2 + 2 gh Þ v 2 = 0 + 2 gh Þ v = 2gh Now, packet will start moving under gravity. Let 't' be the time taken by the food packet to reach the ground. 1 s = ut + at 2 2 1 1 Þ -h = 2 gh t - gt 2 Þ gt 2 - 2 gh t - h = 0 2 2

or, t =

Bus

2´

2 m/sec2

200 m Given, uC = uB = 0, aC = 4 m/s2, aB = 2 m/s2 hence relative acceleration, aCB = 2 m/sec2 1 Now, we know, s = ut + at 2 2 1 200 = ´ 2t 2 Q u = 0 2 Hence, the car will catch up with the bus after time

t = 10 2 second

1 1 15 "escalator" + = 60 40 120 second

2 gh ± 2 gh + 4 ´

1 vp ´ t v = Þ vp = 2 V.t 2 4 m/sec2 Car

So, the person’s speed is

v2 - u2 2 Let v be velocity at mid point. S 2 2 Therefore, vc - u = 2a 2

VMA = –1.8 km/h = –0.5 m/s Vman, B = Vman, A + VA, B = Vman, A + VA – VB = –0.5 + 10 – (–20) = – 0.5 + 30 = 29.5 m/s. 30. (a) vP P 31. (d) Q o 60

32. (c)

1 "escalator" 40 second Walking with the escalator going, the speed add.

or, aS =

VB = -72 km/h = –20 m/s

cos 60° =

Person’s speed walking only is

So, the time to go up the escalator t =

VA = 36 km/h = 10 m/s

72 km/h

1 "escalator" 60 second Standing the escalator without walking the speed is

33. (c)

or, t =

g ´h 2

g 2

2 gh (1 + 2) Þ t = g

or, t = 3.4

2h (1 + 2) g

h g

36. (c) For uniformly accelerated/ deaccelerated motion :

v 2 = u 2 ± 2 gh As equation is quadratic, so, v-h graph will be a parabola

P-24

Physics

v at t= 0, h = d 2

1 ® 2 : V increases downwards h 2 ® velocity changes its direction 2 ® 3 : V decreases upwards

d 1

3 collision takes 2 place

v=

Using, S = ut + Þ S = 0´t +

1 2 gt 2

1 2 gt 2

Þ 200 = gt2

Þt= In last

200 g

[Q 2S = 100m] …(i)

1 s, body travels a distance of 19 m, so in 2

æ 1ö çt – ÷ è 2ø

distance travelled = 81 2

1 æ 1ö g ç t – ÷ = 81 2 è 2ø 2

æ 1ö \ g ç t – ÷ = 81´ 2 è 2ø 81´ 2 æ 1ö Þ çt – ÷ = 2 g è ø

Þ

u + 2 gh H

37. (08.00) Let the ball takes time t to reach the ground

\

u

2

Now, v = u + at

Initially velocity is downwards (–ve) and then after collision it reverses its direction with lesser magnitude, i.e. velocity is upwards (+ve). Note that time t = 0 corresponds to the point on the graph where h = d. Next time collision takes place at 3.

Now,

y1 = 10t – 5t2 ; y2 = 40t – 5t2 y1 = – 240m, t = 8s y2 – y1 = 30t for t < 8s. t > 8s, 1 y2 – y1 = 240 – 40t – gt2 2 40. (c) Speed on reaching ground

39. (b) for \ for

1 1 = ( 200 – 81 ´ 2) 2 g

using (i)

g = 2(10 2 – 9 2)

Þ g =2 2 \ g = 8 m/s2 38. (a) For a body thrown vertically upwards acceleration remains constant (a = – g) and velocity at anytime t is given by V = u – gt During rise velocity decreases linearly and during fall velocity increases linearly and direction is opposite to each other. Hence graph (a) correctly depicts velocity versus time.

Þ

u 2 + 2 gh = -u + gt

Time taken to reach highest point is t = Þt =

u + u 2 + 2 gH

g (from question)

=

u , g

nu g

Þ 2gH = n(n –2)u2 41. (b) For downward motion v = –gt The velocity of the rubber ball increases in downward direction and we get a straight line between v and t with a negative slope. 1 2 Also applying y - y0 = ut + at 2 1 2 1 2 We get y - h = - gt Þ y = h - gt 2 2 The graph between y and t is a parabola with y = h at t = 0. As time increases y decreases. For upward motion. The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here v = u – gt where u is the velocity just after collision. As t increases, v decreases. We get a straight line between v and t with negative slope. 1 2 Also y = ut - gt 2 All these characteristics are represented by graph (b). 42. (d) Initial velocity of parachute after bailing out, u=

2gh

u = 2 ´ 9.8 ´ 50 = 14 5 The velocity at ground, v = 3m/s v2 - u2 32 - 980 » 243 m S= = 2´2 4 Initially he has fallen 50 m. \ Total height from where he bailed out = 243 + 50 = 293 m

50 m v a = - 2 m / s2

3m / s

P-25

Motion in a Straight Line

43. (a)

We have s = ut +

44. (b)

1 2 gt , 2

velocity u

1 Þ h = 0 × T + gT2 2 1 2 Þ h = gT 2

Vertical distance moved in time

Ball A is thrown upwards with

from the building. During its downward journey when it comes back to the point of throw, its speed is equal to the speed of throw (u). So, for the journey of both the balls from point A to B.

T is 3

2

1 æTö 1 gT 2 h h' = g ç ÷ Þ h' = ´ = 2 è 3ø 2 9 9 \ Position of ball from ground = h -

h 8h = 9 9

We can apply v2 – u2 = 2gh. As u, g, h are same for both the balls, vA = vB

A

h B

u u

P-26

3

Physics

Motion in a Plane 5.

TOPIC 1 Vectors 1.

2.

3.

®

A force F = (i$ + 2 $j + 3k$ ) N acts at a point (4$i + 3 $j - k$ ) m. Then the magnitude of torque about the point ($i + 2 $j + k$ ) m will be x N-m. The value of x is ______.

(

[NA 7 Jan. 2020 II] uuur uur uuur A 2 = 5 and A1 + A 2 = 5. The value of uur uuur 3A1 - 2A 2 is : [8 April 2020 II]

)(

(a) – 106.5 4.

(c) 90° 6.

[NA Sep. 05, 2020 (I)] r r r r The sum of two forces P and Q is R such that | R | = r r | P | . The angle q (in degrees) that the resultant of 2 P r r and Q will make with Q is _______. uur Let A1 = 3, uur uuur 2A1 + 3A 2 ·

)

Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle q with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle q is: [10 Jan. 2019 II] (a) 120° (b) 60° (d) 30° ur ur Two vectors A and B have equal magnitudes. The ur ur ur ur magnitude of A + B is ‘n’ times the magnitude of A - B . ur ur The angle between A and B is: [10 Jan. 2019 II]

(

é n 2 - 1ù (a) cos -1 ê 2 ú ë n + 1û

7.

(b) – 99.5

(c) – 112.5 (d) – 118.5 In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be: [10 Jan. 2019 I] 8.

(

)

1 ˆ ˆ a i -k (b) 2

(c)

(

)

(d)

1 a ˆj - iˆ 2

(

)

(

)

1 a ˆj - kˆ 2

9.

(

)

é n - 1ù (b) cos -1 ê ë n + 1úû

é n2 -1ù é n -1ù (c) sin -1 ê 2 ú (d) sin -1 ê n + 1 ë n + 1úû ë û r r Let A = (iˆ + ˆj) and B = (iˆ - ˆj) . The magnitude of a r r r r r r r coplanar vector C such that A.C = B.C = A.B is given by [Online April 16, 2018]

(a)

5 9

(b)

10 9

(c)

20 9

(d)

9 12

ur A vector A is rotated by a small angle Dq radian (Dq > r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed v, the average force experienced by each support after a long time is (assume all collisions are elastic): [Online April 11, 2015]

(a) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1. (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is false. (d) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1. 12. Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? [2010] A

L

B

(a)

mv 2 2(L - nr)

(b)

mv 2 L - 2nr

mv 2 (d) zero L - nr A body of mass 5 kg under the action of constant force r r F = Fxˆi + Fy ˆj has velocity at t = 0 s as v = 6iˆ - 2ˆj m/s

(c) 9.

(

)

r r and at t = 10s as v = +6ˆj m / s . The force F is: [Online April 11, 2014]

æ 3 ˆ 4 ˆö (a) (b) ç - i + j ÷ N è 5 5 ø æ 3ˆ 4 ˆö (d) ç i - j ÷ N (c) 3iˆ - 4ˆj N è5 5 ø 10. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves? [2012]

(

)

F0 mb

F0 mb v (t )

15.

16.

(a) (b) (c) Zero (d) 4.9 ms–2 in vertical direction A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider g = 10 m/s2). [2006] (a) 4 N (b) 16 N (c) 20 N (d) 22 N A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to [2006] (a) 150 N (b) 3 N (c) 30 N (d) 300 N A particle of mass 0.3 kg subject to a force F = – kx with k = 15 N/m . What will be its initial acceleration if it is released from a point 20 cm away from the origin ?[2005] (a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2 A block is kept on a frictionless inclined surface with angle of inclination ‘a’. The incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to [2005]

(b) v (t )

v (t )

t

t

(c)

14.

)

F0 mb

(a)

13.

-3jˆ + 4ˆj N

(

30°

60°

4.9 ms–2 in horizontal direction 9.8 ms–2 in vertical direction

F0 mb

(d) v (t )

t t 11. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. Statement 1: If you push on a cart being pulled by a horse so that it does not move, the cart pushes you back with an equal and opposite force. Statement 2: The cart does not move because the force described in statement 1 cancel each other. [Online May 26, 2012]

a

a

(a) g cosec a (b) g / tan a (c) g tan a (d) g 17. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10m/s2. Then the initial thrust of the blast is [2003] (a) 3.5 ´ 10 5 N (b) 7.0 ´ 10 5 N (c) 14.0 ´ 10 5 N (d) 1.75 ´ 10 5 N 18. Three forces start acting simultaneously on a particle r moving with velocity, v . These forces are represented in magnitude and direction by the three sides of a triangle ABC. The particle will now move with velocity [2003]

P-38

Physics

C

r (a) less than v

A

(a)

2g 3

(b)

g 2

(c)

5g 6

B

r (b) greater than v r (c) v in the direction of the largest force BC (d) vr , remaining unchanged

19. A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling) [2002] (a) solid sphere (b) hollow sphere (c) ring (d) all same

Motion of Connected Bodies, TOPIC 2 Pulley & Equilibrium of Forces 20. A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the midpoint of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g = 10 ms–2 and the rope to be massless) [7 Jan. 2020 II] (a) 100 N (b) 90 N (c) 70 N (d) 75 N 21. An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg. The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g =10 m/s2) must be at least: [7 Jan. 2020 II] (a) 56300 W (b) 62360 W (c) 48000 W (d) 66000 W 22. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45°at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms–2) [9 Jan. 2019 II] (a) 200 N (b) 140 N (c) 70 N (d) 100 N 23. A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the str ing does not slip on the cylinder, with what acceleration will the mass fall or release? [2014]

R m

m (d) g 24. Two blocks of mass M1 = 20 kg and M2 = 12 kg are connected by a metal rod of mass 8 kg. The system is pulled vertically up by applying a force of 480 N as shown. The tension at the mid-point of the rod is : [Online April 22, 2013] 480 N

(a) 144 N

M1

(b) 96 N (c) 240 N (d) 192 N

M2

25. A uniform sphere of weight W and radius 5 cm is being held by a string as shown in the figure. The tension in the string will be : [Online April 9, 2013]

8 cm

W W W W (b) 5 (c) 13 (d) 13 5 12 5 12 26. A spring is compressed between two blocks of masses m1 and m2 placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of v1 and v2 as shown. The blocks travel distances x1 and x2 respectively before coming to rest.

(a)

12

æx ö The ratio ç 1 ÷ is è x2 ø

[Online May 12, 2012] m2

m1

v2

v1

(a)

m2 m1

(b)

m1 m2

(c)

m2 m1

(d)

m1 m2

P-39

Laws of Motion

27. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m. [2007] mF MF (a) (b) M (m + M ) mF (c) ( M + m) F (d) (m + M ) m 28. Two masses m1 = 5g and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move ? [2004] ( g = 9.8m / s 2 )

33. When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed then the acceleration of the particle is [2002] (a) F1/m (b) F2F3 /mF1 (c) (F2 - F3)/m (d) F2 /m. 34. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are [2002] (a) 12 N, 6 N (b) 13 N, 5 N (c) 10 N, 8 N (d) 16N, 2N. 35. A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g/8, then the ratio of the masses is [2002] (a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3 36. Three identical blocks of masses m = 2 kg are drawn by a force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C? [2002] C

(a) 5 m/s2 (b) 9.8 m/s2 (c) 0.2 m/s2 (d) 4.8 m/s2 29. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s2, the reading of the spring balance will be [2003] (a) 24 N (b) 74 N (c) 15 N (d) 49 N 30. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is [2003] Pm Pm PM (a) (b) (c) P (d) M +m M -m M +m 31. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is [2003] (a) both the scales read M kg each (b) the scale of the lower one reads M kg and of the upper one zero (c) the reading of the two scales can be anything but the sum of the reading will be M kg (d) both the scales read M/2 kg each 32. A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively [2002] (a) g, g (b) g – a, g – a (c) g – a, g (d) a, g

B

A

F

(a) 9.2 (b) 3.4 (c) 4 (d) 9.8 37. One end of a massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms-2) can a man of 60 kg climb on the rope? [2002] P C

(a) 16

(b) 6

(c) 4

(d) 8

TOPIC 3 Friction 38. An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is : (g = 10 ms–2) [Sep. 06, 2020 (I)] (a) 0.20 m

(b) 0.45 m

(c) 0.60 m (d) 0.80 m 39. A block starts moving up an inclined plane of inclination 30° with an initial velocity of v0. It comes back to its v initial position with velocity 0 . The value of the 2 coefficient of kinetic friction between the block and the I . The nearest integer to I inclined plane is close to 1000 is _________. [NA Sep. 03, 2020 (II)]

P-40

Physics

40. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F=20 N, making an angle of 30 o with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is m = 0.2. The difference between the accelerations of the block, in case (B) and case (A) will be : (g =10 ms–2) [12 April 2019 II]

44. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is: [2018] T m m2 (a) 18.3 kg m 2 (b) 27.3 kg T

(c) 43.3 kg

m1

(d) 10.3 kg

(a) 0.4 ms–2 (b) 3.2 ms–2 –2 (c) 0.8 ms (d) 0 ms–2 41. Two blocks A and B masses mA=1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : [Take g = 10 m/s2] [10 April 2019 II]

(a) 8 N (b) 16 N (c) 40 N (d) 12 N 42. A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is : [Take g = 10 m/s2] [12 Jan. 2019 II] 10

N

2N 30°

3 3 (b) 2 4 2 1 (c) (d) 3 2 43. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block doesnot move downward? (take g = 10 ms–2) [9 Jan. 2019 I] (a)

m1g

45. A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is :

(a) (c)

1-

n2

[Online April 15, 2018] 1 (b) 1 - 2 n

2

(d)

1

1 2-n

1 1 - n2

46. A body of mass 2kg slides down with an acceleration of 3m/s2 on a rough inclined plane having a slope of 30°. The external force required to take the same body up the plane with the same acceleration will be: (g = 10m/s2) [Online April 15, 2018] (a) 4N (b) 14N (c) 6N (d) 20N 47. A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle q with the horizontal, a point object of mass m is kept. The minimum coefficient of friction mmin between the mass and the inclined surface such that the mass does not move is : [Online April 9, 2016] (a) tan2q (b) tanq (c) 3 tanq (d) 2 tan q 48. Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is: [2015] F

A

B

P

3

N

k 10

(a) 32 N

g

45°

(b) 18 N

(c) 23 N

(d)

25 N

(a) 120 N (c) 100 N

(b) 150 N (d) 80 N

P-41

Laws of Motion

49. A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed n, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely n falling object were to acquire speed after being dropped 10 from height H, then neglecting energy losses and taking g = 10 ms–2, the value of H is close to: [Online April 10, 2015] (a) 0.05 km (b) 0.02 km (c) 0.03 km (d) 0.04 km 50. A block of mass m is placed on a surface with a vertical x3 . If the coefficient of friction 6 is 0.5, the maximum height above the ground at which the block can be placed without slipping is: [2014]

cross section given by y =

1 2 1 1 m m m m (b) (c) (d) 6 3 3 2 51. Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration ‘a’ perpendicular to the axis of the cylinder. What is Ffriction at point P? It is assumed that the cylinder does not slip. [Online April 19, 2014] w v O

(a)

A

h

vo2

L

C 2 vo

(a)

2h + m 2mg

(b)

h + m 2mg

(c)

h v2o + 2m mg

(d)

v2 h + o 2m 2mg

54. A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied to B for the blocks to move together will be: [Online April 9, 2014] (a) 30 N (b) 25 N (c) 27 N (d) 48 N 55. A block is placed on a rough horizontal plane. A time dependent horizontal force F = kt acts on the block, where k is a positive constant. The acceleration - time graph of the block is : [Online April 25, 2013] a a

(a)

P

B

(b)

O a

a

O a

t

t

F friction Ma Ma (d) 3 2 52. A heavy box is to dragged along a rough horizontal floor. To do so, person A pushes it at an angle 30° from the horizontal and requires a minimum force FA, while person B pulls the box at an angle 60° from the horizontal and needs minimum force FB. If the coefficient of friction

(a) Mg

(b) Ma

(c)

between the box and the floor is

(a)

3

(b)

FA 3 , the ratio is FB 5 [Online April 19, 2014]

5 3

2 3 (d) 3 2 53. A small ball of mass m starts at a point A with speed vo and moves along a frictionless track AB as shown. The track BC has coefficient of friction m. The ball comes to stop at C after travelling a distance L which is: [Online April 11, 2014] (c)

(c)

(d)

O O t t 56. A body starts from rest on a long inclined plane of slope 45°. The coefficient of friction between the body and the plane varies as m = 0.3 x, where x is distance travelled down the plane. The body will have maximum speed (for g = 10 m/s2) when x = [Online April 22, 2013] (a) 9.8 m (b) 27 m (c) 12 m (d) 3.33 m 57. A block of weight W rests on a horizontal floor with coefficient of static friction m. It is desired to make the block move by applying minimum amount of force. The angle q from the horizontal at which the force should be applied and magnitude of the force F are respectively. [Online May 19, 2012] mW

(a)

q = tan -1 ( m) , F =

(b)

æ 1ö mW q = tan -1 ç ÷ , F = è mø 1 + m2

1 + m2

P-42

Physics

(c)

q = 0, F = mW

æ m ö mW ,F = (d) q = tan -1 ç ÷ 1+ m è 1 + mø 58. An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a so that the insect does not slip is given by [Online May 12, 2012]

64. A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is 2 (take g = 10 m / s ) [2004] (a) 1.6 (b) 4.0 (c) 2.0 (d) 2.5 65. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is [2003]

a

10N (a) cot a = 3 (b) sec a = 3 (c) cosec a = 3 (d) cos a = 3 59. The minimum force required to start pushing a body up rough (frictional coefficient m) inclined plane is F1 while the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an angle q from the horizontal such that tan q = 2m then the ratio

F1 is F2 [2011 RS]

(a) 1 (b) 2 (c) 3 (d) 4 60. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of length l A : l B = 2 : 3, then the stiffness of spring ‘A’ is given by (a)

[2011 RS]

3k 5

(b)

2k 5

5k 2 61. A smooth block is released at rest on a 45° incline and then slides a distance ‘d’. The time taken to slide is ‘n’ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is [2005]

(c) k

(a)

(c)

(d)

1 mk = 1 – 2 n ms = 1 -

1 n

2

(b)

(d)

mk = 1ms = 1-

(c) 400 m

TOPIC 4

Circular Motion, Banking of Road

67. A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is (g = 10m/s2) [Online April 15, 2018] (a) 0.5 (b) 0.7 (c) 0.3 (d) 0.6 68. A conical pendulum of length 1 m makes an angle q = 45° w.r.t. Z-axis and moves in a circle in the XY plane.The radius of the circle is 0.4 m and its centre is vertically below O. The speed of the pendulum, in its circular path, will be : (Take g = 10 ms–2) [Online April 9, 2017] Z

1

(a) 0.4 m/s

n2

(b) 4 m/s

1

(c) 0.2 m/s

n2

62. The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by [2005] (a) 2 cos f (b) 2 sin f (c) tan f (d) 2 tan f 63. Consider a car moving on a straight road with a speed of 100 m/s . The distance at which car can be stopped is [ m k = 0.5 ] (a) 1000 m (b) 800 m

(a) 20 N (b) 50 N (c) 100 N (d) 2 N 66. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is [2003] (a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06

O q

C

(d) 2 m/s 69. A particle is released on a vertical smooth semicircular track from point X so that OX makes angle q from the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y where OY makes angle f with the horizontal. Then: [Online April 19, 2014] X Y

q

[2005] (d) 100 m

f O

P-43

Laws of Motion

1 cos q 2 2 3 (c) sin f = cos q (d) sin f = cos q 3 4 70. A body of mass ‘m’ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring is 5 cm. The original length of the spring is : [Online April 23, 2013] (a) 15 cm (b) 12 cm (c) 16 cm (d) 10 cm 71. A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly. [2010] y

(a) sin f = cos f

(b)

sin f =

B P(x,y) m 20

O

A

x

(a) 13m/s2 (b) 12 m/s2 (c) 7.2 ms2 (d) 14m/s2 72. For a particle in uniform circular motion, the acceleration r a at a point P(R,q) on the circle of radius R is (Here q is measured from the x-axis) [2010]

(a)

-

n2 n2 cos q iˆ + sin q ˆj R R

(b)

-

n2 n2 sin q iˆ + cos q ˆj R R

(c) -

n2 n2 cos q iˆ sin q ˆj R R

n2 ˆ n2 ˆ i+ j R R 73. An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated F1 on the inner and outer parts of the ring , F is [2005] 2

(d)

(a)

æ R1 ö çè R ÷ø 2

2

(b)

R2 R1

R1 (d) 1 R2 74. Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed ? [2004] (a) The acceleration vector points to the centre of the circle (b) The acceleration vector is tangent to the circle (c) The velocity vector is tangent to the circle (d) The velocity and acceleration vectors are perpendicular to each other. 75. The minimum velocity (in ms-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is [2002] (a) 60 (b) 30 (c) 15 (d) 25

(c)

P-44

1.

Physics

(a) Given r F = k ( v y iˆ + vx ˆj ) \ Fx = kv y iˆ, Fy = kv x ˆj

4.

dv = a = – (g + gv2) dt Let time t required to rise to its zenith (v = 0) so,

v 2y = v x2 + C

\t =

0

ò

dp d (mv) æ dm ö = = vç ÷ dt dt è dt ø

...(i)

dM (t ) = bv 2 (t ) dt Thrust on the satellite,

...(ii)

We have given,

5.

r F = mkv 2 - mg r r F a= = -[ kv 2 + g ] m

Þ v× 0

Þò u

dv = -[kv 2 + g ] dh

v × dv kv 2 + g

h

= ò dh 0

0 1 Þ ln éëkv 2 + g ùû = - h u 2k

g + gv2

[for Hmax, v = 0]

0

æ g v0 ö tan -1 ç ÷ gg è g ø

(d) v2 = u2 – 2gh u 2 - 2 gh

Momentum, P = mv = m u 2 - 2 gh

u2 ,P = 0 f upward direction is positive and downward direction is negative. (b) From Newton’s second law At h = 0, P = mu and at h =

6.

dp = F = kt dt Integrating both sides we get, 3p

òp

T

dp = ò kt dt Þ [ p ] 0

3p p

7.

8.

T

é t2 ù =k ê ú êë 2 úû 0

kT 2 p ÞT =2 2 k R dv R (a) From F < 2 v(t) Þ m < 2 v(t) dt t t dv Rdt > mw2) the relative change in the length of the spring is best given by the option: [9 Jan. 2020 II] (a)

2 æ mw2 ö ç ÷ 3 çè k ÷ø

(b)

2mw2 k

mw2 mw2 (d) k 3k 26. A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed w about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is: [8 Jan. 2020 I]

(c)

(a)

(c)

ml w2 k - wm

(b)

ml w 2 k + mw

(c)

2

(d)

ml w 2 k - mw 2

ml w 2 k + mw

27. A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely? [12 Apr. 2019 II]

(d)

28. A smooth wire of length 2pr is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed w about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of w2 is equal to : [12 Apr. 2019 II]

3g (b) 2 g / (r 3) 2r (d) 2g/r (c) ( g 3) / r 29. A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be : [12 Jan. 2019 II] (a) 2.0 (b) 0.1 (c) 0.4 (d) 1.2 (a)

P-80

Physics

30. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then: [2018] (b) T µ R n /2+1 (c) T µ R (n +1)/2 (d) T µ R n /2 31. The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a : [Online April 9, 2017] (a) T µ R3/2 for any n.

2 kg

F

Fixed pivot x

34. A cubical block of side 30 cm is moving with velocity 2 ms–1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is : [Online April 9, 2016] a = 30 cm

O (a) 13.3 (b) 5.0 (c) 9.4 (d) 6.7 35. Two point masses of mass m1 = fM and m2 = (1 – f) M (f < 1) are in outer space (far from gravitational influence of other objects) at a distance R from each other. They move in circular orbits about their centre of mass with angular velocities w1 for m1 and w2 for m2. In that case [Online May 19, 2012] (a) (1 – f) w1 = fw (b) w1 = w2 and independent of f (c) fw1 = (1 – f)w2 (d) w1 = w2 and depend on f

Movable roller

TOPIC 3

36. Four point masses, each of mass m, are fixed at the corners of a square of side l. The square is rotating with angular frequency w, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is : [Sep. 06, 2020 (I)]

(a) ml 2w

(a)

3g cos q 2l

(b)

2g cos q 3l

3g 2g sin q sin q (d) 2l 2l 33. Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to : (Take the radius of the drum to be 1.25 m and its axle to be horizontal): [Online April 10, 2016] (a) 27.0 (b) 0.4 (c) 1.3 (d) 8.0

(c)

Torque, Couple and Angular Momentum

ax is

(a) constant speed (b) decreasing speed (c) increasing speed 3 (d) speed which is th of that of the roller when the 4 weight is 0.4 m above the ground 32. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle [2017] q with the vertical is

(b) 4 ml2w

(c) 3 ml2w (d) 2 ml2w 37. A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be ______________. [NA Sep. 05, 2020 (II)] 38. A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre __________. [NA Sep. 03, 2020 (I)]

P-81

System of Particles and Rotational Motion

39. A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle q before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of q is approximately: (take g = 10 m/s2)

42. A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is : [Sep. 02, 2020 (I)] F

[Sep. 03, 2020 (I)] O

M, l

q m v

(a) 63° (c) 69° 40.

æ R-aö (a) Mg 1 - ç ÷ è R ø

m

m

(b) 55° (d) 49° w

l q

A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed w the rod makes an angle q with it (see figure). To find q equate the rate of change of angular momentum (direction going into the paper)

ml 2 2 w sin q cos q about the centre of mass (CM) to the 12 torque provided by the horizontal and vertical forces FH and FV about the CM. The value of q is then such that : [Sep. 03, 2020 (II)] (a) cos q = (c) cos q =

2g

(b) cos q =

3l w2 g lw

(d) cos q =

2

g 2l w2 3g 2l w2

41.

A

25

50

75 2m

a 2

æ R ö (b) Mg ç ÷ -1 è R-aø

a a2 (d) Mg 1 - 2 R R 43. Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A mass m moving with velocity v making p angle q = to the rod’s long axis collides with one end of 4 the rod and sticks to it. The angular speed of the rod-mass system just after the collision is: [8 Jan. 2020 I] 3 v 3v (a) (b) 7 2l 7l 4v 3 2v (d) 7l 7 l 44. A particle of mass m is moving along a trajectory given by x = x0 + a cos w1t y = y0 + b cos w2 t The torque, acting on the particle about the origin, at t = 0 is: [10 Apr. 2019 I] 2$ 2$ + my a (a) m ( - x0 b + y0 a ) w1 k (b) 0 w1 k 2 2 $ (c) zero (d) -m( x0 bw2 - y0 aw1 )k 45. The time dependence of the position of a particle of mass r m = 2 is given by r (t ) = 2 t i$ - 3t 2 $j . Its angular momentum, with respect to the origin, at time t = 2 is : [10 Apr. 2019 II] $ $ (a) 48 i + j (b) 36k$

(c)

(

(

)

(c) -34 k$ - $i

0

2

R

(c) Mg

FV FH

O

100 B

Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is : [Sep. 02, 2020 (I)] (a) 0.5 mg (b) 2 mg (c) 0.75 mg (d) 1 mg

)

(d) -48k$ 46. A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5s, is close to : [10 Apr. 2019 II]

P-82

Physics

(a) 4.0×10–6 Nm (b) 1.6×10–5 Nm (c) 7.9×10–6 Nm (d) 2.0×10–5 Nm 47. A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5m. When released, it slips off the table in a very short time t = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to : [8 Apr. 2019 II]

(a) 0.5 (b) 0.3 (c) 0.02 (d) 0.28 48. A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : (Take g = 10 m/s2) [12 Jan. 2019 II] O a = 10 m A

h = 10 m B

(a) 2 kg-m2/s

(b) 8 kg-m2/s

50. The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians): [11 Jan. 2019 II] p p p p (b) (c) (d) 6 3 8 4 51. To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is m, the torque, applied by the machine on the mop is: [10 Jan. 2019 I] (a) m FR/3 (b) m FR/6 2 μ mFR (c) m FR/2 (d) 3 52. A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be: [10 Jan. 2019 II] 2l l

(a)

P

5 Mo

2 Mo

g g (b) 13l 3l g 7g (c) (d) 2l 3l 53. An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is q, then: [9 Jan. 2019 I] (a)

(d) 3 kg-m2/s uur uur 49. A slab is subjected to two forces F1 and F2 of same uur magnitude F as shown in the figure. Force F2 is in XY(c) 6

kg-m2/s

plane while force F1 acts along z-axis at the point r r 2i + 3 j . The moment of these forces about point O will

(

)

be :

[11 Jan. 2019 I] Z

F1 F2

O

30º

6m

(a) (c)

( 3$i - 2 $j + 3k$ ) F ( 3$i + 2 $j – 3k$ ) F

54. (b) (d)

tan q =

(c)

tan q =

y

4m

x

(a)

( 3$i - 2 $j – 3k$ ) F ( 3$i + 2 $j + 3k$ ) F

A

1

(b) tan q =

2 3 2

(d) tan q =

3 B

x

1 2

1 3

P-83

System of Particles and Rotational Motion

A uniform rod AB is suspended from a point X, at a variable distance from x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) values is recorded. The appropriate variable that give a straight line, when plotted, are: [Online April 15, 2018] 1 1 (a) m, (b) m, 2 (c) m, x (d) m, x2 x x 55. A thin uniform bar of length L and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the L L bar after collision at a distance and respectively 3 6 from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be: [Online April 15, 2018] L/6

v

L/3

2v

O

58. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed w rad/s about the vertical. About the point of suspension: [2014] (a) angular momentum is conserved. (b) angular momentum changes in magnitude but not in direction. (c) angular momentum changes in direction but not in magnitude. (d) angular momentum changes both in direction and magnitude. 59. A ball of mass 160 g is th rown up at an an gle of 60° to the horizontal at a speed of 10 ms–1. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly (g = 10 ms–2) [Online April 19, 2014] (a) 1.73 kg m2/s (b) 3.0 kg m2/s (c) 3.46 kg m2/s (d) 6.0 kg m2/s 60. A particle is moving in a circular path of radius a, with a constant velocity v as shown in the figure. The centre of circle is marked by ‘C’. The angular momentum from the origin O can be written as: [Online April 12, 2014] y

v 6v 3v v (b) (c) (d) 6L 5L 5L 5L 56. A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure : [2016]

(a)

O

C

y

D

O

a V

C

a V V a V A B a V 45° R

61.

a

Which of the following statements is false for the angular ur momentum L about the origin? ur éR ù + a ú k$ when the particle is moving from (a) L = mv ê ë 2 û B to C. ur mv Rk$ when the particle is moving from D to A. (b) L = 2 ur mv $ Rk when the particle is moving from A to B. (c) L = – 2 ur éR ù - a ú k$ when the particle is moving from (d) L = mv ê ë 2 û C to D. 57. A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s–1, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is : [Online April 11, 2015] (a) 14.4 kg m2s–1 (b) 8.64 kg m2s–1 (c) 20.16 kg m2s–1 (d) 11.52 kg m2s–1

a

q

62.

63.

64.

x

(a) va (1 + cos 2q) (b) va (1 + cos q) (c) va cos 2q (d) va A particle of mass 2 kg is moving such that at time t, its r position, in meter, is given by r (t ) = 5iˆ - 2t 2 ˆj . The angular momentum of the particle at t = 2s about the origin in kg m–2 s–1 is : [Online April 23, 2013] (a) - 80 k$ (b) (10iˆ - 16 ˆj ) (c) -40k$ (d) 40k$ A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it will be : [Online April 9, 2013] (a) 6.25 rad/sec (b) 0.625 rad/sec (c) 3.35 rad/sec (d) 0.335 rad/sec A stone of mass m, tied to the end of a string, is whirled around in a circle on a horizontal frictionless table. The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn, where A is a constant, r is the instantaneous radius of the circle. The value of n is equal to [Online May 26, 2012] (a) – 1 (b) – 2 (c) – 4 (d) – 3 A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a

P-84

Physics

point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.[2011] (a) continuously decreases (b) continuously increases (c) first increases and then decreases (d) remains unchanged 65. A small particle of mass m is projected at an angle q with the x-axis with an initial velocity n0 in the x-y plane as

n sin q shown in the figure. At a time t < 0 , the angular g

momentum of the particle is

[2010]

y v0

(a) Angular velocity (b) Angular momentum (c) Moment of inertia (d) Rotational kinetic energy r 70. Let F be the force r acting on a particle having position r vector r , and T be the torque of this force about the origin. Then [2003] r r r r (a) r .Tr = 0 and Fr .Tr ¹ 0 r (b) r .T ¹ 0 and F .T = 0 r r r r (c) r .T ¹ 0 and F .T ¹ 0 r r r r (d) r .T = 0 and F .T = 0 71. A particle of mass m moves along line PC with velocity v as shown. What is the angular momentum of the particle about P? [2002] C L

q

P

x

(a)

-mg n0t 2 cos q ˆj

1 (c) - mg n 0 t 2 cos q kˆ 2

(b)

mg n 0t cos q kˆ

(d)

1 mgn 0 t 2 cos q iˆ 2

where iˆ, ˆj and kˆ are unit vectors along x, y and z-axis respectively. 66. Angular momentum of the particle rotating with a central force is constant due to [2007] (a) constant torque (b) constant force (c) constant linear momentum (d) zero torque 67. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity w. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity w' = [2006] w(m - 2M ) w (m + 2 M ) (b) (a) (m + 2M ) m wm wm (c) (d) (m + M ) (m + 2M ) 68. A force of – Fkˆ acts on O, the origin of the coordinate system. The torque about the point (1, –1) is [2006] Z

O

Y

X

(a) F (iˆ - ˆj ) (b) - F (iˆ + ˆj ) ˆ ˆ (c) F (i + j ) (d) - F (iˆ - ˆj ) 69. A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected ? [2004]

l

(a) mvL

TOPIC 4

O

(b) mvl

(c) mvr

(d) zero.

Moment of Inertia and Rotational K.E.

72. Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is : [Sep. 06, 2020 (I)] R

H

MR 2 M (R2 + H 2 ) (b) 2 4 MH 2 MR 2 (c) (d) 3 3 73. The linear mass density of a thin rod AB of length L varies xö æ from A to B as l ( x ) = l 0 ç1 + ÷ , where x is the distance è Lø from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is : [Sep. 06, 2020 (II)]

(a)

5 7 ML2 (b) ML2 12 18 2 3 (c) ML2 (d) ML2 5 7 74. A wheel is rotating freely with an angular speed w on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel

(a)

P-85

System of Particles and Rotational Motion

of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is : [Sep. 05, 2020 (I)] (a)

5 6

(b)

1 4

3 4 75. ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I0. If part ADE is removed, the moment of

(c) 0

(d)

inertia of the remaining part about the same axis is

NI0 16

where N is an integer. Value of N is _______. [NA Sep. 04, 2020 (I)] A

79. Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is æ R 2 L2 ö + ÷÷ . If such a cylinder is to be made for a I = M çç è 4 12 ø given mass of a material, the ratio L/R for it to have minimum possible I is : [Sep. 03, 2020 (I)]

(a)

2 3

(b)

3 2 (d) 2 3 80. An massless equilateral triangle EFG of side 'a' (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the N ma 2 line EX perpendicular to EG in the plane of EFG is 20 where N is an integer. The value of N is _________. [Sep. 03, 2020 (II)]

(c)

X E

D

3 2

F

G B

C

76. A circular disc of mass M and radius R is rotating about its axis with angular speed w1. If another stationary disc R having radius and same mass M is dropped co-axially 2 on to the rotating disc. Gradually both discs attain constant angular speed w2. The energy lost in the process is p% of the initial energy. Value of p is ___________. [NA Sep. 04, 2020 (I)] 77. Consider two uniform discs of the same thickness and different radii R1 = R and R2 = aR made of the same material. If the ratio of their moments of inertia I1 and I2, respectively, about their axes is I1 : I2 = 1 : 16 then the value of a is : [Sep. 04, 2020 (II)] (a) 2 2 (c) 2 78. y

O

(b) 2 (d) 4 O'

80 cm

x 60 cm For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is : [Sep. 04, 2020 (II)] (a) 2/3 (b) 1/4 (c) 1/8 (d) 1/2

E

a

G

(b)

20 J 3

81. Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are 0.1 kg-m2 and 10 rad s–1 respectively while those for the second one are 0.2 kg-m 2 and 5 rad s–1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is : [Sep. 02, 2020 (II)] (a)

10 J 3

5 2 J J (d) 3 3 82. Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio I0 of moment of inertia I0 of the system about an axis IA passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is: [9 Jan. 2020 I] 13 (a) 23 15 (b) 13 23 (c) 13 13 (d) 15

(c)

P-86

Physics

83. One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30° from the horizontal (see figure). Its angular speed when it hits the table is given as ns -1 , where n is an integer. The value of n is ________ . [9 Jan. 2020 I]

84. A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses m1 and m2 (m1 > m2) are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when m1 descents by a distance h is: [9 Jan. 2020 II]

87. Mass per unit area of a circular disc of radius a depends on the distance r from its centre as s(r) = A + Br. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is: [7 Jan. 2020 II]

aB ö 4æ A (a) 2pa çè + ÷ø 4 5

Bö 4 æ A aB ö 4æA (c) pa çè + ÷ø (d) 2pa çè + ÷ø 4 5 4 5 88. A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as æ s0 ö ç ÷ , then the radius of gyration of the disc about its axis è r ø passing through the centre is : [12 Apr. 2019 I]

1/ 2

(a)

a2 + b2 + ab 2

(b)

a+b 2

1/ 2

(c)

a2 + b2 + ab 3

(d)

a+b 3

é 2(m1 - m2 ) gh ù ú (a) ê 2 ëê (m1 + m2 )R + 1 ûú é 2(m1 + m2 ) gh ù ú (b) ê 2 êë (m1 + m2 )R + 1 úû

89. Two coaxial discs, having moments of inertia I1 and

1/2

é (m1 - m2 ) ù ú (c) ê 2 ëê (m1 + m2 )R + 1 ûú

gh

1/2

é ù m1 + m2 ú gh (d) ê 2 êë (m1 + m2 )R + 1 úû 85. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be: [7 Jan. 2020 I]

1 4 gh r 3

(b) r

3 2 gh

w1 , 2 about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef – Ei) is : [10 Apr. 2019 I] 3 2 I w2 I1w12 I w2 (b) 1 1 (c) I1w1 (d) - 1 1 8 6 12 24 90. A thin disc of mass M and radius R has mass per unit area s(r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is : [10 Apr. 2019 I]

(a) -

3 1 2 gh r (d) 4 gh r 3 86. The radius of gyration of a uniform rod of length l, about l an axis passing through a point away from the centre 4 of the rod, and perpendicular to it, is: [7 Jan. 2020 I]

(a)

1 l 4

(b)

1 l 8

(c)

7 l 48

(d)

3 l 8

MR 2 3

(b)

MR 2 6

(d)

2MR 2 3

MR 2 2 91. A solid sphere of mass M and radius R is divided into two (c)

(c)

I1 , are 2

rotating with respective angular velocities w1 and

(a) (a)

Bö 4 æ aA + ÷ (b) 2pa çè 4 5ø

7M and is 8 converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is given by : [10 Apr. 2019 II] (a) 185 (b) 140 (c) 285 (d) 65

unequal parts. The first part has a mass of

P-87

System of Particles and Rotational Motion

92. A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of q, where q is the angle by which it has rotated, is given as kq2. If its moment of inertia is I then the angular acceleration of the disc is: [9 April 2019 I] k k k 2k q q (c) 2 I q q (a) (b) (d) 4I I I 93. Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of: [9 Apr. 2019 II] (a) 2.5s (b) 2s (c) 5s (d) 3s 94. A thin smooth rod of length L and mass M is rotating freely with angular speed w0 about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be: [9 Apr. 2019 II] M w0 M w0 (a) (b) M +m M + 3m M w0 M w0 (d) (c) M + 6m M + 2m 95. A thin circular plate of mass M and radius R has its density varying as r(r) = r0 r with r0 as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = a MR2. The value of the coefficient a is: [8 April 2019 I] 3 (a) 1 2 (b) 3 5 (c) 8 5 (d) 2 96. Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be 1. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is: [12 Jan. 2019 I] (a) 12 cm (b) 16 cm (c) 14 cm (d) 18 cm 97. The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is ‘I(x)’. Which one of the graphs represents the variation of I(x) with x correctly? [12 Jan. 2019 II]

I(x)

I(x)

(a)

(b)

x

O

x

O

I(x)

I(x)

(c)

(d) O

x

O

x

98. An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then : [11 Jan. 2019 I] A

D

E G

B

C

F

15 3 I0 (b) I = I0 16 4 9 I0 (c) I = I 0 (d) I = 16 4 99. a string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) [11 Jan. 2019 II] 40 N

(a)

I=

(a) 20 rad/s2 (b) 16 rad/s2 2 (c) 12 rad/s (d) 10 rad/s2 100. A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO’, passing through the centre of D1, as shown in the figure, will be : [11 Jan. 2019 II] O'

D2

O

D3

D1

(a) MR2 (b) 3MR2 4 2 MR 2 MR 2 (d) (c) 5 3 101. Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is: [10 Jan. 2019 II]

P-88

Physics

(a) (c)

137 MR 2 15

(b)

209 MR 2 15

(d)

17 MR 2 15 152 MR 2 15

m are connected at the two ends of 2 a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system (see figure). Because of torsional constant k, the restoring toruque is t= kq for angular displacement q. If the rod is rotated by q0 and released, the tension in it when it passes through its mean position will be: [9 Jan. 2019 I]

102. Two masses m and

105. From a uniform circular disc of radius R and mass 9 M, a R is removed as shown in the figure. 3 The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing, through centre of disc is : [2018]

small disc of radius

2R 3 R

40 37 MR2 (c) 10 MR2 (d) MR2 9 9 106. A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z¢ axes will be [Online April 16, 2018]

(a) 4 MR2 (b)

z

z¢

O

3k q0 2 2k q0 2 k q0 2 k q0 2 (b) (c) (d) l l l 2l 103. A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rads–1) will be (g = 10 ms–2) [9 Jan. 2019 II]

x

(a)

(a) 1 : 2 (b) 1 : 4 (c) 1 : 3 (d) 1 : 5 107. A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle a with the horizontal, will be proportional to: (see figure) [Online April 15, 2018] M

30°

(a)

30 7

(b)

30

20 30 (d) 3 2 104. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is: [2018] (c)

P O

(a)

19 MR 2 2

(b)

55 MR 2 (c) 2

73 181 MR 2 (d) MR2 2 2

y

a

N

(a) (b) cosa cos a (c) sin a (d) sin a 108. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum ? [2017] 3 3 3 (a) 1 (b) (c) (d) 2 2 2 109. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing thr ough its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is : [Online April 8, 2017] (a)

7 Io 8

(b)

15 Io 16

(c)

3Io 4

(d)

31I o 32

C

E

F O A

D

B

P-89

System of Particles and Rotational Motion

R is made in a thin uniform 4 disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is : [Online April 9, 2017] 219 MR 2 (a) 256

110. A circular hole of radius

R 237 MR 2 R/4 512 o' O 19 MR 2 (c) 3R/4 512 197 MR 2 (d) 256 111. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is : [2015] 4MR 2 4MR 2 (a) (b) 9 3p 3 3p 2 MR MR 2 (c) (d) 32 2p 16 2p 112. Consider a thin uniform square sheet made of a rigid material. If its side is ‘a’ mass m and moment of inertia I about one of its diagonals, then :[Online April 10, 2015]

(b)

(a)

I>

ma 2 12

(b)

ma 2 ma 2 IB (c) IA = IB

126. A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is [2003] L (a) (b) 2 L 4 L (c) 4 L (d) 2 127. Moment of inertia of a circular wire of mass M and radius R about its diameter is [2002] (a) MR2/2 (b) MR 2 (c) 2MR 2 (d) MR2/4 128. Initial angular velocity of a circular disc of mass M is w 1. Then two small spheres of mass m are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc? [2002] æ M + mö æ M + mö (a) çè (b) çè ÷w ÷w M ø 1 m ø 1

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is m = 0.4, the maximum possible value b for box not to topple before moving is a ________. [NA 7 Jan. 2020 II] 131.A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio

of 100 ×

hsph hcyl

is given by :

[8 Apr. 2019 II]

ΙY = 16 Ι X

(d) ΙY = 64 Ι X (a)

2 5

(b) 1

(c)

14 15

(d)

4 5

P-91

System of Particles and Rotational Motion

132.The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius

R 2

R and (iii) a solid sphere of radius . If, in each case, the 4 speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is: [9 April 2019 I] (a) 4 : 3 : 2 (b) 10 : 15 : 7 (c) 14 : 15 : 20 (d) 2 : 3 : 4 133. A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is: [10 Jan. 2019 I]

(a)

3F 2mR

(b)

F 3m R

(c)

F 2mR

(d)

2F 3m R

134. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD, which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and Cd (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to: [2016] B D

O

Figure). If they roll on the incline without slipping such sin qc that their accelerations are the same, then the ratio sin q s is: [Online April 9, 2014] MC

A

M

S

B

qC qS

C

(a)

8 7

D

15 14

(b)

8 15 (d) 7 14 137. A loop of radius r and mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero.What will be the velocity of the centre of the hoop when it ceases to slip ? [2013]

(c)

(a)

rw0 4

(b)

rw0 3

rw0 (d) rw0 2 138. A tennis ball (treated as hollow spherical shell) starting from O rolls down a hill. At point A the ball becomes air borne leaving at an angle of 30° with the horizontal. The ball strikes the ground at B. What is the value of the distance AB ? (Moment of inertia of a spherical shell of mass m and radius 2 R about its diameter = mR 2 ) 3 [Online April 22, 2013]

(c)

O

A

C

(a) go straight. (b) turn left and right alternately. (c) turn left. (d) turn right. 135. A uniform solid cylindrical roller of mass ‘m’ is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is ‘a’ and it is rolling without slipping then the value of ‘F’ is: [Online April 10, 2015] 5 ma (a) ma (b) 3 3 ma (c) (d) 2 ma 2 136. A cylinder of mass Mc and sphere of mass Ms are placed at points A and B of two inclines, respectively (See

2.0 m 30° 0.2 m

A

B

(a) 1.87 m (b) 2.08 m (c) 1.57 m (d) 1.77 m 139. A thick-walled hollow sphere has outside radius R0. It rolls down an incline without slipping and its speed at the bottom is v0. Now the incline is waxed, so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be 5v0/4. The radius of gyration of the hollow sphere about an axis through its centre is [Online May 26, 2012] (a) 3R0/2 (b) 3R0/4 (c) 9R0 /16 (d) 3R0

P-92

Physics

140. A solid sphere is rolling on a surface as shown in figure, with a translational velocity v ms–1. If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is [Online May 12, 2012]

h

v

(a)

(b)

2gh

7 gh 5

7 10 gh gh (d) 2 7 141. A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. Then its acceleration is [2007]

(c)

(a) (c)

g sin q 2

1 - MR / I g sin q

1 + MR 2 / I

(b) (d)

g sin q 1 + I / MR 2 g sin q 1 - I / MR 2

P-93

System of Particles and Rotational Motion

1.

(3) Centre of mass of solid hemisphere of radius R lies at a distance

2.

L

Læ æ aL2 bL2 ö bx 3 ö xdm = ax + dx = + ç ÷ ç ÷ ò òç ç 2 4 ÷ø L2 ÷ø è 0 0è

3R above the centre of flat side of hemisphere. 8

3R 3 ´ 8 \ hcm = = = 3 cm 8 8 (23.00) Let s be the mass density of circular disc.

\

X CM

Original mass of the disc, m0 = pa 2 s

æ aL2 bL2 ö + çç ÷ 2 4 ÷ø è = bL aL + 3

Þ X CM =

a2 s Removed mass, m = 4

4.

3L æ 2a + b ö ç ÷ 4 è 3a + b ø

(b)

æ a2 ö æ 4p - 1ö Remaining, mass, m ' = ç pa 2 - ÷ s = a 2 ç s è 4 ÷ø 4 è ø Y

a 2 a 2

For given Lamina x y m1 = 1, C1 = (1.5, 2.5) m2 =3, C2 = (0.5, 1.5) m x + m2 x2 1.5 + 1.5 = = 0.75 X cm = 1 1 m1 + m2 4

X 1

New position of centre of mass X CM

a2 a 2 m0 x0 - mx pa ´ 0 - 4 ´ 2 = = m0 - m a2 pa 2 4

=

3.

a -a 3 / 8 -a -a = = =1 ö 2 2(4p - 1) 8p - 2 23 æ çè p - ÷ø a 4

\ x = 23 (b) Given,

æxö Linear mass density, r( x ) = a + b ç ÷ èLø ò xdm X CM = ò dm L

ò dm = ò r( x)dx 0

Lé

æxö = ò êa + b ç ÷ èLø 0 êë

2ù

bL ú dx = aL + 3 úû

m1 y1 + m2 y2 2.5 + 4.5 = = 1.75 m1 + m2 4 \ Coordinate of centre of mass of flag shaped lamina (0.75, 1.75) (a) Mass of sphere = volume of sphere x density of sphere Ycm =

2

5.

=

4 3 pR r 3

Mass of cavity M cavity =

4 p(1)3 r 3

Mass of remaining 4 4 M (Remaining) = pR3r – p(1)3 r 3 3 Centre of mass of remaining part, M r + M 2 r2 X COM = 1 1 M1 + M 2 é4 3 ù é4 ù 3 êë 3 pR rúû 0 + êë 3 p(1) (– r) úû [ R –1] Þ –(2 – R ) = 4 3 4 pR r + p(1)3 (–r) 3 3

P-94

Physics

Þ

Þ

( R – 1) 3

( R –1)

= 2– R

=

( R –1) ( R –1)( R 2 + R + 1)

=2–R

b 12 So CM coordinates one

Þ (R2 + R + 1) (2 – R) = 1 6.

M a ´ 4 4 =- a M 12 M4

M ´0-

and yCM = -

(d)

a a 5a - = 2 12 12

x0 =

and y0 = Xcm =

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

b b 5b - = 2 12 12

2m (L,L)

10. (a)

Lö æ mç 2L, ÷ 2ø è

1 ´ 0 + 1.5 ´ 3 + 2.5 ´ 0 1.5 ´ 3 X cm = = = 0.9cm 1 + 1.5 + 2.5 5 Y

1 ´ 0 + 1.5 ´ 0 + 2.5 ´ 4 2.5 ´ 4 = = 2cm 1 + 1.5 + 2.5 5 Hence, centre of mass of system is at point (0.9, 2) Ycm =

7.

(c) x cm =

3L

2L m

m y + m2 y2 + m3 y3 Ycm = 1 1 m1 + m2 + m3

æ 5L ö ç ,0 ÷ è 2 ø

X

x-coordinate of centre of mass is 5mL 2 = 13 L Xcm = 4m 8 y-coordinate of centre of mass is

50 ´ 0 + 100 ´1 + 150 ´ 0.5 7 = m 50 + 100 + 150 12

2mL + 2mL +

æ Lö 2m ´ L + m ´ ç ÷ + m ´ 0 è 2ø 5L = Ycm = 4m 8 11. (c) To produce maximum moment of force line of action of force must be perpendicular to line AB. A

y cm = 8.

50 ´ 0 + 100 ´ 0 + 150 ´ 50 + 100 + 150

q

3 m 4

(a) Acceleration of centre of mass (acm) is given by r r m a + m2 a2 + ........ r \ acm = 1 1 m1 + m2 + ........

(2m)ajˆ + 3m ´ aiˆ + ma (-iˆ) + 4m ´ a(- ˆj ) 2m + 3m + 4m + m 2aiˆ - 2ajˆ a ˆ ˆ = = (i - j ) 10 5 (d) With respect to point q, the CM of the cut-off portion =

9.

3 2 =

æa bö ç , ÷ . è 4 4ø Using, xCM =

4m q 2m

2 1 = 4 2 12. (c) According to principle of moments when a system is stable or balance, the anti-clockwise moment is equal to clockwise moment. i.e., load × load arm = effort × effort arm When 5 mg weight is placed, load arm shifts to left side, hence left arm becomes shorter than right arm. \ tan q =

13. (c) Centre of mass xcm MX - mx M -m

B

Þ

1 y y2 + = 2 x x2

x = 2

æ x ö1 (rx ) ç ÷ + ry y /2 è2ø2 r( x + y )

P-95

System of Particles and Rotational Motion

x £1 L With increase in the value of n, the centre of mass shift towards the end x = L This is satisfied by only option (a).

A

Here

x

L

L

ò xdm

C

y

B (0,0)

xCM = 0

Bc y 1 + 3 = = 1.37 AB x 2 14. (d) Let density of cone = r.

h

ò ypr =

2

dyr

0

1 2 pR hr 3

h 2

=

ò0 r

=

dy h

R

C

For a cone, we know that r y y = \r= R R n n

L

xcm =

ò (ax + 0

L

0

n +1

L

é kx ù ê nú ëê (n + 1) L ûú 0

=

L(n + 1) n+2

L ; n = 1, 2

2R h

é y4 ù 3ê ú 3 ë 4 û0 = ycm = 0 3 = h 3 4 h h 15. (b) Centre of mass of the rod is given by: 3 ò 3 y dy

n

æxö ò k çè L ÷ø dx

2L 3L ; n = 2, xCM = ;.... 3 4 For n ® ¥xcm = L Moment of inertia of a square plate about an axis through its centre and perpendicular to its plane is. 19. (b) Let s be the mass per unit area of the disc. Then the mass of the complete disc = s(p(2R)2 )

r

h

0 L

xCM =

a

B

é x n+ 2 ù kê nú êë (n + 2) L úû 0

For n = 0 , xCM =

A y

0

n

æ xö ò k çè L÷ø .xdx

L

...(i)

1 2 R h 3

ò l dx

0

ò ydm ò dm

ydy

=

L

ò dm

\

Centre of mass, ycm =

0

=

L

L

ò x (l dx)

bx2 ) dx L bx

ò (a + L )dx 0 2

aL bL2 L æ a + b ö + ç ÷ 3 = è 2 3ø = 2 bL b aL + a+ 2 2 a b + 7L 2 3 = Now b 12 a+ 2 On solving we get, b = 2a 16. (c) 17. (d)

æ xö 18. (a) The linear mass density l = k ç ÷ è Lø

O

The mass of the removed disc = s (pR 2 ) = psR 2 Let us consider the above situation to be a complete disc of radius 2R on which a disc of radius R of negative mass is superimposed. Let O be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as : 2 R 4ps R 2 O –ps R

( 6p( 2R) ) ´ 0 + ( -6( pR )) R = 2

xc.m

\ xc.m

2

4psR2 - psR2

\ xc.m =

n

R

-psR2 ´ R

3psR2 R 1 == aR Þ a = 3 3

20. (c) Initially,

P-96

Physics

m1 ( - x1 ) + m2 x2 Þ m1 x1 = m2 x2 ...(1) m1 + m2 Let the particles is displaced through distanced away from centre of mass x1– d x2– d ¢ d d¢ m2 m1 O 0=

\0 =

24. (c) Here, rdr w2 r = rgdh R

h

0

0

Þ w2 ò rdr = g ò dh w dh

m1 (d - x1 ) + m2 ( x2 - d ') m1 + m2

dr

Þ 0 = m1d - m1 x1 + m2 x2 - m2 d ' m1 d [From (1).] m2 21. (a) The centre of mass of bodies B and C taken together does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes into play while breaking. l 22. (d) B A y1 Þd'=

Þ

w2 R 2 = gh 2

(Given R = 5 cm)

w2 R 2 25w2 = 2g 2g 25. (c) Free body diagram in the frame of disc \h =

kx

® mw2 (l0 + x ) ¬¾¾ m ¾¾

F

y2

P

\ mw2 (l 0 + x ) = kx

2l y

Þ x=

(0, 0) C To have translational motion without rotation, the force uur F has to be applied at centre of mass. i.e. the point ‘P’has to be at the centre of mass Taking point C at the origin position, positions of y, and y2 are r1 = 2l, r2 = l and ml = m and m2 = 2m m y + m2 y2 m ´ 2 l + 2m ´ l 4l = = y= 1 1 m1 + m2 3m 3

k – mw2 For k >> mw2 Þ

Fradial =

(Q r = l + x here) kx = mlw2 + mxw2

2

N a

\x=

mw acosq q P(a, b) mg

mgsinq

For steady circular motion 2

mw a cos q = mg sin q Þw=

g tan q a

\w =

g ´ 8aC = 2 2 gC a

q

mv 2 = mr w2 r

\ kx = m(l + x)w2

dy = tan q = 8Cx dx At P, tan q = 8Ca w

x mw2 = l0 k

26. (b) At elongated position (x),

23. (a) y = 4Cx 2 Þ y

ml 0 w2

mlw2 k – mw2

T

x

0

l

2 27. (d) ò (- dT ) = ò (dm)w x 2

mw a x

mgcosq x

æm ö 2 – T = ò ç l dx ÷w x è ø l

mw2 2 (l - x 2 ) l It is a parabola between T and x.

or T =

P-97

System of Particles and Rotational Motion

28. (b) N sin q = mw2 (r/2)

...(i)

The rotational speed of the drum v g 10 < < R R 1.25 The maximum rotational speed of the drum in revolutions per minute Þω
K3 A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetics energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g – ºC) (1 cal = 4.2 × 107 ergs) close to : [Sep. 05, 2020 (I)] (a) 87.5ºC (b) 83.3ºC (c) 119.2ºC (d) 38.4ºC The specific heat of water = 4200 J kg–1 K–1 and the latent heat of ice = 3.4 × 105 J kg–1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) : [Sep. 04, 2020 (I)] (a) 61.7 (b) 63.8 (c) 69.3 (d) 64.6 A calorimter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of 'm' is close to (Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 °C–1) [Sep. 03, 2020 (II)] (a) 2 (b) 4 (c) 3.2 (d) 2.6 Three containers C1, C2and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process) [8 Jan. 2020 II] C1

C2

C3

T

1l

2l

—

60°C

–

1l

2l

30°C

2l

—

1l

60°C

1l

1l

1l

q

The value of q (in °C to the nearest integer) is______. 20. M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/ g and heat of fusion of ice is 80 cal/g], the value of M is ________ [NA 7 Jan. 2020 II] o 21. When M1 gram of ice at –10 C (Specific heat = 0.5 cal g–1 oC–1) is added to M2 gram of water at 50oC, finally no ice is left and the water is at 0oC. The value of latent heat of ice, in cal g–1 is: [12 April 2019 I]

P-157

Thermal Properties of Matter

(a)

50 M 2 -5 M1

5M1 (b) M - 50 2

(c)

50M 2 M1

(d)

5 M2 -5 M1

22. A massless spring (K = 800 N/m), attached with a mass (500 g) is completely immersed in 1kg of water. The spring is stretched by 2cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K) [9 April 2019 II] (a) 10–4 K (b) 10–5 K (c) 10–1 K (d) 10–3 K 23. Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘q2’ and ‘q1’ respectively, (q2 > q1). The temperature at the interface is: [9 April 2019 II]

q1 9q 2 q 2 + q1 + (b) 10 10 2 q1 5q2 q 2q + (d) 1 + 2 (c) 6 6 3 3 24. A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is: [12 Jan. 2019 I] K1 + K 2 (a) (b) K1 + K 2 2 2K1 + 3K 2 K1 + 3K 2 (c) (d) 5 4 25. Ice at –20°C is added to 50 g of water at 40°C, When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to [11 Jan. 2019 I] (Specific heat of water = 4.2J/g/°C Specific heat of Ice = 2.1 J/g/°C Heat of fusion of water at 0°C = 334J/g) (a) 50g (b) 100 g (c) 60 g (d) 40 g 26. When 100 g of a liquid A at 100°C is added to 50 g of a

(a)

liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be : [11 Jan. 2019 II] (a) 85°C (b) 60°C (c) 80°C (d) 70°C

27. A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg– 1K–1 and 400 Jkg–1 K–1 ] [11 Jan. 2019 II] (a) 15% (b) 30% (c) 25% (d) 20% 28. A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK–1 m–1, the energy flux through it in the steady state is: [10 Jan. 2019 I] (a) 90 Wm–2 (b) 120 Wm–2 (c) 65 Wm–2 (d) 200 Wm–2 29. An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stablizes at 21.5°C. (Specific heat of brass is 394 J kg–1 K–1) [10 Jan. 2019 II] –1 –1 (a) 458 J kg K (b) 1232 J kg–1 K–1 (c) 916 J kg–1 K–1 (d) 654 J kg–1 K–1 30. Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent

3L , is 2 connected across AB (See figure). In steady state, temperature difference between P and Q will be close to: [9 Jan. 2019 I]

rod PQ, of same cross-section as AB and length

L 4 A L 2

P

L

Q

B

(a) 45°C (b) 75°C (c) 60°C (d) 35°C 31. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C [2017] (a) 1250°C (b) 825°C (c) 800°C (d) 885° C 32. In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150°C. Immediately, it is put into water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40°C. The specific heat of aluminium is : (take 4.2 Joule=1 calorie) [Online April 8, 2017] (a) 378 J/kg – °C (b) 315 J/kg – °C (c) 476 J/kg – °C (d) 434 J/kg – °C

P-158

Physics

33. An experiment takes 10 minutes to raise the temperature of water in a container from 0ºC to 100ºC and another 55 minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be 1 cal / g ºC, the heat of vapourization according to this experiment will come out to be : [Online April 11, 2015] (a) 560 cal/ g (b) 550 cal/ g (c) 540 cal/ g (d) 530 cal/ g 34. Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure. Area of cross - section of each rod = 4 cm2. End of copper rod is maintained at 100ºC where as ends of brass and steel are kept at 0ºC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:[2014] (a) 1.2 cal/s (b) 2.4 cal/s (c) 4.8 cal/s (d) 6.0 cal/s 35. A black coloured solid sphere of radius R and mass M is inside a cavity with vacuum inside. The walls of the cavity are maintained at temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as aT3 per unit mass with the temperature T of the sphere, where a is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be (s is Stefan Boltzmann constant) [Online April 19, 2014] (a)

(c) 36.

37.

Ma

æ3ö In ç ÷ 4pR s è 2 ø

(b)

2

Ma

æ 16 ö In ç ÷ 16pR s è 3 ø 2

(d)

Ma

æ 16 ö In ç ÷ 4pR s è 3 ø 2

Ma

æ3ö In ç ÷ 16pR s è 2 ø 2

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27°C to 77°C? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible). [Online April 9, 2014] (a) 8 min 20 s (b) 6 min 2 s (c) 7 min (d) 14 min Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged.What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is r and L is its latent heat of vaporization. [2013] (a) rL/T

(b)

T / rL (c) T/rL

(d) 2T/rL

38. A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from 30°C to 25°C. A mass of 100g of another liquid in an identical vessel with identical surroundings takes the same time to cool from 30° C to 25° C. The specific heat of the liquid is : (The water equivalent of the vessel is 30g.) [Online April 25, 2013] (a) 2.0 kcal/kg (b) 7 kcal/kg (c) 3 kcal/kg (d) 0.5 kcal/kg 39. 500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100°C is added to it. Then water in the calorimeter is : (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/ g) [Online April 23, 2013] (a) 580 g (b) 590 g (c) 600 g (d) 610 g 40. Given that 1 g of water in liquid phase has volume 1 cm 3 and in vapour phase 1671 cm3 at atmospheric pressure and the latent heat of vaporization of water is 2256 J/g; the change in the internal energy in joules for 1 g of water at 373 K when it changes from liquid phase to vapour phase at the same temperature is : [Online April 22, 2013] (a) 2256 (b) 167 (c) 2089 (d) 1 41. A large cylindrical rod of length L is made by joining two

L identical rods of copper and steel of length æç ö÷ each. è 2ø The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at 100°C and that of steel at 0°C then the temperature of junction is (Thermal conductivity of copper is 9 times that of steel) [Online May 19, 2012] (a) 90°C (b) 50°C (c) 10°C (d) 67°C 42. The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K is (s = 5.7 × 10–8 W m–2 K–4) [Online May 7, 2012] (a) 1.7 × 1010 J (b) 1.1 × 1012 J (c) 2.8 × 108 J (d) 4.6 × 106 J 43. 100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K): [2011] (a) 8.4 kJ (b) 84 kJ (c) 2.1 kJ (d) 4.2 kJ 44. The specific heat capacity of a metal at low temperature (T) is given as 3

æ T ö C p (kJK -1kg -1 ) = 32 ç è 400 ÷ø A 100 gram vessel of this metal is to be cooled from 20ºK to 4ºK by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is [2011 RS] (a) greater than 0.148 kJ (b) between 0.148 kJ and 0.028 kJ (c) less than 0.028 kJ (d) equal to 0.002 kJ

P-159

Thermal Properties of Matter

45. A long metallic bar is carrying heat from one of its ends to the other end under steady–state. The variation of temperature q along the length x of the bar from its hot end is best described by which of the following figures?[2009]

q

q

(a)

r1

(b)

x

x

q

r2

q

(c)

One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of length l1 and l2 and thermal conductivities K1 and K2 respectively. The temperature at the interface of the two section is [2007] T1 l1

K1

47.

x

l2

T2

K2

(a)

( K1l1T1 + K 2l2T2 ) ( K1l1 + K 2l2 )

(b)

( K 2l2T1 + K1l1T2 ) ( K1l1 + K 2 l2 )

(c)

( K 2l1T1 + K1l2T2 ) ( K 2 l1 + K1l2 )

(d)

( K1l2T1 + K 2l1T2 ) ( K1l2 + K 2 l1 )

(a) (c)

r02 R2 s

T

T4 r

2

(b)

4

(b)

(r2 - r1 ) (r1 r2 )

(c)

( r2 - r1 )

(d)

r1 r2 (r2 - r1 )

50. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be [2004] (a) 32 (b) 16 (c) 4 (d) 64 51. The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1 ) . The rate of heat transfer through the slab, in a steady state is æ A(T2 - T1 ) K ö çè ÷ø f , with f equal to x

(d)

R2 s

T

r

7 T0 3 5 (d) T f = T0 2

(b) T f =

[2004]

4x

2K

T1

4

temperature T0, while Box contains one mole of helium at æ 7ö temperature çè ÷ø T0 . The boxes are then put into thermal 3 contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf in terms of T0 is [2006] 3 T0 7 3 (c) T f = T0 2

K

2

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at

(a) T f =

x

T4

4pr r2 where r0 is the radius of the Earth and s is Stefan's constant. 48.

2

pr02 R2 s

T2

ær ö ln ç 2 ÷ è r1 ø

Assuming the Sun to be a spherical body of radius R at a temperature of TK, evaluate the total radiant powerd incident of Earth at a distance r from the Sun [2006]

4pr02 R 2s

T1

(a) (d)

x 46.

49. The figure shows a system of two concentric spheres of radii r1 and r2 are kept at temperatures T1 and T2, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to [2005]

2 1 1 (b) (c) 1 (d) 3 2 3 52. The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by [2003] (a) Rayleigh Jeans law (b) Planck’s law of radiation (c) Stefan’s law of radiation (d) Wien’s law 53. Heat given to a body which raises its temperature by 1°C is [2002] (a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient

(a)

P-160

Physics

62. A hot body, obeying Newton’s law of cooling is cooling down from its peak value 80°C to an ambient temperature of 30°C. It takes 5 minutes in cooling down from 80°C to 40°C. How much time will it take to cool down from 62°C to 32°C? (Given In 2 = 0.693, In 5 = 1.609) [Online April 11, 2014] (a) 3.75 minutes (b) 8.6 minutes (c) 9.6 minutes (d) 6.5 minutes 63. If a piece of metal is heated to temperature q and then allowed to cool in a room which is at temperature q0, the graph between the temperature T of the metal and time t will be closest to [2013]

T (a)

(c)

T q0

(a)

60.

(d)

T q0

(b)

0

t

t

(d)

A body takes 10 minutes to cool from 60°C to 50°C. The temperature of surroundings is constant at 25°C. Then, the temperature of the body after next 10 minutes will be approximately [Online April 15, 2018] (a) 43°C (b) 47°C (c) 41°C (d) 45°C 61. Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is: [Online April 12, 2014] (a) 25°C (b) 10°C (c) 15°C (d) 20°C

0

(d)

t

loge (q – q0)

loge (q – q0)

(b)

(c) (c)

t

O

t t O O 64. A liquid in a beaker has temperature q(t) at time t and q0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge(q – q0) and t is : [2012]

0

(a)

T q0

loge (q – q0)

A metallic sphere cools from 50°C to 40°C in 300 s. If atmospheric temperature around is 20°C, then the sphere's temperature after the next 5 minutes will be close to : [Sep. 03, 2020 (II)] (a) 31°C (b) 33°C (c) 28°C (d) 35°C 59. Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8 × 102 kg/m3 and specific heat of 2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3 and specific heat of 4000 J kg–1 K–1. Which of the following best describes their temperature versus time graph schematically ? (assume the emissivity of both the beakers to be the same) [8 April 2019 I]

t

O

TOPIC 3 Newton's Law of Cooling 58.

(b)

loge (q – q0)

54. Infrared radiation is detected by [2002] (a) spectrometer (b) pyrometer (c) nanometer (d) photometer 55. Which of the following is more close to a black body? [2002] (a) black board paint (b) green leaves (c) black holes (d) red roses 56. If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should[2002] (a) increase (b) remain unchanged (c) decrease (d) first increase then decrease 57. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is [2002] (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9.

0

t

65. According to Newton’s law of cooling, the rate of cooling of a body is proportional to (Dq)n , where Dq is the difference of the temperature of the body and the surroundings, and n is equal to [2003] (a) two (b) three (c) four (d) one

P-161

Thermal Properties of Matter

1.

(a) Let L'1 and L'2 be the lengths of the wire when temperature is changed by DT °C . At T °C, Leq = L1 + L2

4.

Volume increase by 0.06% therefore density decrease by 0.06%. (60.00) Volume, V = Ibh

5.

DV Dl Db Dh = + + V l b h (g = coefficient of volume expansion) Þ g = 5 × 10–5 + 5 × 10–6 + 5 × 10–6 = 60 × 10–6/°C \ Value of C = 60.00 (Bonus) Dtemp = Dload and A = pr2 = p(10–3)2 = p × 10–6 \ g=

At T + D°C L'eq = L'1 + L'2 \ Leq (1 + a eq DT ) = L1 (1 + a1DT ) + L2 (1 + a 2 DT )

[Q L ' = L (1 + a DT )] Þ ( L1 + L2 )(1 + a eq DT ) = L1 + L2 + L1a1DT + L2a 2 DT

Þ a eq =

2.

a1 L1 + a 2 L2 L1 + L2

Vb1 = Vm + Vm g m DT

f = 2p = 6.28 kg g (a) Change in length in both rods are same i.e. Dl1 =D l 2

6.

Unfilled volume (V0 - Vm ) = (Vb - Vm1 )

la1D q1 = la 2 Dq 2

Þ V0 g beaker = Vm g M

a1 Dq 2 = a 2 Dq1

V0 g beaker gM

or, Vm =

500 ´ 6 ´ 10 -6

q = 230°C 7.

(a) Let required temperature = T°C

M.P. o

o

0C

Dl = l aDT Here, a = Coefficient of linear expansion Here, Dl = 0.02%, DT = 10ºC

TC

B.P.

x0 x0 2

Þ a = 2 ´ 10-5

x0 6

-5

M Qr = V

DV -5 -2 ´ 100 = gDT = (6 ´ 10 ´ 10 ´ 100) = 6 ´ 10 V

x 0 x 0 x0 – = 2 3 6 x ö æ & ç x 0 – 0 ÷ = (100 – 0°C) 3 ø è

Þ T° C =

o

100 C

x0 3

Dl 0.02 = l DT 100 ´ 10

Volume coefficient of expansion, g = 3a = 6 ´ 10

é a1 4 ù êQ = ú ë a2 3 û

4 q – 30 = 3 180 – 30

= 20 cc. 1.5 ´ 10-4 (a) Change in length of the metal wire (Dl) when its temperature is changed by DT is given by

\a =

( p ´ 10 -6 ) ´ 1011

\ F = 20p N \ m =

When beaker is partially filled with Vm volume of mercury,

\Vm =

F ´ 0.2

or 0.2 × 10–5 × 20 =

(20.00) Volume capacity of beaker, V0 = 500 cc Vb = V0 + V0 g beaker DT

3.

FL AY

L a DT =

P-162

Physics

2x 0 300 =100 Þ x 0 = 3 2 x 0 150 Þ T° C = = = 25° C 6 6

340 - 273 °Y - ( -160) = 373 - 273 -50 - ( -160)

Þ

8.

Þ

F/ A stress = A Dl / l ( ) strain Using, coefficient of linear expansion,

(a) Young’s modulus Y =

a=

14.

(c) K=

Y=

F A ( aDT ) As we know, Bulk modulus DP æ -DV ö çè ÷ V ø

Þ

Y=

Þ

DV P = V K g=3 a

F æ DL ö = a D q÷ çèQ ø A. a. Dq L Force developed in the rail F = YAaDt = 2 × 1011 × 40 × 10–4 × 1.2 × 10–5 × 10 = 9.6 × 104 = 1 × 105 N 11. (c) Due to thermal exp., change in length (Dl) = l a DT ... (i) Normal stress Young’s modulus (Y) = Longitudinal strain FA Dl F Y= Þ = Dl l l AY Fl AY

From eqn (i),

F = AY a DT 12. (b) When there is no change in liquid level in vessel then g¢real = g¢ vessel Change in volume in liquid relative to vessel

DVapp = Vg 'app Dq = V(g 'real - g 'vessel ) 13.

Reading on any scale – LFP (c) UFP - LFP = constant for all scales

… (ii)

F Þ F = Y .S.aDT S .aDT \ The ring is pressing the wheel from both sides, Thus Fnet = 2F = 2YSaDT (a) As the rods are identical, so they have same length (l) and area of cross-section (A). They are connected in series. So, heat current will be same for all rods. Y =

15.

æ DQ ö æ DQ ö æ DQ ö Heat current = ç ÷ =ç ÷ =ç ÷ D t D t è ø AB è ø BC è Dt øCD Þ

(100 - 70) K1 A (70 - 20) K 2 A (20 - 0) K 3 A = = l l l

Þ K1 (100 - 70) = K 2 (70 - 20) = K3 (20 - 0) Þ K1 (30) = K 2 (50) = K3 (20) Þ

Fl = l a DT AY

DR R DT

R 1 DR = = a.DT Þ DR aDT R From equation (i) and (ii)

P P P = gDt Þ Dt = = K gK 3aK Thermal stress F A (b) Young's modulus = = DL L Strain Y=

…(i)

Þ

\

Dl =

F ´ 2 pR S2 pDR FR Y= S.DR

The coefficient of linear expansion a =

V = V0 (1 + gDt) DV = gDt V0

10.

\ Y = – 86.3° Y (d) The Young modulus is given as stress F/S = strain DL / L Here, DL = 2p DR L = 2p R

Dl Dl Þ =aDT lD T l

\Y =

9.

67 y + 160 = 100 110

K1 K 2 K3 = = 10 6 15

Þ K1 : K 2 : K3 = 10 : 6 :15 Þ K1 : K3 = 2 : 3. 16. (a) According to question, one half of its kinetic energy is converted into heat in the wood. 1 2 1 mv ´ = ms DT 2 2

v2 210 ´ 210 = = 87.5°C 4 ´ s 4 ´ 4.2 ´ 0.3 ´ 1000 17. (a) Here ice melts due to water. Let the amount of ice melts = mice Þ DT =

mw sw Dq = mice Lice

P-163

Thermal Properties of Matter

\ mice = =

18.

mw sw Dq Lice

22. (b)

0.2 ´ 4200 ´ 25 3.4 ´ 105

= 0.0617 kg = 61.7 g

\ DT = 1 ´ 10 -5 K

(a) Heat given by water = mw Cw (Tmix - Tw ) = 200 ´1´ (31 - 25) Heat taken by steam = m Lstem + m Cw (Ts – Tmix) = m × 540 + m (1) × (100 –31) = m × 540 + m (1) × (69) From the principal of calorimeter, Heat lost = Heat gained \ (200)(31 - 25) = m ´ 540 + m(1)(69)

Þ 1200 = m(609) Þ m » 2. 19. (50.00) Let Q1, Q2, Q3 be the temperatures of container C1,C2 and C3 respectively. Using principle of calorimetry in container C1, we have (q1 – 60) = 2 ms(60 – q) Þ q1 – 60 = 120 – 2q Þ q1 = 180 – 2q ...(i) For container C2 ms (q2 – 30) = 2ms (30 – q) Þ q2= 90 – 2q3 ...(ii) For container C3 2ms (q1 – 60) = ms (60 – q) Þ 2q1–120 = 60 – q Þ 2q1 + q = 180 ...(iii) Also, q1 + q2 + q3 = 3q ...(iv) Adding (i), (ii) and (iii) 3q1 + 3q2 + 3q3 = 450 Þ q1 + q2 + q3 = 150 Þ 3q = 150 Þ q = 50 ºC 20. (40) Using the principal of calorimetry Mice Lf + mice (40 – 0) Cw = mstream Lv + mstream (100 – 40) Cw Þ M (540) + M × 1 × (100 – 40) = 200 × 80 + 200 × 1 × 40 Þ 600 M = 24000 Þ M = 40g 21. (a) M1Cice × (10) + M1L = M2Cw (50) or M1 × Cice (=0.5) × 10 + M1L = M2 × 1 × 50 50M 2 Þ L = M -5 1

1 2 .kx = mC (DT ) + mw Cw DT 2 1 2 or ´ 800 ´ 0.02 = 0.5 ´ 400 ´ DT + 1 ´ 4184 ´ DT 2 d

23. (a) H1 = H2 q 2 3k

q 3d

k q1

æ q2 - q ö æ q - q1 ö = kA ç or (3k ) A ç ÷ è d ø è 3d ÷ø æ q + 9q 2 ö or q = ç 1 è 10 ÷ø 24. (d) Effective thermal conductivity of system K eq = =

K1A1 + K 2 A 2 A1 + A 2

A2

K1pR 2 + K 2 [p (2R)2 - pR 2 ]

K1A1

p(2R)2

K1(pR 2 ) + K 2 (3pR 2 )

K2

K1 + 3K 2 4 4pR 25. (d) Let m gram of ice is added. From principal of calorimeter heat gained (by ice) = heat lost (by water) \ 20 × 2.1 × m + (m – 20) × 334 = 50 × 4.2 × 40 376 m = 8400 + 6680 m = 40.1 26. (c) Heat loss = Heat gain = mSDq So, mASADqA= mBSBDqB Þ 100 × SA × (100 – 90) = 50 × SB × (90 – 75) =

2

=

3 S 4 B Now, 100 × SA × (100 – q) = 50× SB × (q – 50)

2SA = 1.5SB Þ SA =

æ 3ö 2 × çè ÷ø × (100 – q) = (q – 50) 4 300 – 3q = 2q – 100 400 = 5q Þ q = 80°C 27. (d) Assume final temperature = T°C Heat lass = Heat gain = msDT ÞmB sB DTB = mw swDTw 0.1 × 400 × (500 – T) = 0.5 × 4200 × (T – 30) + 800 (T – 30) Þ 40 (500 – T) = (T – 30) (2100 + 800) Þ 20000 – 40T = 2900 T– 30 × 2900 Þ 20000 + 30 × 2900 = T(2940) T = 30.4°C

P-164

Physics

and P × 55 × 60 = mL ...(ii) Dividing equation (i) by (ii) we get 10 C ´100 = 55 L \ L = 550 cal./g. 34. (c) Rate of heat flow is given by,

DT 6.4 ´100 = ´100 = 21%, T 30 so the closest answer is 20%. Temp. of Temp. of heat source heat reservoir 1m 28. (a) 3 2 10 K 10 K

Q=

æ dQ ö kADT çè ÷ø = dt l

Where, K = coefficient of thermal conductivity l = length of rod and A = area of cross-section of rod

1 æ dQ ö kD T Energy flux, çè ÷ø = A dt l

29.

100°C

( 0.1)( 900)

= 90 W/m 2 1 (c) Let specific heat of unknown metal be ‘s’ According to principle of calorimetry, Heat lost = Heat gain m × sDq = m1sbrass (Dq1 + m2 swater + Dq2) Þ 192 × S × (100 – 21.5) = 128 × 394 × (21.5 – 8.4) Solving we get,+ 240 × 4200 × (21.5 – 8.4) S = 916 Jkg–1k–1 =

30. (a)

Copper B

120

R/2

R/4

32.

33.

Brass 0°C

0.92 ´ 4(100 - T ) 46 0.26 ´ 4 ´ (T - 0) 0.12 ´ 4 ´ (T - 0) + 13 12 Þ 200 – 2T = 2T + T Þ T = 40°C

=

R

P

R/4

L/4 Q

R/2

B

120 ´ 5 3 360 DTPQ = = 45°C × R= 8R 5 8 (d) According to principle of calorimetry, Heat lost = Heat gain 100 × 0.1(T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45 10 T– 750 = 450 + 7650 = 8100 Þ T – 75 = 810 T = 885°C (d) According to principle of calorimetry, Qgiven = Qused 0.2 × S × (150 – 40) = 150 × 1 × (40 – 27) + 25 × (40 – 27) 0.2 × S × 110 = 150 × 13 + 25 × 13 Specific heat of aluminium 13 ´ 25 ´ 7 S= = 434 J/kg-°C 0.2 ´ 110 (b) As Pt = mCDT So, P × 10 × 60 = mC 100 ...(i)

0.92 ´ 4 ´ 60 = 4.8 cal/s 46 (c) In the given problem, fall in temperature of sphere,

\

O

In steady state temperature difference between P and Q,

31.

Steel

If the junction temperature is T, then QCopper = QBrass + QSteel

L L/4

T

0°C

DTAB 120 120 ´ 5 = = 8 R AB 8R R 5

A

KA(q1 - q 2 ) l

35.

QCopper =

dT = ( 3T0 - 2T0 ) = T0 Temperature of surrounding, Tsurr = T0 Initial temperature of sphere, Tinitial = 3T0 Specific heat of the material of the sphere varies as, c = aT 3 per unit mass (a = a constant) Applying formula,

(

dT sA 4 4 = T - Tsurr dt McJ

Þ

)

T0 s 4pR 2 é = ( 3T )4 - ( T0 )4 ùûú dt Ma ( 3T )3 J ëê 0 0

Þ dt =

Ma 27T04 J

s 4 pR 2 ´ 80T04

Solving we get, Time taken for the sphere to cool down temperature 2T 0,

t=

Ma

æ 16 ö ln ç ÷ 16pR s è 3 ø 2

P-165

Thermal Properties of Matter

36. (a) From question, In 1 sec heat gained by water = 1 KW – 160 J/s = 1000 J/s – 160 J/s = 840 J/s Total heat required to raise the temperature of water (volume 2L) from 27°c to 77°c = mwater ×sp. ht × Dq = 2 × 103 × 4.2 × 50 [Q mass = density × volume] And, 840 × t = 2 × 103 × 4.2 × 50 2 ´103 ´ 4.2 ´ 50 840 = 500 s = 8 min 20s 37. (d) When radius is decrease by DR,

or, t =

4pR 2 DRrL = 4pT[R 2 - (R - DR) 2 ] Þ rR2 DRL = T[R 2 - R 2 + 2RDR - DR 2 ] Þ rR 2 DRL = T2RDR

[ DR is very small]

2T rL 38. (d) As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from 30°C to 25°C) is same 2 minutes, therefore ÞR=

æ dQ ö æ dQ ö =ç ÷ çè ÷ø dt water è dt ø liquid (mw C w + W)DT (ml Cl + W) D T = t t (W = water equivalent of the vessel) or , m w C w = m l C l

=

L L + k ´ 0´ 2 2 L L 9k ´ + k ´ 2 2

9k ´ 100 ´

=

900 kL = 2 = 90°C 10kL 2 42. (a) According to Stefan’s law E = s T4 Heat radiated per unit area in 1 hour (3600s) is = 5.7× 10–8 × (300)4 × 3600 = 1.7 × 1010 J 43. (a) DU = DQ = mcDT 100 × 4184 (50 – 30) » 8.4 kJ 1000 44. (d) Required work = energy released

=

ò

Here, Q = mc dT 4

æ T3 ö = ò 0.1 ´ 32 ´ ç dT = 3÷ è 400 ø 20 4

òT

3

4

3.2

ò 64 ´ 106 T

3

dT

20

dT = 0.002kJ

Therefore, required work = 0.002 kJ 45. (a) Let Q be the temperature at a distance x from hot end of bar. Let Q is the temperature of hot end. The heat flow rate is given by

mWCW ml

dQ kA(q1 - q) = dt x

39. (b) As 1g of steam at 100°C melts 8g of ice at 0°C. 10 g of steam will melt 8× 10 g of ice at 0°C Water in calorimeter = 500 + 80 + 10g = 590g 40. (c) 41. (a) L Copper

K copper lsteel + Ksteel lcopper

20

50 ´ 1 = 0.5 kcal / kg 100

100°C

K copper qcopper lsteel + K steel qsteel lcopper

q=

= 5 ´ 10 –8

or,

\ Specific heat of liquid , Cl =

From formula temperature of junction;

Steel

0°C

L/2 L/2 Let conductivity of steel Ksteel = k then from question Conductivity of copper Kcopper = 9k qcopper = 100°C qsteel = 0°C L lsteel = lcopper = 2

x dQ kA dt Thus, the graph of Q versus x is a straight line with a positive intercept and a negative slope. The above equation can be graphically represented by option (a). 46. (d) Let T be the temperature of the interface. In the steady state, Q1 = Q2 Þ q - q = x dQ 1 kA dt

T1

\

Þ q = q1 -

1

2

K1

K2

T2

K1 A(T1 - T ) K 2 A(T - T2 ) , = l1 l2

P-166

Physics

where A is the area of cross-section. Þ

K1 A(T1 - T )l 2 = K 2 A(T - T2 )l1

Þ

K1T1l 2 - K1T l 2 = K 2T l1 - K 2T2 l1

Þ

( K 2 l1 + K1l 2 )T = K1T1l 2 + K2T2 l1 K1T1l 2 + K 2T2 l1 K 2 l1 + K1l 2

Þ T = =

K1l 2T1 + K 2 l1T2 . K1l 2 + K 2 l1

47. (b) From stefan's law, total power radiated by Sun, E = sT4 × 4pR2 The intensity of power Per unit area incident on earth's surface =

4

sT ´ 4pR

2

E

× Cross – Section area of earth facing the

4pr 2

4 2

sun =

dQ KA[(T - dT ) - T ] - KAdT = = dt dr dr

dT (Q A = 4pr 2 ) dr Since the area of the surface through which heat will flow is not constant. Integrating both sides between the limits of radii and temperatures of the two shells, we get 2 = -4pKr

æ dQ ö çè ÷ dt ø æ dQ ö çè ÷ dt ø

r2

1

T2

ò r 2 dr = -4pK ò dT

r1

T1

r2

T2

r1

T1

-2 ò r dr = -4pK

ò dT

2

4pr Total power received by Earth

E' =

The radial rate of flow of heat through this elementary shell will be

sT R 2

dQ é 1 1 ù ê - ú = -4pK [T2 - T 1 ] dt ë r1 r2 û

or

(pr02 )

r 48. (c) When two gases are mixed to gether then Heat lost by He gas = Heat gained by N2 gas

n1Cv1 DT1 = n2Cv2 DT2

\

dQ -4pKr1r2 (T2 - T1 ) = dt (r2 - r1 ) r r dQ µ 1 2 dt (r2 - r1 )

50. (d) From stefan's law, energy radiated by sun per second E = sAT 4 ;

3 é7 ù 5 R T0 - T f ú = R éëT f - T0 ùû 2 ëê 3 û 2

\ A µ R2

7T0 - 3T f = 5T f - 5T0

\ E µ R 2T 4

Þ 12T0 = 8T f Þ T f =

Þ Tf =

12 T0 8

\

3 T0 .. 2

E2 R22 T24 = E1 R12 T14

put R2 = 2R, R1 = R ; T2 = 2T, T1 = T

49. (d)

T - dT dr ·

T1

r1

r

T2 r2

Consider a thin concentric shell of thickness (dr) and of radius (r) and let the temperature of inner and outer surfaces of this shell be T and (T – dT) respectively.

E (2R )2 (2T ) 4 Þ 2 = = 64 E1 R 2T 4

51. (d) The thermal resistance is given by x 4x x 2 x 3x + = + = KA 2KA KA KA KA Amount of heat flow per second, (T - T1 ) KA dQ D T = = 2 x 3 dt 3x KA

=

1 ì A(T2 - T1 ) K ü í ý 3î x þ

\f =

1 3

P-167

Thermal Properties of Matter

52. (d) Wein’s law correctly explains the spectrum 53. (b) Heat required for raising the temperature of a body through 1ºC is called its thermal capacity. 54. (b) Pyrometer is used to detect infra-red radiation. 55. (a) Black body is one which absorb all incident radiation. Black board paint is quite approximately equal to black bodies. 56. (c) When water is cooled at 0°C to form ice, energy is released from water in the form of heat. As energy is equivalent to mass, therefore, when water is cooled to ice, its mass decreases. 57. (a) From stefan's law, the energy radiated per second is given by E = esT4 A Here, T = temperature of the body A = surface area of the body For same material e is same. s is stefan's constant Let T1 and T2 be the temperature of two spheres. A1 and A2 be the area of two spheres. \

=

E1 T14 A1 T14 4pr12 = = E2 T24 A2 T24 4pr22 (4000)4 ´12 (2000)4 ´ 42

=

Þ 40 - T = \T =

T Þ 200 - 5T = T 5

200 = 33.3°C 6

æ dT ö 4 59. (b) Rate of Heat loss = mS çè ÷ø = esAT dt

æ dT ö æ dT ö Þ ç- ÷ > ç- ÷ è dt ø A è dt ø B

So, A cools down at faster rate. 60. (a) According to Newton’s law of cooling,

æ q1 - q2 ö æ q1 + q2 ö - q0 ÷ çè ÷ø = K çè ø t 2 æ 60 - 50 ö æ 60 + 50 ö - 25÷ çè ÷ø = K çè ø 10 2

..... (i)

..... (ii)

q1 - q2 é q + q2 ù = -K ê 1 - q0 ú t 2 ë û

where q0 is the temperature of surrounding. Now, hot water cools from 60°C to 50°C in 10 minutes, ...(i)

Let T be the temperature of sphere after next 5 minutes. Then

40 - T 40 + T - 40 T = = 10 50 + 40 - 40 50

rB S B 103 4000 ´ = ´ r A S A 8 ´ 102 2000

61. (b) By Newton’s law of cooling

Here, T1 = 50°C, T2 = 40°C an d To = 20°C, t = 600S = 5 minutes

Dividing eqn. (ii) by (i), we get

æ dT ö çè - ÷ø dt B

=

10 60 = Þ q = 42.85°C @ 43°C (50 - q) q

T1 - T2 éT + T ù = K ê 1 2 - T0 ú t 2 ë û

40 - T æ 40 + T ö = Kç - 20 ÷ 5 è 2 ø

æ dT ö çè - ÷ø dt A

Dividing eq. (i) by (ii),

1 1

50 - 40 æ 50 + 40 ö = Kç - 20 ÷ 5 Min è 2 ø

dT es ´ A ´ T 4 dT 1 = Þµ dt r ´ Vol. ´ S dt rS

æ 50 - q ö æ 50 + q ö - 25÷ and, çè ÷ø = K çè ø 10 2

58. (b) From Newton's Law of cooling,

Þ

-

...(ii)

60 - 50 é 60 + 50 ù = -K ê - q0 ú 10 2 ë û

...(i)

Again, it cools from 50°C to 42°C in next 10 minutes.

50 - 42 é 50 + 42 ù = -K ê - q0 ú 10 ë 2 û

...(ii)

Dividing equations (i) by (ii) we get 55 - q0 1 = 0.8 46 - q0 10 55 - q 0 = 8 46 - q 0

460 - 10q0 = 440 - 8q0 2q0 = 20 q0 = 10°C

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Physics

62. (b) From Newton’s law of cooling, t=

63.

æ q - q0 ö 1 log e ç 2 ÷ k è q1 - q0 ø

64.

1 (40 - 30) log e k (80 - 30)

And, t =

1 (32 - 30) log e (62 - 30) k

...(1)

Þ

...(2)

Þ

Dividing equation (2) by (1), 1 log e t k = 5 1 log e k

(a) According to newton's law of cooling dq = - k(q - q0 ) dt

From question and above equation, 5=

(c) According to Newton’s law of cooling, the temperature goes on decreasing with time non-linearly.

(32 - 30) (62 - 30) (40 - 30) (80 - 30)

On solving we get, time taken to cool down from 62°C to 32°C, t = 8.6 minutes.

dq = - kdt (q - q0 ) q

ò

q0

t

dq = - k ò dt (q - q 0 ) q

Þ

log(q - q0 ) = -kt + c Which represents an equation of straight line. Thus the option (a) is correct. 65. (d) From Newton’s law of cooling -

dQ µ (Dq) dt

11

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Thermodynamics

Thermodynamics TOPIC 1 First Law of Thermodynamics 1.

A gas can be taken from A to B via two different processes ACB and ADB.

(c) 5.

6.

2.

3.

4.

When path ACB is used 60 J of heat flows into the system and 30J of work is done by the system. If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB is : [9 Jan. 2019 I] (a) 40 J (b) 80 J (c) 100 J (d) 20 J 200g water is heated from 40°C to 60°C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kgK): [Online April 9, 2016] (a) 167.4 kJ (b) 8.4 kJ (c) 4.2 kJ (d) 16.7 kJ A gas is compressed from a volume of 2m3 to a volume of 1m3 at a constant pressure of 100 N/m2. Then it is heated at constant volume by supplying 150 J of energy. As a result, the internal energy of the gas: [Online April 19, 2014] (a) increases by 250 J (b) decreases by 250 J (c) increases by 50 J (d) decreases by 50 J An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be [2008] (a)

T1T2 ( PV 1 1 + P2V2 ) PV 1 1T2 + P2V2T1

(b)

PV 1 1T1 + P2V2T2 PV 1 1 + P2V2

PV 1 1T2 + P2V2T1 PV 1 1 + P2V2

(d)

T1T2 ( PV 1 1 + P2V2 ) PV T 1 1 1 + P2V2T2

When a system is taken from state i to state f along the path iaf, it is found that Q =50 cal and W = 20 cal. Along the path ibf Q = 36 cal. W along the path ibf is [2007] a f

i b (a) 14 cal (b) 6 cal (c) 16 cal (d) 66 cal A system goes from A to B via two processes I and II as shown in figure. If DU1 and DU2 are the changes in internal energies in the processes I and II respectively, then [2005] p II A

B

I v

(a) relation between DU1 and DU 2 can not be determined

7.

(b)

DU1 = DU 2

(c)

DU 2 < DU1

(d) DU 2 > DU1 Which of the following is incorrect regarding the first law of thermodynamics? [2005] (a) It is a restatement of the principle of conservation of energy (b) It is not applicable to any cyclic process (c) It does not introduces the concept of the entropy (d) It introduces the concept of the internal energy

TOPIC 2 8.

Specific Heat Capacity and Thermodynamical Processes

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are lebelled as A ® B, A ® C and A ® D. The change

P-170

Physics

in internal energies during these process are taken as EAB, EAC and EAD and the workdone as WAB, WAC and WAD. The correct relation between these parameters are : [5 Sep. 2020 (I)] D T1>T2 C P B T1 A T2 V

9.

(a) EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0 (b) EAB = EAC = EAD, WAB > 0, WAC = 0, WAD > 0 (c) EAB < EAC < EAD, WAB > 0, WAC > WAD (d) EAB > EAC > EAD, WAB < WAC < WAD In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is : [5 Sep. 2020 (II)]

1 32 10. Match the thermodynamic processes taking place in a system with the correct conditions. In the table : DQ is the heat supplied, DW is the work done and DU is change in internal energy of the system. [4 Sep. 2020 (II)] Process Condition (I) Adiabatic (A) DW = 0 (II) Isothermal (B) DQ = 0

(a) 32

(III) Isochoric

(b) 326

(c) 128

(d)

(a)

(b)

(c)

(d)

14. Starting at temperature 300 K, one mole of an ideal diatomic gas (g = 1.4) is first compressed adiabatically from volume V1 . It is then allowed to expand isobarically to 16 volume 2V2. If all the processes are the quasi-static then the final temperature of the gas (in °K) is (to the nearest integer) ______. [9 Jan. 2020 II] 15. A thermodynamic cycle xyzx is shown on a V-T diagram.

V1 to V2 =

(C) DU ¹ 0, DW ¹ 0,

DQ ¹ 0 (IV) Isobaric (D) DU = 0 (a) (I)-(A), (II)-(B), (III)-(D), (IV)-(D) (b) (I)-(B), (II)-(A), (III)-(D), (IV)-(C) (c) (I)-(A), (II)-(A), (III)-(B), (IV)-(C) (d) (I)-(B), (II)-(D), (III)-(A), (IV)-(C) 11. A balloon filled with helium (32°C and 1.7 atm.) bursts. Immediately afterwards the expansion of helium can be considered as : [3 Sep. 2020 (I)] (a) irreversible isothermal (b) irreversible adiabatic (c) reversible adiabatic (d) reversible isotherm7al 12. An engine takes in 5 mole of air at 20°C and 1 atm, and compresses it adiabaticaly to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is ________. [NA 2 Sep. 2020 (I)] 13. Which of the following is an equivalent cyclic process corresponding to the thermodynamic cyclic given in the figure? where, 1 ® 2 is adiabatic. (Graphs are schematic and are not to scale) [9 Jan. 2020 I]

The P-V diagram that best describes this cycle is: (Diagrams are schematic and not to scale) [8 Jan. 2020 I]

(a)

(b)

(c)

(d)

16. A litre of dry air at STP expands adiabatically to a volume of 3 litres. If g = 1.40, the work done by air is: (31.4 = 4.6555) [Take air to be an ideal gas] [7 Jan. 2020 I] (a) 60.7 J (b) 90.5 J (c) 100.8 J (d) 48 J

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Thermodynamics

17. Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from t1 to t2. If

Cp Cv

= g for this

t2 gas then a good estimate for t is given by: 1 [7 Jan. 2020 I]

(a) 2

(b)

1 2

(c)

æ 1ö çè ÷ø 2

g

(d)

æ 1ö çè ÷ø 2

g +1 2

18. A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is – 180 J, The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work down by the gas along the path abc is: [12 Apr. 2019 I]

(a) 120 J (b) 130 J (c) 100 J (d) 140 J 19. A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20oC is : [Given that R = 8.31 J mol – 1 K – 1] [10 Apr. 2019 I] (a) 350 J (b) 374 J (c) 748 J (d) 700 J

(a) DQA < DQB, DUA < DUB (b) DQA > DQB, DUA > DUB (c) DQA > DQB, DUA = DUB (d) DQA = DQB; DUA = DUB 23. A thermally insulted vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closed to: (Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 10 5 J kg–1) [8 April 2019 I] (a) 150 g (b) 20 g (c) 130 g (d) 35 g 24. The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by : [8 Apr. 2019 II]

(a) a d b c (b) d a c b (c) a d c b (d) d a b c 25. For the given cyclic process CAB as shown for gas, the work done is: [12 Jan. 2019 I] 6.0 5

20. n moles of an ideal gas with constant volume heat capacity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is: [10 Apr. 2019 I] nR CV + nR

nR (b) C - nR V

4nR (c) CV - nR

4nR (d) C + nR V

(a)

21. One mole of an ideal gas passes through a process where é 1 æ V ö2 ù P = P ê1 - ç 0 ÷ ú . 0 pressure and volume obey the relation êë 2 è V ø úû

Here Po and Vo are constants. Calculate the charge in the temperature of the gas if its volume changes from Vo to 2Vo. [10 Apr. 2019 II] 1 Po Vo 5 Po Vo 3 Po Vo 1 Po Vo (b) (c) (d) 2 R 4 R 4 R 4 R 22. Following figure shows two processes A and B for a gas. If DQA and DQB are the amount of heat absorbed by the system in two cases, and DUA and DUB are changes in internal energies, respectively, then: [9 April 2019 I]

(a)

A

C

4 p(Pa) 3 2 B

1 1

2

3

4

5 V(m3)

(a) 30 J (b) 10 J (c) 1 J (d) 5 J 26. A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is: [11 Jan. 2019 I] 3 2 2 5 (b) (c) (d) 5 5 3 3 27. Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by gas is close to: (Gas constant R = 8.31 J/mol-K) [10 Jan. 2019 II] (a) 581 J (b) 291 J (c) 146 J (d) 73 J 28. One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by [Online April 16, 2018]

(a)

P-172

Physics V

P 3P0

b C

P0 A V0

2V0

P

P d

(a)

9P0 V0 9P0 V0 9P0 V0 3P0 V0 (b) (c) (d) 2nR nR 4nR 2nR 31. The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is : [Online April 9, 2016]

(a)

2 3 3 2 (a) (b) (c) (d) 5 2 5 3 32. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q

Cp ö æ is : ç g = [2015] ÷ Cv ø è 3g + 5 3g - 5 g +1 g -1 (a) (b) (c) (d) 6 6 2 2 33. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u = U 1æUö µ T 4 and pressure p = ç ÷ . If the shell now V 3è V ø undergoes an adiabatic expansion the relation between T and R is : [2015]

Tµ

1

(c) T µ e–R (d) T µ e–3R R3 34. An ideal gas goes through a reversible cycle a®b®c®d has the V - T diagram shown below. Process d®a and b®c are adiabatic.

a

c a

(b)

b

b c

d

V

V P

P d

c

a

b

B V0 2V0 V

(b)

T

(c)

P0

a

The corresponding P - V diagram for the process is (all figures are schematic and not drawn to scale) : [Online April 10, 2015]

A

2P0

1 R

d

V

(a) 25 P0 V0 (b) 25 P0 V0 (c) 25 P0 V0 (d) 5 P0V0 8 R 4 R 16 R 8 R 29. One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be: [Online April 15, 2018] (a) 300R ln 6 (b) 300R (c) 300R ln 7 (d) 300R ln 2 30. 'n' moles of an ideal gas undergoes a process A ® B as shown in the figure. The maximum temperature of the gas during the process will be : [2016] P

(a) T µ

c

B

(d) V

a

b d

c V

35. One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement: [2014]

B

800 K

P

A

400 K

600 k C

V (a) The change in internal energy in whole cyclic process is 250 R. (b) The change in internal energy in the process CA is 700 R. (c) The change in internal energy in the process AB is 350 R. (d) The change in internal energy in the process BC is – 500 R. 36. An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section 8.0 × 10–3 m2. Initially the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by 0.1 m. The force constant of the spring is 8000 N/m and the atmospheric pressure is 1.0 × 105 N/m2. The cylinder and

P-173

Thermodynamics

the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be: (Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible). [Online April 11, 2014]

41. Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure. The efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas) [2012] B

2P0

(a) 15.4 % (b) 9.1 %

P0

Po To R (a) Po - a

Po To R (b) Po + a

(c) PoToRIn 2 (d) PoToR 39. A certain amount of gas is taken through a cyclic process (A B C D A) that has two isobars, one isochore and one isothermal. The cycle can be represented on a P-V indicator diagram as : [Online April 22, 2013] P

(a)

A

B

C

B

P

D

B

A

A

)

(

B

C

(d) A

(b)

)

D

V V

40. An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes 32 times of its initial value. If the final pressure of gas is 128 atmospheres, the value of ‘g’of the gas is : [Online April 22, 2013] (a) 1.5 (b) 1.4 (c) 1.3 (d) 1.6

)

(

a 2V 2 2 m -1 2

(

)

a 2 aV 2 2 m -1 (d) m -1 2 2 44. n moles of an ideal gas undergo a process A ® B as shown in the figure. Maximum temperature of the gas during the process is [Online May 12, 2012]

(c)

A

2P0

D

C

D

(

aV 2 m -1 2

B

P0

V

P

(c)

(a)

C

V

P

(d) 12.5 % 2V0 V0 42. An ideal monatomic gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure Pi. If the same gas is expanded adiabatically to a volume 2V, the final pressure is Pa. The P ratio a is [Online May 26, 2012] Pi (a) 2–1/3 (b) 21/3 (c) 22/3 (d) 2–2/3 43. The pressure of an ideal gas varies with volume as P = aV, where a is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes ‘m’ times its initial volume. The work done by the gas in the process is [Online May 19, 2012]

P

(b)

D

A

(c) 10.5%

(a) 300 K (b) 800 K (c) 500 K (d) 1000 K 37. During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly: (R = 8.3 JK–1 mol–1) [Online April 11, 2014] (a) 40 K (b) 33 K (c) 20 K (d) 14 K 38. The equation of state for a gas is given by PV = nRT + aV, where n is the number of moles and a is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are To and Po respectively. The work done by the gas when its temperature doubles isobarically will be: [Online April 9, 2014]

C

V0 2V 0

3P0V0 9 P0V0 9 P0V0 (b) (c) (d) nR 2 nR 2 nR 4 nR 45. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. Statement 1: In an adiabatic process, change in internal energy of a gas is equal to work done on/by the gas in the process. Statement 2: The temperature of a gas remains constant in an adiabatic process. [Online May 7, 2012] (a) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1. (b) Statement 1 is true, Statement 2 is false.

(a)

9P0V0

V

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Physics

(c) Statement 1 is false, Statement 2 is true. (d) Statement 1 is false, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1. 46. A container with insulating walls is divided into equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure P and temperature T, whereas the other part is completly evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be : [2011 RS] T P P T ,T (b) P, T (c) P, (d) , 2 2 2 2 Directions for questions 47 to 49: Questions are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram. [2009]

(a)

5

2 × 10

A

B

D

C

300K

500K

P (Pa) 1 × 10

5

T

47. Assuming the gas to be ideal the work done on the gas in taking it from A to B is (a) 300 R (b) 400 R (c) 500 R (d) 200 R 48. The work done on the gas in taking it from D to A is (a) + 414 R (b) – 690 R (c) + 690 R (d) – 414 R 49. The net work done on the gas in the cycle ABCDA is (a) 279 R (b) 1076 R (c) 1904 R (d) zero 50. The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is [2006] (R = 8.3 J mol–1 K–1) (a) diatomic (b) triatomic (c) a mixture of monoatomic and diatomic (d) monoatomic 51. Which of the following parameters does not characterize the thermodynamic state of matter? [2003] (a) Temperature (b) Pressure (c) Work (d) Volume

Carnot Engine, Refrigerators TOPIC 3 and Second Law of Thermodynamics 52. An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close is ______. [NA 6 Sep. 2020 (II)] 3 PO

B

C

A

D

2 PO PO

VO

2VO

53. If minimum possible work is done by a refrigerator in converting 100 grams of water at 0°C to ice, how much heat (in calories) is released to the surroundings at temperature 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer? [NA 3 Sep. 2020 (II)] 54. A heat engine is involved with exchange of heat of 1915 J, – 40 J, +125 J and – Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is : [2 Sep. 2020 (II)] (a) 640 J (b) 40 J (c) 980 J (d) 400 J 1 55. A Carnot engine having an efficiency of is being used 10 as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is: [8 Jan. 2020 II] (a) 99 J (b) 100 J (c) 1 J (d) 90 J 56. A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle, is _______. [NA 7 Jan. 2020 I] 57. Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T1 and T2. The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1 and T2, if both the engines perform equal amount of work ? [7 Jan. 2020 II] 2T1T2 T1 + T2 (a) T = T + T (b) T = 2 1 2 (c) T = T1T2

(d) T = 0

58. A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62oC, its efficiency is doubled. The temperatures of the source and the sink are, respectively. [12 Apr. 2019 II] o o o (a) 62 C, 124 C (b) 99 C, 37oC o o (c) 124 C, 62 C (d) 37oC, 99oC 59. Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2 and T3, as shown, with T1 > T2 > T3 > T(4) The three engines are equally efficient if: [10 Jan. 2019 I]

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Thermodynamics

( ) = (T T ) ; T = (T T ) = (T T ) ; T = (T T ) = ( T T ) ; T = (T T )

(a) T2 = ( T1 T4 )

1/2

(b) T2 (c) T2 (d) T2

2 1 4

2 1 4

3 1 4

; T3 = T12 T4

1/3

3

2 1 4

3

2 1

4

3

1

3 4

1/3

1/3

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.

1/3

In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is :

1/3

1/4

1/4

60. Two Carnot engines A and B are operated in series. The first one, A receives heat at T1 (= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and in turn, rejects to a heat reservoir at T3 (= 400 K). Calculate the temperature T2 if the work outputs of the two engines are equal: [9 Jan. 2019 II] (a) 600 K (b) 400 K (c) 300 K (d) 500 K 61. A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature.The amount of work done in each cycle to operate the refrigerator is: [Online April 15, 2018] (a) 420 J (b) 2100 J (c) 772 J (d) 2520 J 62. Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100K. If the efficiencies of the two engines A and B are represented h by hA and hB respectively, then what is the value of A hB [Online April 15, 2018] 12 12 5 7 (b) (c) (d) 7 5 12 12 63. An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take Cv =1.5 R, where R is gas constant) [Online April 8, 2017]

(a)

2P0

(a) 0.24 (b) 0.15

B

C

P P0

A

D

(c) 0.32 V0

2V0

V (d) 0.08 64. A Carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of ice is 336 × 103 J kg–1. If 5 kg of water at 0°C is converted into ice at 0°C by the freezer, then the energy consumed by the freezer is close to : [Online April 10, 2016] (a) 1.51 × 105 J (b) 1.68 × 106 J (c) 1.71 × 107 J (d) 1.67 × 105 J 65. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways : [2015] (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.

(a) ln2, 2ln2 (b) 2ln2, 8ln2 (c) ln2, 4ln2 (d) ln2, ln2 66. A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be: [Online April 12, 2014] (a) 20% and – 43°C (b) 40% and – 33°C (c) 50% and – 20°C (d) 70% and – 10°C 67.

p 2p0 p0

A D v0

B C 2v0 v

The above p-v diagramrepresents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is [2013] (a) p 0 v 0

æ 13 ö (b) ç ÷ p0 v0 è2ø

æ 11 ö (c) ç ÷ p0 v0 è2ø

(d) 4p0v0

68. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be : [2012] (a) efficiency of Carnot engine cannot be made larger than 50% (b) 1200 K (c) 750 K (d) 600 K 69. The door of a working refrigerator is left open in a well insulated room. The temperature of air in the room will [Online May 26, 2012] (a) decrease (b) increase in winters and decrease in summers (c) remain the same (d) increase 70. This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. Statement 1: An inventor claims to have constructed an engine that has an efficiency of 30% when operated between the boiling and freezing points of water. This is not possible.

P-176

Physics

Statement 2: The efficiency of a real engine is always less than the efficiency of a Carnot engine operating between the same two temperatures. [Online May 19, 2012] (a) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1. (b) Statement 1 is true, Statement 2 is false. (c) Statement 1 is false, Statement 2 is true. (d) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1. 71. A Carnot engine operating between temperatures T1 and T2 has efficiency increases to

75.

76.

1 . When T2 is lowered by 62 K its efficiency 6

1 . Then T1 and T2 are, respectively: 3

[2011]

(a) 372 K and 310 K (b) 330 K and 268 K (c) 310 K and 248 K (d) 372 K and 310 K 72. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is [2010] (a) 0.5 (b) 0.75 (c) 0.99 (d) 0.25 73. A Carnot engine, having an efficiency of h = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [2007] (a) 100 J (b) 99 J (c) 90 J (d) 1 J 74. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is [2005] T

2T0 T0 S0

2S0

S

1 2 1 1 (b) (c) (d) 3 2 4 3 Which of the following statements is correct for any thermodynamic system ? [2004] (a) The change in entropy can never be zero (b) Internal energy and entropy are state functions (c) The internal energy changes in all processes (d) The work done in an adiabatic process is always zero. “Heat cannot by itself flow from a body at lower temperature to a body at higher temperature” is a statement or consequence of [2003] (a) second law of thermodynamics (b) conservation of momentum (c) conservation of mass (d) first law of thermodynamics A Carnot engine takes 3 × 106 cal of heat from a reservoir at 627°C, and gives it to a sink at 27°C. The work done by the engine is [2003] (a) 4.2 × 106 J (b) 8.4 × 106 J (c) 16.8 × 106 J (d) zero Which statement is incorrect? [2002] (a) All reversible cycles have same efficiency (b) Reversible cycle has more efficiency than an irreversible one (c) Carnot cycle is a reversible one (d) Carnot cycle has the maximum efficiency in all cycles Even Carnot engine cannot give 100% efficiency because we cannot [2002] (a) prevent radiation (b) find ideal sources (c) reach absolute zero temperature (d) eliminate friction

(a)

77.

78.

79.

P-177

Thermodynamics

1.

2.

5.

(a)

DU remains same for both paths ACB and ADB DQACB = DWACB + DUACB Þ 60 J = 30 J + DUACB Þ UACB = 30 J \ DUADB = DUACB = 30 J DQADB = DUADB + DWADB = 10 J + 30 J = 40 J (d) Volume of water does not change, no work is done on or by the system (W = 0) According to first law of thermodynamics

Q < ΧU ∗ W

3.

For Isochoric process Q < ΧU DU = mcdT = 2 × 4184 × 20 = 16.7 kJ. (a) As we know, (Ist law of thermodynamics) DQ = D u + D w

6.

7. 8.

Þ DQ = D u + P D v or 150 = Du + 100 (1 - 2 )

4.

= Du - 100 \ Du = 150 + 100 = 250J Thus the internal energy of the gas increases by 250 J (a) Here Q = 0 and W = 0. Therefore from first law of thermodynamics DU = Q + W = 0 Internal energy of first vessle + Internal energy of second vessel = Internal energy of combined vessel n1Cv T1 + n2 Cv T2 = (n1 + n2 )Cv T \T =

n1T1 + n2 T2 n1 + n2

For first vessel n1 =

PV 1 1 and for second vessle RT1

PV n2 = 2 2 RT2 PV 1 1 ´ T + P2V2 ´ T 1 2 RT1 RT2 \T= PV 1 1 + P2V2 RT1 RT2

=

T1T2 ( PV 1 1 + P2V2 ) PV T 1 1 2 + P2V2T1

(b) For path iaf, Q1 = 50 cal, W1 = 20 cal By first law of thermodynamics, a f DU = Q1 – W1 = 50 – 20 = 30 cal. For path ibf Q2 = 36 cal i b W2 = ? DUibf = Q2 – W2 Since, the change in internal energy does not depend on the path, therefore DUiaf = DUibf DUiaf = DUibf Þ 30 = Q2 – W2 Þ W2 = 36 – 30 = 6 cal. (b) Change in internal energy is independent of path taken by the process. It only depends on initial and final states i.e., DU1 = DU2 (b, c) First law is applicable to a cyclic process. Concept of entropy is introduced by the second law of thermodynamics. (b) Temperature change DT is same for all three processes A ® B; A ® C and A ® D DU = nCv DT = same E AB = E AC = E AD Work done, W = P ´ DV AB ® volume is increasing Þ WAB > 0 AD ® volume is decreasing Þ WAD < 0

9.

AC ® volume is constant Þ WAC = 0 (c) In adiabatic process PV g = constant g

æ mö \ P ç ÷ = constant è rø As mass is constant

æ mö çèQ V = r ø÷

\ P µ rg If Pi and Pf be the initial and final pressure of the gas and ri and r f be the initial and final density of the gas. Then g

ærf ö = ç ÷ = (32)7 / 5 Pi è ri ø nP Þ i = (25 )7 /5 = 27 Pi Pf

Þ n = 27 = 128. 10. (d) (I) Adiabatic process : No exchange of heat takes place with surroundings.

Þ DQ = 0 (II) Isothermal process : Temperature remains constant

P-178

Physics

f nR DT Þ DU = 0 2 No change in internal energy [DU = 0]. (III) Isochoric process volume remains constant \ DT = 0 Þ DU =

DV = 0 Þ W = ò P × dV = 0 Hence work done is zero. (IV) In isobaric process pressure remains constant. W = P × DV ¹ 0 DU =

f f nR DT = [ P DV ] ¹ 0 2 2

\ DQ = nC p DT ¹ 0 11. (b) Bursting of helium balloon is irreversible and in this process DQ = 0 , so adiabatic. 12. (46) For adiabatic process, TV g -1 = constant or, T1V1g -1 = T2V2g -1 T1 = 20°C + 273 = 293 K , V2 =

æV ö T1 (V1 ) g -1 = T2 ç 1 ÷ è 10 ø æ 1ö Þ 293 = T2 ç ÷ è 10 ø

7 V1 and g = 5 10

g -1

2/5

14. (1818) For an adiabatic process, TVg–1 = constant \ T1V1g –1 = T2V2g –1 1.4 -1

æ ö çV ÷ Þ T2 = (300) ´ ç 1 ÷ çç V1 ÷÷ è 16 ø 0.4 Þ T2=300×(16) Ideal gas equation, PV = nRT nRT \ V= P Þ V = kT (since pressure is constant for isobaric process) So, during isobaric process V2 = kT2 ...(i) 2V2 = kTf ...(ii) Dividing (i) by (ii) 1 T2 = 2 Tf

Tf = 2T2 = 300 × 2 × (16)0.4 =1818 K 15. (a) From the corresponding V-T graph given in question, Process xy ® Isobaric expansion, Process yz ® Isochoric (Pressure decreases) Process zx ® Isothermal compression Therefore, corresponding PV graph is as shown in figure

Þ T2 = 293(10) 2/ 5 ; 736 K

DT = 736 - 293 = 443 K During the process, change in internal energy DU = NCV DT = 5 ´

5 ´ 8.3 ´ 443 ; 46 ´ 103 J = X kJ 2

\ X = 46 . 13. (c) For process 3 ® 1 volume is constant \ Graph given in option (d) is wrong. And process 1 ® 2 is adiabatic \ graph in option (1) is wrong Q v = constant P ­, T ­ For Process 2 ® 3 Pressure constant i.e., P = constant \ V¯T¯ Hence graph (c) is the correct V – T graph of given P – V graph

V

2 3

16. (b) Given, V1 = 1 litre, P1 = 1 atm V2 = 3 litre, g = 1.40, g g Using, PVr = constant Þ PV 1 1 = P2V2

1.4

æ1ö Þ P2 = P1 ´ ç ÷ è 3ø

T

1 atm 4.6555

PV – P V \ Work done, W = 1 1 2 2 g –1 1 æ ö ´ 3 ÷1.01325 ´ 105 ´ 10 –3 ç1´1 – 4.6555 ø è = = 90.1 J 0.4 Closest value of W = 90.5 J 17. (Bonus) We know that Relaxation time,

Tµ

1

=

V

...(i) T Equation of adiabatic process is TVg–1 = constant

Þ

1 T µ g-1 V

P-179

Thermodynamics

Þ T µ V 1+ Þ T µV Þ

g –1 2

using (i)

1+g 2

Tf

1+g 2

æ 2V ö =ç ÷ Ti è V ø

= (2)

1+g 2

18. (b) DUac = – (DUca) = – (– 180) = 180 J Q = 250 + 60 = 310 J Now Q = DU + W or 310 = 180 + W or W = 130 J 19. (c) As the process is isochoric so, 67.2 3R ´ ´ 20 = 90R = 90 ´ 8.31 ; 748 j. 22.4 2 20. (a) At constant volume Work done (W) = nRDT Heat given Q = CvDT + nRDT Q = nc v DT =

So, \

W nRDT nR = = Q Cv DT + nRDT C V + nR

21. (b) We have given, é 1 æ V ö2 ù P = P0 ê1 - ç 0 ÷ ú êë 2 è V ø úû When V1 = V0

é 1 ù P0 Þ P1 = P0 ê1 - ú = ë 2û 2 When V2 = 2V0 é 1 æ 1 ö ù æ 7 P0 ö Þ P2 = P0 ê1 - ç ÷ ú = ç ÷ ë 2 è 4 øû è 8 ø PV P V é PV ù DT = T2 - T1 = 1 1 - 2 2 êQ T = nR nR ë nR úû æ 1 ö æ 1 ö æ P0V0 7 P0V0 ö DT = ç 1 1 - P2V2 ) = ç ÷ ( PV ÷ç 4 ÷ø è nR ø è nR ø è 2

5P V 5P V = 0 0 = 0 0 (Q n = 1) 4 nR 4R 22. (c) Internal energy depends only on initial and final state So, DUA = DUB Also DQ = DU + W As WA > WB Þ DQA > DQB 23. (b) Suppose amount of water evaporated be M gram. Then (150 – M) gram water converted into ice. so, heat consumed in evoporation = Heat released in fusion M × Lv = (150 – M) × Ls M × 2.1 × 106 = (150 – M) × 3.36 × 105 Þ M – 20 g

24. (d) a ® Isobasic, b ® Isothermal, c ® Adiabatic, d ® Isochoric 25. (b) Total work done by the gas during the cycle is equal to area of triangle ABC. 1 \ DW = ´ 4 ´ 5 = 10 J 2 26. (b) Equation of adiabatic change is TVg-1 = constant 7 7 Put g = , we get: g - 1 = - 1 5 5 2 \x = 5 27. (b) Work done, 1 W = PDV = nRDT = ´ 8.31´ 70 ; 291J 2 28. (a) Equation of the BC 2P P = P0 - 0 (V - 2V0 ) V0 using PV = nRT 2P V 2 + 4P0 V P0 V - 0 V0 Temperature, T = 1´ R (Q n = 1 mole given) P é 2V 2 ù T = 0 ê5V ú F ëê V0 ûú dT 4V 5 =0Þ5= 0 Þ V = V0 dV V0 4 P é 5V 2 25 2 ù 25 P0 V0 T = 0 ê5 ´ 0 ´ V0 ú = Rë 4 V0 16 û 8 R æ pf ö 29. (d) Work done on gas = nRT ln çç ÷÷ = R(300) ln(2) è p1 ø æ Pf ö = 2 given ÷÷ = 300 Rln2 ççQ è pi ø 30. (c) The equation for the line is P

3Po c 2Po Po

q Po

q Vo Vo

2Vo

V

-P0 - P0 P = V V + 3P [slope = V , c = 3P0] 0 0 PV0 + P0V = 3P0V0 But pV = nRT nRT \P= V nRT From (i) & (ii) V0 + P0V = 3P0V0 V

...(i) ...(ii)

P-180

Physics

nRT V0 + P0V = 3P0V0 V 2 \ nRT V0 + P0V = 3P0V0V ...(iii)

From (i) & (ii)

dT For temperature to be maximum =0 dV Differentiating e.q. (iii) by ‘V’ we get dT nRV0 + P0(2V) = 3P0V0 dV dT \ nRV0 = 3P0V0 – 2 P0V dV dT 3P0 V0 - 2P0 V =0 = dV nRV0

3V0 V= 2

\

3P P= 0 2

[From (i)]

9P0 V0 [From (iii)] 4nR (a) Efficiency of heat engine is given by C w R R 2 h = = 1- V = = = 5R Q CP Cp 5 2 (Q Cp – Cv = R) 5 For monoatomic gas C P = R . 2 1 (a) t = æ N ö 3RT 2pd2 ç ÷ èVø M

\ Tmax =

8.

9.

t µ

V

T As, TVg–1 = K So, t µ Vg + 1/2 Therefore, q =

g+ 1 2

1æU ö 10. (a) As, P = ç ÷ 3èV ø U = KT 4 But V 1 4 So, P = KT 3 uRT 1 = KT 4 [As PV = u RT] or V 3 4 3 3 p R T = constant 3 1 Therefore, Tµ R 11. (b) In VT graph ab-process : Isobaric, temperature increases. bc process : Adiabatic, pressure decreases. cd process : Isobaric, volume decreases. da process : Adiabatic, pressure increases. The above processes correctly represented in P-V diagram (b).

12. (d) In cyclic process, change in total internal energy is zero. DUcyclic = 0 5R DT 2 Where, Cv = molar specific heat at constant volume. For BC, DT = –200 K \ DUBC = –500R 13. (c) 14. (c) Given : work done, W = 830 J No. of moles of gas, m = 2 For diatomic gas g = 1.4 Work done during an adiabatic change

DUBC = nCv DT = 1 ´

W=

mR (T1 - T2 ) g -1

Þ 830 = Þ DT =

2 ´ 8.3( DT ) 2 ´ 8.3(DT ) = 1.4 - 1 0.4

830 ´ 0.4 = 20 K 2 ´ 8.3

15. (a) 16. (c) P-V indicator diagram for isobaric P

slope

dP =0 dV

V P-V indicator diagram for isochoric process P slope dP =¥ dV V P-V indicator diagram for isothermal process P slope dP -P = = dV V V 17. (b) Volume of the gas m v= and d Using PV g = constant g

P' V æ d'ö = =ç ÷ P V' è d ø or 128 = (32)g 7 \ g = = 1.4 5 18. (a) The efficiency output work h= heat given to the system

P-181

Thermodynamics

3 3 3 = n RDT = V0 DP = P0V0 2 2 2 n n Wi = ( P0V0 ) + (2 P0V0 ) + 2 P0V0 2 2 Heat given in going B to C = nCpDT 5 æ5 ö = n ç R ÷ DT = (2 P0 )DV 2 è2 ø = 5P0V0 and W0 = area under PV diagram P0V0 2 W PV h= = 0 0 = 13 Q P0V0 13 2 Efficiency in % 2 200 ´ 100 = ; 15.4% 13 13 42. (d) For isothermal process : PV = Pi .2V P = 2Pi ...(i) For adiabatic process PVg = Pa (2V)g h=

(Q for monatomic gas g= 5 3 ) 5

or,

5

[From (i)]

2Pi V 3 = Pa (2V) 3

Pa 2 = 5 Pi 2 3-2 Pa =2 3 Þ Pi 43. (d) Given P = aV

ò

PdV

V

=

ò

aVdV =

V

Vacuum

It is the free expansion \ So, T remains constant Þ PV 1 1 = P2V2 Þ P

V = P2 (V ) 2

æ Pö P2 = ç ÷ è 2ø 47. (b) The process A ® B is isobaric. \ work done WAB = nR(T2 – T1) = 2R (500 - 300) = 400 R 48. (a) The process D to A is isothermal as temperature is constant. P Work done, WDA = 2.303nRT log10 D PA

= 2.303 ´ 2 R ´ 300 log10

1 ´ 105

– 414R. 2 ´ 105 Therefore, work done on the gas is +414 R. 49. (a) The net work in the cycle ABCDA is W = W AB + WBC + WCD + WDA

PB PC

= 2.303 ´ 2R ´ 500log

mV

mV

P, T

= 400R + 2.303nRT log

Þ

Work done, w =

46. (d)

aV 2 (m 2 - 1) . 2

44. (b) Work done during the process A ® B = Area of trapezium (= area bounded by indicator diagram with V-axis) 1 3 2 P0 + P0 ) ( 2V0 - V0 ) = P0V0 ( 2 2 Ideal gas eqn : PV = nRT

=

PV 3P0V0 = nR 2nR 45. (b) In an adiabatic process, dH = 0 And according to first law of thermodynamics dH = dU + W \ W = – dU

+ (-400R) - 414R

2 ´ 105 1 ´ 105

- 414 R

= 693.2 R – 414 R = 279.2 R 50. (a) Work done in adiabatic compression is given by nRDT W= 1- g 1000 ´ 8.3 ´ 7 Þ -146000 = 1- g 58.1 58.1 or 1 - g = Þ g = 1+ = 1.4 146 146 Hence the gas is diatomic. 51. (c) Work is not a state function. The remaining three parameters are state function. P 52. (19) 3P0

B

C

Þ T=

P0

A V0

D

2V0

V

P-182

Physics

From the figure, Also,

Work, W = 2 P0V0 Heat given, Qin = WAB + WBC = n × CV DTAB + nCP DTBC 3R n5 R =n (TB - TA ) + (TC - TB ) 2 2 3R 5R ö æ and CP = çèQ Cv = ÷ 2 2ø 3 5 = ( PBVB - PAV A ) + ( PCVC - PBVB ) 2 2 3 5 = ´ [3P0V0 - P0V0 ] + [6 P0V0 - 3P0V0 ] 2 2 15 21 = 3 P0V0 + P0V0 = P0V0 2 2 2 P0V0 W 4 = = 21 21 Qin P0V0 2 400 h% = » 19. 21 53. (8791) Given, Heat absorbed, Q2 = mL = 80 × 100 = 8000 Cal Temperature of ice, T2 = 273 K Temperature of surrounding, T1 = 273 + 27 = 300 K

1 w = 10 Q1 Þ Q1 = w × 10 = 100 J So, Q1 – Q2 = w Þ Q2 = Q1– w Þ 100 – 10 = Q2 = 90 J Þ

56. (600.00) Given; T1 = 900 K, T2 = 300K, W = 1200 J Using, 1 – Þ 1–

Efficiency, h =

Efficiency = Þ

w Q1 - Q2 T1 - T2 300 - 273 = = = Q2 Q2 T2 273

Q1 - 8000 27 = Þ Q1 = 8791 Cal 8000 273

54. (c) Efficiency, h = =

Work done W = Heat absorbed SQ

Q1 + Q2 + Q3 + Q4 = 0.5 Q1 + Q3

Here, Q1 = 1915 J, Q2 = – 40 J and Q3 = 125 J \

1915 - 40 + 125 + Q4 = 0.5 1915 + 125

Þ

Þ Q4 = -Q = -980 J

Þ Q = 980 J 55. (d) For carnot refrigerator Efficiency =

Q1 – Q2 Q1

Where, Q1 = heat lost from sorrounding Q2 = heat absorbed from reservoir at low temperature.

T2 W = T1 Q1

300 1200 = 900 Q1

2 1200 = Þ Q1 = 1800 3 Q1

Therefore heat energy delivered by the engine to the low temperature reservoir, Q2 = Q1 – W = 1800 – 1200 = 600.00 J 57. (b) Let QH = Heat taken by first engine QL = Heat rejected by first engine Q2 = Heat rejected by second engine Work done by 1st engine = work done by 2nd engine W = QH – QL = QL – Q2 Þ 2QL = QH + Q2 qH q2 + qL qL Let T be the temperature of cold reservoir of first engine. Then in carnot engine. QH T1 Q T = and L = QL T Q2 T2 2=

Þ 2= Þ

T1 T2 + T T

2T = T1 + T2

58. (b) Using, n = 1 -

Þ 1915 - 40 + 125 + Q4 = 1020 Þ Q4 = 1020 - 2000

Q1 – Q2 w = Q1 Q1

n=

using (i) Þ T=

T1 + T2 2

T2 T1

T2 1 = 1- T 6 1

T2 - 62 T and 3 = 1 - T 1 On solving, we get T1 = 99°C and T2 = 37°C 59. (b) According to question, h1 = h2 = h3 T2 T T = 1– 3 =1– 4 T1 T2 T3 [Q Three engines are equally efficient] \1–

P-183

Thermodynamics

Þ

T2 T3 T4 = = T1 T2 T3

Thermal efficiency of engine (h) = 64. (d) DH = mL = 5 × 336 × 103 = Qsink

Þ T2 = T1T3

...(i)

T3 = T2 T4 From (i) and (ii)

...(ii)

T2 = (T12 T4 )

1

2 T3 = (T1 T4 )

1

60. (d) hA =

Qsink T < sink Qsource Tsource Tsource ´ Qsink Tsink Energy consumed by freezer [ Qsource
l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by: (R is universal gas constant and g is the acceleration due to gravity) [12 Jan. 2019 II] (a)

RT é l1 - 3l2 ù ê ú ng ë l1 I 2 û

RT é 2l1 + l2 ù (b) g ê l I ú ë 1 2 û

nRT é l1 - l2 ù nRT é 1 1 ù ê ú (d) ê + ú g ë l1 l2 û g ë l2 l1 û The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be : [2017] (a) 2.5 × 1025 (b) –2.5 × 1025 (c) –1.61 × 1023 (d) 1.38 × 1023 For the P-V diagram given for an ideal gas,

(c)

5.

6.

1 P P=

Constant V

2 V

out of the following which one correctly represents the T-P diagram ? [Online April 9, 2017] 2 2 T (a) T (b)

1

1 P

P

T

T 2 (c) 7.

1

1

2

(d)

P Chamber I ideal gas 1

P Chamber II real gas 2

3

4

P-187

Kinetic Theory

8.

There are two identical chambers, completely thermally insulated from surroundings. Both chambers have a partition wall dividing the chambers in two compartments. Compartment 1 is filled with an ideal gas and Compartment 3 is filled with a real gas. Compartments 2 and 4 are vacuum. A small hole (orifice) is made in the partition walls and the gases are allowed to expand in vacuum. Statement-1: No change in the temperature of the gas takes place when ideal gas expands in vacuum. However, the temperature of real gas goes down (cooling) when it expands in vacuum. Statement-2: The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential. [Online April 9, 2013] (a) Statement-1 is false and Statement-2 is true. (b) Statement-1 and Statement-2 both are true. Statement-2 is the correct explanation of Statement-1. (c) Statement-1 is true and Statement-2 is false. (d) Statement-1 and Statement-2 both are true. Statement-2 is not correct explanation of Statement-1. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will [2002] (a) increase (b) decrease (c) remain same (d) decrease for some, while increase for others

(a) 104 N/m2 (b) 108 N/m2 3 2 (c) 10 N/m (d) 1016 N/m2 13. The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to : [8 April 2019 II] –23 [Boltzmann Constant kB = 1.38 × 10 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth : 6.4 × 106 m Gravitational acceleration on Earth = 10 ms–2] (a) 800 K (b) 3 × 105 K 4 (c) 10 K (d) 650 K 14. A mixture of 2 moles of helium gas (atomic mass = 4u), and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds

é Vrms ( helium ) ù ê ú is close to : [9 Jan. 2019 I] ë Vrms ( argon ) û (a) 3.16 (b) 0.32 (c) 0.45 (d) 2.24 15. N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ? [Online April 9, 2017] 1 nRT (b) 0 2 3 5 nRT nRT (c) (d) 2 2 16. In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is: [Online April 10, 2015] 1 (a) (b) 2 2 1 (c) 1 (d) 4 17. A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0°C. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. (kB is Boltzmann constant). [Online April 19, 2014] 273k B (a) 0 (b) 2Mg

(a)

TOPIC 2

Speed of Gas, Pressure and Kinetic Energy

Number of molecules in a volume of 4 cm 3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2, density of mercury = 13.6 g/cm3) [Sep. 05, 2020 (I)] (a) 4.0 × 1018 (b) 4.0 × 1016 16 (c) 5.8 × 10 (d) 5.8 × 1018 10. Nitrogen gas is at 300°C temperature. The temperature (in K) at which the rms speed of a H2 molecule would be equal to the rms speed of a nitrogen molecule, is _____________. (Molar mass of N2 gas 28 g); [NA Sep. 05, 2020 (II)] 11. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be: [9 April 2019 I] (a) 100 m/s (b) 80 5 m/s (c) 100 5 m/s (d) 80 m/s 12. If 1022 gas molecules each of mass 10–26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104 m/s, the pressure exerted by the gas molecules will be of the order of : [8 April 2019 I] 9.

819k B 546k B (d) 2Mg 3Mg 18. At room temperature a diatomic gas is found to have an r.m.s. speed of 1930 ms–1. The gas is: [Online April 12, 2014] (a) H2 (b) Cl2 (c) O2 (d) F2 (c)

P-188

Physics

19. In the isothermal expansion of 10g of gas from volume V to 2V the work done by the gas is 575J. What is the root mean square speed of the molecules of the gas at that temperature? [Online April 25, 2013] (a) 398m/s (b) 520m/s (c) 499m/s (d) 532m/s 20. A perfect gas at 27°C is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to [Online May 7, 2012] (a) 327°C (b) 200°C (c) 54°C (d) 300°C 21. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g. It is moving with speed v and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by: [2011] (a)

22.

23.

24.

25.

( g - 1) Mv 2 K 2 gR

The total internal energy, U of a mole of this gas, and the æ Cp ö value of g ç = ÷ are given, respectively, by: è Cv ø [Sep. 06, 2020 (I)] (a) U =

( g - 1)

( g - 1) Mv 2 K 2R

(a)

n1T1 + n2T2 + n3T3 n1 + n2 + n3

(b)

(c)

n12T12 + n22T22 + n32T32 n1T1 + n2T2 + n3T3

(d)

(b) U = 5RT and g =

7 5

5 7 6 RT and g = (d) U = 5RT and g = 2 5 5 27. In a dilute gas at pressure P and temperature T, the mean time between successive collisions of a molecule varies with T is : [Sep. 06, 2020 (II)]

(c) U =

(a) T

gM 2v (b) K 2R

(c)

5 6 RT and g = 2 5

(b)

1 T

1 (d) T T Match the Cp/Cv ratio for ideal gases with different type of molecules : [Sep. 04, 2020 (I)] Column-I Column-II Molecule Type Cp/Cv (A) Monatomic (I) 7/5 (B) Diatomic rigid molecules (II) 9/7 (C) Diatomic non-rigid molecules(III) 4/3 (D) Triatomic rigid molecules (IV) 5/3 (a) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to _________. [NA Sep. 04, 2020 (I)]

(c)

2

(d) 2( g + 1) R Mv K Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. The masses of molecules are m1, m2 and m3 and the number of molecules are n1, n2 and n3 respectively. Assuming no loss of energy, the final temperature of the mixture is : [2011]

28.

n1T12 + n2T22 + n3T32 n1T1 + n2T2 + n3T3

(T1 + T2 + T3 ) 3

One kg of a diatomic gas is at a pressure of 8 × 104N/m2. The density of the gas is 4kg/m3. What is the energy of the gas due to its thermal motion?[2009] (a) 5 × 104 J (b) 6 × 104 J 4 (c) 7 × 10 J (d) 3 × 104 J The speed of sound in oxygen (O2) at a certain temperature is 460 ms–1. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) [2008] –1 –1 (a) 1421 ms (b) 500 ms (c) 650 ms–1 (d) 330 ms–1 At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C? [2002] (a) 80 K (b) –73 K (c) 3 K (d) 20 K

29.

30.

Consider a gas of triatomic molecules. The molecules are assumed to be triangular and made of massless rigid rods whose vertices are occupied by atoms. The internal energy of a mole of the gas at temperature T is : [Sep. 03, 2020 (I)]

Degree of Freedom, Specific TOPIC 3 Heat Capacity, and Mean Free Path 26.

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T.

(a)

5 RT 2

(b)

3 RT 2

9 RT (d) 3RT 2 31. To raise the temperature of a certain mass of gas by 50°C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant

(c)

P-189

Kinetic Theory

32.

33.

34.

35.

36.

volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)? [Sep. 03, 2020 (II)] (a) 5 (b) 6 (c) 3 (d) 7 A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is : [Sep. 02, 2020 (I)] (a) 15 (b) 13 (c) 20 (d) 11 An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true? [Sep. 02, 2020 (II)] (1) The mean free path of the molecules decreases (2) The mean collision time between the molecules decreases (3) The mean free path remains unchanged (4) The mean collision time remains unchanged (a) (2) and (3) (b) (1) and (2) (c) (3) and (4) (d) (1) and (4) Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional m vibrational mode, and have a mass . The ratio of the 4 B specific heats (CA V and CV ) of gas A and B, respectively is: [9 Jan 2020 I] (a) 7 : 9 (b) 5 : 9 (c) 3 : 5 (d) 5 : 7 Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to: [9 Jan 2020 II] (a) 3.67 (b) 1.83 (c) 2.3 (d) 4.67 The plot that depicts the behavior of the mean free time t (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale) [8 Jan. 2020 I]

37. Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CP/CV value will be: [8 Jan. 2020 II] (a) 19/13 (b) 67/45 (c) 40/27 (d) 23/15 Cp 5 38. Two moles of an ideal gas with C = are mixed with 3 3 V

Cp 4 moles of another ideal gas with C = . The value of 3 V Cp

39.

40.

41.

42.

for the mixture is: [7 Jan. 2020 I] CV (a) 1. 45 (b) 1.50 (c) 1.47 (d) 1.42 Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K) [12 April 2019 I] (a) 19.7 J/mol L (b) 15.7 J/mol K (c) 17.4 J/mol K (d) 21.6 J/mol K A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process ? [12 April 2019 II] (a) 25 J (b) 35 J (c) 30 J (d) 40 J A 25×10 – 3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300 K) . The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule? [10 April 2019 I] (a) ~1012 (b) ~1011 (c) ~1010 (d) ~1013 When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by DT. The heat required to produce the same change in temperature, at a constant pressure is : [10 April 2019 II] (a)

2 Q 3

(b)

5 Q 3

7 3 Q Q (d) 5 2 An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v , m is its mass and k B is Boltzmann constant, then its temperature will be: [9 April 2019 I]

(c)

(a)

43.

t

t

(b) 1 T

T

t

(a)

t

(c)

(d) 1 T

T

mv 2 6kB

mv 2 (c) 7k B

(b)

mv 2 3k B

mv 2 (d) 5k B

P-190

44.

45.

46.

47.

48.

49.

50.

Physics

The specific heats, C p and Cv of a gas of diatomic molecules, A, are given (in units of J mol–1 k–1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then: [9 April 2019 II] (a) A is rigid but B has a vibrational mode. (b) A has a vibrational mode but B has none. (c) A has one vibrational mode and B has two. (d) Both A and B have a vibrational mode each. An ideal gas occupies a volume of 2 m 3 at a pressure of 3 × 106 Pa. The energy of the gas: [12 Jan. 2019 I] (a) 9 × 106 J (b) 6 × 104 J (c) 108 J (d) 3 × 102 J An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 × 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions wiil be close to: [12 Jan. 2019 II] –7 –8 (a) 2 × 10 s (b) 4 × 10 s (c) 0.5 × 10–8 s (d) 3 × 10–6 s

51. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (1) the final temperature of the gas and (2) change in its internal energy. [2018] (a) (1) 189 K (2) 2.7 kJ (b) (1) 195 K (2) –2.7 kJ (c) (1) 189 K (2) –2.7 kJ (d) (1) 195 K (2) 2.7 kJ 52. Two moles of helium are mixed with n with moles of 3 C hydrogen. If P = for the mixture, then the value of n CV 2 is [Online April 16, 2018] (a) 3/2 (b) 2 (c) 1 (d) 3 53. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is : [2017] (a) a = 14 b (b) a = 28 b

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is : [11 Jan. 2019 I] (a) 15 RT (b) 12 RT (c) 4 RT (d) 20 RT In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by DT. The amount of heat absorbed by gas is (R is gas constant) : [11 Jan. 2019 II] 1 1 KRΔT (a) RΔT (b) 2 2 2K 3 ΔT RΔT (c) (d) 3 2 Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion? [10 Jan 2019 II] 3 5 (a) 10 J (b) 10 J (c) 104 J (d) 106 J A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole] [9 Jan. 2019 II] (a) 0.9 kJ (b) 6 kJ (c) 10 kJ (d) 14 kJ

54.

1 b (d) a = b 14 An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (C p) and at constant volume (Cv) is : [Online April 8, 2017] 7 (a) 6 (b) 2 7 5 (c) (d) 5 2 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) : [2016] CP - C C - CV (a) n = (b) n = C - CV C - CP C – CP CP (c) n = (d) n = C – CV CV Using equipartition of energy, the specific heat (in J kg–1 K–1) of aluminium at room temperature can be estimated to be (atomic weight of aluminium = 27) [Online April 11, 2015] (a) 410 (b) 25 (c) 1850 (d) 925 Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 10–15 atm. at room temperature (300 K). Taking R = 8.0 JK–1 mole–1, 1 atm = 105 Pa and NAvogadro = 6 × 1023 mole–1, the mean distance between molecules of gas in an evacuated vessel will be of the order of: [Online April 9, 2014] (a) 0.2 mm (b) 0.2 mm (c) 0.2 cm (d) 0.2 nm

(c) a =

55.

56.

57.

P-191

Kinetic Theory

58.

Figure shows the variation in temperature (DT) with the amount of heat supplied (Q) in an isobaric process corresponding to a monoatomic (M), diatomic (D) and a polyatomic (P) gas. The initial state of all the gases are the same and the scales for the two axes coincide. Ignoring vibrational degrees of freedom, the lines a, b and c respectively correspond to : [Online April 9, 2013]

oxygen. The ratio

a b

Q

c DT

59.

(a) P, M and D

(b) M, D and P

(c) P, D and M

(d) D, M and P

A given ideal gas with g =

Cp

= 1.5 at a temperature T. If Cv the gas is compressed adiabatically to one-fourth of its initial volume, the final temperature will be [Online May 12, 2012]

(a) 2 2T (c) 2 T

60. If CP and CV denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then [2007] (a) CP – CV = 28R (b) CP – CV = R/28 (c) CP – CV = R/14 (d) CP – CV = R 61. A gaseous mixture consists of 16 g of helium and 16 g of

(b) 4 T (d) 8 T

Cp Cv

of the mixture is

[2005]

(a) 1.62 (b) 1.59 (c) 1.54 (d) 1.4 62. One mole of ideal monatomic gas (g = 5/3) is mixed with one mole of diatomic gas (g = 7/5). What is g for the mixture? g Denotes the ratio of specific heat at constant pressure, to that at constant volume [2004] (a) 35/23 (b) 23/15 (c) 3/2 (d) 4/3 63. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP/CV for the gas is [2003] (a)

4 3

(b) 2

(c)

5 3

(d)

3 2

P-192

1.

Physics

(5) Using ideal gas equation, PV = nRT

æ 1

Þ PV 1 1 = nR ´ 250

[Q T1 = 250 K]

...(i)

5n R ´ 2000 4 Dividing eq. (i) by (ii),

[Q T2 = 2000 K]

...(ii)

P2 (2V1 =

nRT æ l1 – l 2 ö

5.

P1 P 1 4 ´ 250 = Þ 1 = 2 P2 5 ´ 2000 P2 5 P2 = 5. P1 (150) In first case, From ideal gas equation PV = nRT \

2.

DP V ...(i) P In second case, using ideal gas equation again P DV = -nR DT DV = -

nR DT P Equating (i) and (ii), we get DV = -

6.

8.

4

0

4.

r

òr

2

4

(e -ar )dr

0

µ n0 a–3/4 (d) Clearly from figure, P2A = P1A + mg

or,

7.

DT 300 K V = = = 150 K/atm nR DP 2 atm

-ar ´ 4pr 2 dr = 4p n = ò n0 e 0

9.

nRT × A nRT × A + mg Al 2 = Al1

l1

n

l2

n

T

T

1 ´ 105 ´ 30 1 ö æ 1 ´ 6.023 ´ 10 23 ç ÷ 8.314 300 290 è ø = – 2.5 × 1025 (c) From P-V graph,

1 , T = constant and Pressure is increasing from 2 V to 1 so option (3) represents correct T-P graph. (a) In ideal gases the molecules are considered as point particles and for point particles, there is no internal excitation, no vibration and no rotation. For an ideal gas the internal energy can only be translational kinetic energy and for real gas both kinetic as well as potential energy. (c) The centre of mass of gas molecules also moves with lorry with uniform speed. As there is no relative motion of gas molecule. So, kinetic energy and hence temperature remain same. 3 kT = 4 ´ 10 -14 2 P = 2 cm of Hg, V = 4 cm3

(c) Given : K.E.mean =

N=

PV PrgV 2 ´ 13.6 ´ 980 ´ 4 = ; 4 ´ 1018 8 KT KT ´ 10 -14 3

10. (41) Room mean square speed is given by

P1A P2A

P0V0 æ 1 1 ö - ÷N ç R è T f Ti ø 0

Pµ

(a) N = ò r(dv) r

PV (N ) RT 0

=

...(ii)

Comparing the above equation with | DT | = C | DP | , we have

3.

n=

\ nf – ni =

DP nR DT V =V Þ DT = DP P P nR

C=

\ m = g ç l ×l ÷ è 1 2 ø (b) Given: Temperature Ti = 17 + 273 = 290 K Temperature Tf = 27 + 273 = 300 K Atmospheric pressure, P0 = 1 × 105 Pa Volume of room, V0 = 30 m3 Difference in number of molecules, nf – ni = ? Using ideal gas equation, PV = nRT(N0), N0 = Avogadro's number Þ

(As temperature is constant)

P D V + V DP = 0

1ö

Þ nRT ç l – l ÷ = mg è 2 1ø

mg

vrms =

3RT M

P-193

Kinetic Theory

Here, M = Molar mass of gas molecule T = temperature of the gas molecule We have given vN2 = vH2 3RTN2

\

M N2

Þ 11.

TH2 2

=

3RTH 2 M H2

573 Þ TH2 = 41 K 28

(c) Vrms = v1 = v2

=

3RT M

T1 (273 + 127) = = T2 (273 + 237)

400 = 500

4 2 = 5 5

5 5 v1 = ´ 200 = 100 5 m/s. 2 2 (Bouns) Rate of change of momentum during collision \ v2 =

12.

=

mv – (– mv ) 2mv = N Dt Dt

so pressure P = N ´ (2mv) Dt ´ A 1022 ´ 2 ´ 10–26 ´ 104 = 2 N / m2 1´ 1 (c) vrms = ve =

13.

3RT M 3 ´ 8.314 ´ 300 2 (1930 ) = M 3 ´ 8.314 ´ 300 M= » 2 ´10 -3 kg 1930 ´1930 The gas is H2.

18. (a) Q

3RT = 11.2 ´ 103 M

or

or

14.

3kT = 11.2 ´ 103 m

3 ´ 1.38 ´ 10-23 T 2 ´ 10-3

V1rms (a) Using V = 2rms

= 11.2 ´ 10

3

\ v = 104 K

M2 M1

Vrms ( He )

15.

M Ar 40 = M He = 4 = 3.16 Vrms ( Ar ) (a) Energy associated with N moles of diatomic gas, 5 Ui = N RT 2 Energy associated with n moles of monoatomic gas 3 = n RT 2 Total energy when n moles of diatomic gas converted into 3 5 monoatomic (Uf) = 2n RT + (N - n) RT 2 2

=

1 5 nRT + NRT 2 2

Now, change in total kinetic energy of the gas 1 DU = Q = nRT 2 1 mN 2 V rm s 16 . (c) Pressure, P = 3 V (mN )T or, P = V If the gas mass and temperature are constant then P µ (Vrms)2 µ T So, force µ (Vrms)2 µ T i.e., Value of q = 1 17. (d) Kinetic energy of each molecule, 3 K.E. = K B T 2 In the given problem, Temperature, T = 0°C = 273 K Height attained by the gas molecule, h = ? 819K B 3 K.E. = K B ( 273) = 2 2 K.E. = P.E. 819K B = Mgh Þ 2 819K B or h = 2Mg C=

3rv mass of the gas 20. (a) Given, V1 = V V2 = 2V

19. (c) v rms =

T1 = 27° + 273 = 300 K T2 = ? From charle’s law V1 V2 = Q Pressure is constant ) T1 T2 ( V 2V or, 300 = T 2 \ T2 = 600 K = 600 – 273 = 327°C 21. (c) As, work done is zero. So, loss in kinetic energy = heat gain by the gas 1 2 R mv = nCv DT = n DT 2 g -1 1 2 m R mv = DT 2 M g -1

P-194

Physics

\ DT = 22.

Mv 2 ( g - 1) K 2R

And g = n1

(a) Number of moles of first gas = N A

But vrms =

n1 + n2 + n3 RTmix NA

(a) Given, mass = 1 kg Density = 4 kg m–3

g=

\

vO2 vHe

=

M O2

´

\ g = 1+

2 9 = 7 7 (D) Triatomic rigid molecules, f = 6 \ g = 1+

gRT M

M He g He

1.4 4 ´ = 0.3237 32 1.67 vO 2 460 \ vHe = = = 1421 m / s 0.3237 0.3237 (d) RMS velocity of a gas molecule is given by 3RT M Let T be the temperature at which the velocity of hydrogen molecule is equal to the velocity of oxygen molecule.

2 4 = 6 3 29. (266.67) Here work done on gas and heat supplied to the gas are zero. Let T be the final equilibrium temperature of the gas in the vessel. Total internal energy of gases remain same. \ g = 1+

i.e., u1 + u2 = u '1 + u '2 or, n1Cv DT1 + n2Cv DT2 = (n1 + n2 )CvT Þ (0.1)Cv (200) + (0.05)Cv (400) = (0.15)CvT

Vrms =

3RT 3R ´ (273 + 47) = 2 32 Þ T = 20K (c) Total degree of freedom f = 3 + 2 = 5

\

26.

Total energy, U =

nfRT 5RT = 2 2

2 , where f = degree of freedom f

2 7 = 5 5 (C) Diatomic non-rigid molecules, f = 7

[Q PM = dRT ]

=

25.

Cv

= 1+

2 5 = 3 3 (B) Diatomic rigid molecules, f = 5

[As R and T is constant] g O2

Cp

\ g = 1+

(a) The speed of sound in a gas is given by v = g M

3kT R

(A) Monatomic, f = 3

=

\ vµ

vrms

l 1 Þtµ 3kT T m 28. (c) As we know,

mass 1 = m3 Volume = density 4 Internal energy of the diatomic gas

24.

l

\t =

n1T1 + n2T2 + n3T3 n1 + n2 + n3

5 5 1 PV = ´ 8 ´ 104 ´ = 5 ´ 10 4 J 2 2 4 Alternatively: 5 5m 5 m PM RT = ´ K.E = nRT = d 2 2M 2M 4 5 mP 5 1 ´ 8 ´ 10 = = ´ = 5 ´ 104 J 2 d 2 4

1

But, mean time of collision, t =

n1 n n RT1 + 2 RT2 + 3 RT3 NA NA NA

23.

2 2 7 = 1+ = f 5 5

2pnd 2 where, d = diameter of the molecule n = number of molecules per unit volume

If there is no loss of energy then P1V1 + P2V2 + P3V3 = PV

Tmix =

= 1+

Cv

27. (b) Mean free path, l =

n2 Number of moles of second gas = N A n3 Number of moles of third gas = N A

=

Cp

800 = 266.67 K 3 (d) Here degree of freedom, f = 3 + 3 = 6 for triatomic nonlinear molecule. Internal energy of a mole of the gas at temperature T, \T =

30.

f 6 nRT = RT = 3RT 2 2 31. (b) Let Cp and Cv be the specific heat capacity of the gas at constant pressure and volume. U=

P-195

Kinetic Theory

At constant pressure, heat required

DQ1 = nC p DT Þ 160 = nC p × 50

...(i)

At constant volume, heat required DQ2 = nCv DT Þ 240 = nCv ×100

...(ii)

Dividing (i) by (ii), we get Cp 4 160 C p 50 = × Þ = Cv 3 240 Cv 100 g=

Cp Cv

=

4 2 = 1+ 3 f

(Here, f = degree of freedom)

32.

Þ f = 6. (a) Total energy of the gas mixture,

33.

f1n1 RT1 f 2 n2 RT2 + Emix = 2 2 5 5 = 3 ´ RT + ´ 3RT = 15RT 2 2 (a) As we know mean free path 1 æ Nö 2 ç ÷ pd 2 èV ø

l=

Here,

N = no. of molecule V = volume of container d = diameter of molecule

But PV = nRT = nNKT N P Þ = =n V KT 1 KT l= 2 pd 2 P For constant volume and hence constant number density P is constant. T So mean free path remains same. As temperature increases no. of collision increases so relaxation time decreases. (d) Specific heat of gas at constant volume

n of gas molecules

34.

5 R 5 7 = 2 = B B C = R \ Hence C 7 v 7 v 2 R 2 CvA

35. (Bonus) Mean free path of a gas molecule is given by 1 l= 2pd 2 n Here, n = number of collisions per unit volume d = diameter of the molecule If average speed of molecule is v then l Mean free time, t = v M 1 1 Þ t= = 2 2 3RT 2pnd v 2pnd æ 3RT ö çQ v = M ÷ è ø \ tµ

2 M \ t1 = M1 ´ d2 t2 M2 d12 d2

2

=

40 æ 0.1 ö ´ç ÷ = 1.09 140 è 0.07 ø

36. (c) Relaxation time (t ) µ

mean free path 1 Þ tµ speed v

and, v µ T 1 \tµ T Hence graph between t v/s

1

is a straight line which T is correctly depicted by graph shown in option (c). 37. (a) Helium is a monoatomic gas and Oxygen is a diatomic gas. For helium, CV1 =

3 5 R and CP1 = R 2 2

For oxygen, CV2 = g=

5 7 R and CP = R 2 2 2

N1CP1 + N 2 CP2 N1CV1 + N 2 CV2

Cv =

1 fR; f = degree of freedom 2 For gas A (diatomic) f = 5 (3 translational + 2 rotational)

5 7 n. R + 2n. R 2 = 19nR ´ 2 g= 2 Þ 3 5 2(13nR ) n. R + 2n. R 2 2

5 R 2 For gas B (diatomic) in addition to (3 translational + 2 rotational) 2 vibrational degree of freedom.

æC ö 19 \ç P ÷ = C è V ø mixture 13

\C A = v

P-196

38.

Physics

(d) Using, gmixture=

n1C p + n2C p 1

n1Cv + n2Cv 1

Þ Þ

2

2

n1 n n +n + 2 = 1 2 g 1 –1 g 2 –1 g m –1 3

2

5 = g m –1

+ 4 5 –1 –1 3 3 9 2´3 5 5 Þ + = Þ g m –1 = g m –1 1 2 12

17 = 1.42 12 n1[Cv1 ] + n2 [Cv2 ] (c) [Cv]min = n1 + n2 Þ gm =

39.

3R 5R ù é ê2´ 2 + 3´ 2 ú ú = êê 2+3 ú ë û = 2.1 R = 2.1 × 8.3 = 17.4 J/mol–k

40.

Cv 1 1 5 (b) F = C = r = (7 / 5) = p 7 or

W 5 2 = 1- = Q 7 7

7 7 ´ 10 W = = 35 J 2 2 (c) V = 25 × 10–3 m3, N = 1 mole of O2 T = 300 K Vrms = 200 m/s \ l=

1

2Npr 2 1 = 200.Npr 2 . 2 Average time = t l

42.

23

2 ´ 200 ´ 6.023 ´ 10

.p ´ 10-18 ´ 0.09 25 ´ 10 -3 The closest value in the given option is = 1010 (c) Amount of heat required (Q) to raise the temperature at constant volume Q = nCvDT ...(i) Amount of heat required (Q1) at constant pressure Q1 = nCPDT ...(ii) Dividing equation (ii) by (i), we get =

\

Q1 C p = Q Cv

æ 7ö Þ Q1 = (Q) ç ÷ è 5ø

or T =

mv 2 6k B

gA =

44. (b)

CP 29 = = 1.32 < 1.4 (diatomic) Cv 22

30 10 = = 1.43 > 1.4 21 7 Gas A has more than 5-degrees of freedom. 45. (a) Energy of the gas, E f f = nRT = PV 2 2 f = (3 ´ 106 )(2) = f ´ 3 ´ 106 2 Considering gas is monoatomic i.e., f = 3 Energy, E = 9 × 106 J 1 46. (b) Using, t= 2npd 2 Vavg

and g B =

\t µ

or,

T P

no.of molecules ù é êë\n = úû Volume

t1 500 P = ´ » 4´10–8 –8 6 ´10 2P 300

f1 f n1RT + 2 n 2RT 2 2 Considering translational and rotational modes, degrees of freedom f1 = 5 and f2 = 3

47. (a) U =

or Q =

41.

1 mv 2 = 3k BT 2

43. (a)

C p 7ö æ çQ g = C = 5 ÷ è ø v

5 3 \ u = (3RT) + ´ 5RT 2 2 U = 15RT

48. (a) According to question VT = K we also know that PV = nRT

æ PV ö ÞT = ç è nR ÷ø æ PV ö 2 ÞVç ÷ = k Þ PV = K è nR ø R Q C= + CV (For polytropic process) 1– x R 3R R + = 1– 2 2 2 \DQ = nC DT R = ´DT [here, n = 1 mole] 2 49. (c) Thermal energy of N molecule C=

æ3 ö = N ç kT ÷ è2 ø

P-197

Kinetic Theory

50.

=

N 3 3 3 RT = ( nRT ) = PV NA 2 2 2

=

3 æmö 3 æ 2ö Pç ÷= Pç ÷ 2 è r ø 2 è8ø

3 2 4 4 = ´ 4´10 ´ = 1.5 ´10 J 2 8 therefore, order = 104 J (c) Heat transferred, Q = nCv DT as gas in closed vessel To double the rms speed, temperature should be 4 times

i.e., T' = 4T as vrms =

3RT / M

15 5 ´ R ´ ´ ( 4T - T ) 28 2 7 é CP ù êë\ CV = γ diatomic = 5 & C p - Cv = R úû or, Q = 10000 J = 10 kJ (c) In an adiabatic process TVg–1 = Constant or, T1V1g–1 = T2V2g–1 5 For monoatomic gas g = 3 300 (300)V2/3 = T2(2V)2/3 Þ T2 = (2) 2/3 T2 = 189 K (final temperature) f Change in internal energy DU = n R DT 2 æ 3 öæ 25 ö = 2 ç ÷ç ÷ (-111) = -2.7 kJ è 2 øè 3 ø (b) Using formula, n1g 1 n 2 g 2 + æ Cp ö g1 - 1 g 2 - 1 g mixture = ç ÷ = n1 n è Cv ø mix + 2 g1 - 1 g 2 - 1

54. (d) The ratio of specific heats at constant pressure (C p) and constant volume (Cv) Cp æ 2ö = g = ç1 + ÷ è fø Cv where f is degree of freedom Cp æ 2 ö 7 = ç1 + ÷ = Cv è 5 ø 5 55. (d) For a polytropic process R R C = Cv + \ C - Cv = 1- n 1- n R R \ 1- n = \ 1=n C - Cv C - Cv

\Q=

51.

52.

æ Cp ö 3 Putting the value of n1 = 2, n2 = n, ç ÷ = è C v ø mix 2

53.

5 7 g 1 = , g 2 = and solving we get, n = 2 3 5 (a) As we know, Cp – Cv = R where Cp and Cv are molar specific heat capacities

or, Cp – Cv =

R M

For hydrogen (M = 2) Cp – Cv = a =

R 2

For nitrogen (M = 28) Cp – Cv = b =

R 28

\

a = 14 b

or, a = 14b

\ n=

C - Cp

(Q C p - C v = R ) C - Cv (d) Using equipartition of energy, we have =

56.

C - C v - R C - C v - Cp + C v = C - Cv C - Cv

6 KT = mCT 2

C=

3 ´1.38 ´10 –23 ´ 6.02 ´1023

27 ´10–3 \ C = 925 J/kgK 57. (b) 58. (b) On giving same amount of heat at constant pressure, there is no change in temperature for mono, dia and polyatomic. æ No. of molecules ö ( DQ) P = mC p DT ç m = è Avogedro 's no. ÷ø

or

DT µ

1 no. of molecules

59. (c) TV g -1 = constant g -1

T1V1g -1 = T2V2 Þ T (V )

1

2

æV ö =T2 ç ÷ è 4ø

1

2

Vù é êëQ g = 1.5, T1 = T ,V1 = V and V2 = 4 úû 1

æ 4V ö 2 \ T2 = ç ÷ T = 2T èV ø 60. (b) According to Mayer's relationship CP – CV = R, as per the question (CP – CV) M = R Þ CP – CV = R/28 Here M = 28 = mass of 1 unit of N2 61. (a) For mixture of gas specific heat at constant volume

Cv =

n1Cv1 + n2Cv2 n1 + n2

P-198

Physics

No. of moles of helium, mHe 16 n1 = M =4 = 4 He Number of moles of oxygen, 16 1 = n2 = 32 2 3 1 5 5 4 ´ R + ´ R 6R + R 2 2 2 4 = = 9 1ö æ \ Cv çè 4 + ÷ø 2 2

29 R ´ 2 29 R = and 9´ 4 18 Specific heat at constant pressure =

Cp =

n1C p1 + n2C p2 (n1 + n2 )

7 10 R + R 4 = 47 R = 9 18 2

5R 1 7 R + ´ 4´ 2 2 2 = 1ö æ çè 4 + ÷ø 2

Þ \

Cp Cv

=

47 R 18 ´ = 1.62 18 29 R

5 7 g = 3 2 5 n1 = 1, n2 = 1

62. (c) g 1 =

n1 + n2 n n = 1 + 2 g -1 g1 - 1 g 2 - 1 1 +1 1 1 3 5 = + = + =4 g - 1 5 - 1 7 -1 2 2 3 5 2 3 \ =4 Þ g= g -1 2 Þ

63. (d) P µ T 3 Þ PT -3 = constant ....(i) But for an adiabatic process, the pressure temperature relationship is given by P1-g T g = constant

Þ PT

g 1-g

= constt.

From (i) and (ii)

....(ii)

g 3 = -3 Þ g = -3 + 3g Þ g = 1- g 2

13

P-199

Oscillations

Oscillations TOPIC 1 1.

2.

3.

Displacement, Phase, Velocity and Acceleration in S.H.M.

5.

The position co-ordinates of a particle moving in a 3-D coordinate system is given by [9 Jan 2019, II] x = a coswt y = a sinwt and z = awt The speed of the particle is: (a) (c) 2 aw (b) aw 3 aw (d) 2aw Two simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures. x(t) = A sin (at + d) y(t) = B sin (bt) Identify the correct match below [Online April 15, 2018] p (a) Parameters: A = B, a = 2b; d = ; Curve: Circle 2 p (b) Parameters: A = B, a = b; d = ; Curve: Line 2 p (c) Parameters: A ¹ B, a = b; d = ; Curve: Ellipse 2 (d) Parameters: A ¹ B, a = b; d = 0; Curve: Parabola The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s–1. At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is p [Online April 8, 2017] 4 (a)

500 m/s2

(b)

500

2

m/s2

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement -A and they are moving towards each other. If they cross 2 each other at time t, then t is: [Online April 9, 2016]

(a) 6.

7.

5T 6

(b)

T 3

(c)

T 4

(d)

T 6

A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to : [Online April 10, 2015] 1 1 (a) sin t + cos 2t (b) cos t - sin 2t 2 2 1 1 (c) sin t - sin 2t (d) sin t + sin 2t 2 2 x and y displacements of a particle are given as x(t) = a sin wt and y (t) = a sin 2wt. Its trajectory will look like : [Online April 10, 2015] y

y

x

(a)

750 m/s2

4.

(c) (d) 750 2 m/s2 A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a distance

x

(b)

2A from equilibrium position. The new amplitude of the 3 motion is : [2016] 7A A 41 (d) 3A (a) A 3 (b) (c) 3 3

y

y

(c)

x

(d)

x

P-200

8.

9.

10.

Physics

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves form equilibrium position to half of its amplitude. [Online April 19, 2014] (a) 4 cm/s (b) 6 cm/s (c) 12 cm/s (d) 16 cm/s Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants? [Online April 12, 2014] 2 (a) a + bx – cx (b) bx2 (c) a – bx + cx2 (d) – bx A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x = a1 cos wt and y = a2 cos 2 wt traces a curve given by: [Online April 9, 2014]

(a)

y

y

a2

a2

a1

x

O

(b)

12. Two particles are executing simple harmonic motion of the same amplitude A and frequency w along the x-axis. Their mean position is separated by distance X0(X0 > A). If the maximum separation between them is (X0 + A), the phase difference between their motion is: [2011] (a)

(c)

11.

x

O

(d)

a1

y

(c)

(d)

[2011] 1

(a)

M +m M

(c)

æ M + mö 2 çè ÷ M ø

(b)

æ M ö2 çè ÷ M + mø

(d)

M M +m

1

is written as a = A cos(wt + d ) ,then

a2 x

t

y

t

p 2

[2007]

A = x0 w 2 , d = 3p / 4 (b) A = x0, d = -p / 4 2

O

(b)

(d)

æ A1 ö

(c) A = x0 w 2 , d = p / 4 (d) A = x0 w , d = -p / 4 15. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency w. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time (a) at the mean position of the platform [2006] g (b) for an amplitude of 2 w (c) for an amplitude of 16.

t

p 6

law x = x0 cos(wt - p / 4) . If the acceleration of the particle

The displacement y(t) = A sin (wt + f) of a pendulum for 2p is correctly represented by f= 3 [Online May 19, 2012] y y

(a)

(c)

of ç A ÷ is: è 2ø

x

y

a2

p 4

13. A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio

(a)

a1

(b)

14. A point mass oscillates along the x-axis according to the

O a1

y

p 3

t

g2

w2 (d) at the highest position of the platform The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is [2006] (a) 0.01 s (b) 10 s (c) 0.1 s (d) 100 s

17. The function sin 2 (wt ) represents [2005] (a) a periodic, but not simple harmonic motion with a p period w (b) a periodic, but not simple harmonic motion with a 2p period w (c) a simple harmonic motion with a period

p w

(d) a simple harmonic motion with a period

2p w

P-201

Oscillations

18.

Two simple harmonic motions are represented by the pö æ equations y1 = 0.1 sin ç100pt + ÷ and y 2 = 0.1 cos pt . 3ø è

19.

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is [2005] p -p p -p (a) (b) (c) (d) 6 6 3 3 Two particles A and B of equal masses are suspended from two massless springs of spring constants k 1 and k 2, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of A and B is [2003] k2 k1 k1 k2 (b) k (c) (d) k k2 k1 1 2 The displacement of a particle varies according to the

(a) 20.

relation x = 4(cos pt + sin pt ). The amplitude of the particle is [2003] (a) – 4

TOPIC 2

(d) 8

Energy in Simple Harmonic Motion

The displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale)

displacement

21.

(c) 4 2

(b) 4

O

2T 4 T 4

3T T 4

time (s)

5T 4

Which of the following statements is/are true for this motion? [Sep. 02, 2020 (II)] 3T 4 (2) The acceleration is maximum at t = T

(1) The force is zero at t =

(3) The speed is maximum at t =

T 4

(4) The P.E. is equal to K.E. of the oscillation at t =

22.

T 2

(a) (1), (2) and (4) (b) (2), (3) and (4) (c) (1), (2) and (3) (d) (1) and (4) A particle undergoing simple harmonic motion has time

pt . The 90 ratio of kinetic to potential energy of this particle at t = 210s will be : [11 Jan 2019, I]

23. A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. [11 Jan 2019, II] (a) K2 = 2K1

24.

(a)

(b) 1

(c) 2

(d)

1 3

K2 =

K1 2

K1 (d) K2 = K1 4 A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be: [9 Jan 2019, II]

(c)

K2 =

(a)

A 2

(b)

A 2 2

A (d) A 2 25. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like: [2017] (c)

(a)

(b)

(c)

(d)

26. A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm–1 and oscillates in a medium of constant 10–2 kg s–1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to : [Online April 9, 2017] (a) 2 s (b) 3.5 s (c) 5 s (d) 7 s 27. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale) [2015] E

KE

E PE

(a)

d

dependent displacement given by x ( t ) = A sin

1 9

(b)

(b)

KE

PE

E

(c)

E

KE PE

(d) d

PE KE

d

P-202

28.

29.

30.

A pendulum with time period of 1s is losing energy. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s–1) is : [Online April 11, 2015] 1 1 1 ln3 (c) 2 ln3 (a) (b) (d) 30 15 2 This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs S1 and S2 of force constants k1 and k 2 respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2. Statement 1 : If stretched by the same amount work done on S1 Statement 2 : k1 < k2 [2012] (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is false. (c) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1 (d) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1 A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is [2007] 2 2 2 2 2 2 (a) 2p ma n (b) p ma n 1 2 2 ma n (d) 4p 2 ma 2 n2 4 Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? [2006]

(c)

31.

1 1 1 1 s s s s (b) (d) (c) 6 3 4 12 The total energy of a particle, executing simple harmonic motion is [2004] (a) independent of x (b) µ x2 (c) µ x (d) µ x1/2 where x is the displacement from the mean position, hence total energy is independent of x. A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and total energy (T.E) are measured as a function of displacement x. Which of the following statements is true ? [2003] (a) K.E. is maximum when x = 0 (b) T.E is zero when x = 0 (c) K.E is maximum when x is maximum (d) P.E is maximum when x = 0 In a simple harmonic oscillator, at the mean position [2002] (a) kinetic energy is minimum, potential energy is maximum (b) both kinetic and potential energies are maximum (c) kinetic energy is maximum, potential energy is minimum (d) both kinetic and potential energies are minimum

(a)

32.

33.

34.

Physics

Time Period, Frequency, TOPIC 3 Simple Pendulum and Spring Pendulum 35. An object of mass m is suspended at the end of a massless wire of length L and area of cross-section, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is: [Sep. 06, 2020 (I)] (a)

f =

1 mL 2p YA

(b) f =

1 YA 2p mL

1 mA 1 YL (d) f = 2p YL 2p mA 36. When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by y (t) = y0 sin2wt, where ‘y’ is measured from the lower

(c)

f =

end of unstretched spring. Then w is : [Sep. 06, 2020 (II)] (a)

1 2

g y0

(b)

g 2 y0

(d)

g y0

2g y0 37. A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is : [Sep. 03, 2020 (II)] 1 1 (a) (b) 1 (c) (d) 2 2 2 38. A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude q0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l (l w2 Thus (k1 > k2) 30. (b) The kinetic energy of a particle executing S.H.M. at any instant t is given by 1 2 2 2 ma w sin wt 2 where, m = mass of particle a = amplitude w = angular frequency t = time

K=

The average value of sin 2wt over a cycle is

1 . 2

1ö æ 1ö æ 1 2 \KE = mw2a2 çè ÷ø çè Q < sin q > = ÷ø 2 2 2 =

1 1 2 2 mw a = ma2 (2pn)2 4 4 2

(Q w = 2pn)

2 2

or, < K > = p ma n 31. (a) K.E. of a body undergoing SHM is given by, 1 2 2 2 ma w cos wt 2 Here, a = amplitude of SHM w = angular velocity of SHM K .E. =

1 2 2 Total energy in S.H.M = ma w 2 Given K.E. = 75% T.E. 1 75 1 ma 2 w 2 cos 2 w t = ´ ma 2 w 2 2 100 2 p 2 Þ 0.75 = cos wt Þ wt = 6 p p´2 1 Þt= Þt= Þt= s 6´w 6 ´ 2p 6 32. (a) At any instant the total energy in SHM is 1 2 kA = constant, 2 0 where A0 = amplitude k = spring constant hence total energy is independent of x. 33. (a) K.E. of simple harmonic motion 1 = mw2 (a 2 - x 2 ) 2

P-212

Physics

34.

(c) The kinetic energy (K. E.) of particle in SHM is given by, 1 K.E = k ( A2 - x 2 ) ; 2 1 2 Potential energy of particle in SHM is U = kx 2 Where A = amplitude and k = mw2 x = displacement from the mean position At the mean position x = 0 1 2 \ K.E. = kA = Maximum 2 and U = 0 35. (b) An elastic wire can be treated as a spring and its spring constant.

é F YA k= êQ Y = A L ë Frequency of oscillation, 1 2p

f =

36.

Dl ù l0 úû

l g When immersed non viscous liquid

39. (c) T = 2p

g ö 15 g æ amt = ç g - ÷ = 16 ø 16 è l

T ¢ = 2p

0net

Now

l t = -2Kx cos q 2

x=

k 1 YA = m 2p mL

y0 (1 - cos 2wt ) 2

1 - cos 2wt ö æ 2 çèQ sin wt = ÷ø 2

y0 - y0 = cos 2wt 2 2 Þ y = A cos 2wt y0 2 Angular velocity = 2w

37.

f=

2

2g 1 2g Þw= = y0 2 y0

g 2 y0

l 2

1 2 kx 2 Here, x = distance of block from mean position, k = spring constant

1 C 1 = 2p I 2p

Kl2 2 = 1 6K Ml 2 2p M 12

41. (b)

q0 l

(a) Potential energy of spring =

1 At mean position, potential energy = kA2 2 At equilibrium position, half of the mass of block breaks off, so its potential energy becomes half.

Remaining energy =

1 æ 1 2ö 1 2 ç kA ÷ø = kA ' 2è2 2

Here, A' = New distance of block from mean position

Þ A' = 38.

T

q

ky0 k 2g = mg Þ = m y0 2

Also, spring constant k = m(2w) k = m

15

Kx æ Kl 2 ö é Kl 2 ù Þ t=ç q = -Cq êlet C = ú ÷ 2 ûú è 2 ø ëê Þ So, frequency of resulting oscillations

\ Amplitude =

Þ 2w =

4

q

Þ y-

For equilibrium of mass,

15 g 16

=

40. (c) Net torque due to spring force:

(c) y = y0 sin 2 wt Þy=

l

= 2p

(b)

mM Velocity before colision Velocity after colision v1 = 2gl(1 - cos q1)

Using momentum conservation mv = MVm – mV1 m 2gl(1 - cos q0 ) = MVm - m 2gl(1 - cos q) Þ m 2gl

{

A 2

v = 2gl(1 - cos q0 )

and e = 1 =

}

1 - cos q0 + 1 - cos q1 = MVm

Vm + 2gl(1 - cos q1) 2gl(1 - cos q0 )

P-213

Oscillations

( 2gl (

2gl m

... (i) ) 1 - cos q1 ) = MVM ... (ii)

1 - cos q0 - 1 - cos q1 = Vm 1 - cos q 0 +

Dividing (ii) by (i) we get

( (

)=M 1 - cos q1 ) m

Þ w A 2 – x2 = w2 x Þ A 2 – x 2 = w2 x 2 Þ 5 2 – 4 2 = w2 ( 4 2 )

1 - cos q0 + 1 - cos q1

3 = w × 4 Þ w=

1 - cos q0 -

\ T = 2p/w =

By componendo and dividendo rule

46. (Bonus)

æq ö sin ç 1 ÷ 1 - cos q1 è 2ø m-M = = m+M æ q0 ö 1 - cos q0 sin ç ÷ è 2ø

2 p 8p = 3/ 4 3

B0– B )

x

q -q M q0 - q1 Þ = ÞM=m 0 1 m q0 + q1 q0 + q1

42.

pö

mg

Extra boyant force = rAxg B0 + B = mg + ma \ B = ma = rAxg = (pr2rg)x

ø

\ Amplitude = 10 cm Time period, T =

2p 2p 2 = = s w 3p 3

( pr rg ) x 2

(a) Angular frequency of pendulum w = \ relative change in angular frequency

g l

Dw 1 Dg [as length remains constant] = w 2 g D g = 2Aws2 [ws = angular frequency of support and, A = amplitude]

a=

m using, a = w2x

pr 2rg m W ;7.95 rads–1 47. (c) Þw=

L/2

Dw 1 2Aws2 = ´ w 2 g

Mp æ R e = ç g e M e çè R p

L– X–L

Tp 1 g Þ = e= 3 gp g Te

(c) Velocity, v = w A 2 – x 2 acceleration, a = –w2x and according to question, | v | =| a |

f1 =

1 C 2p 1

=

1 3C 2 ML2

2

2 ö æ1ö 1 ÷÷ = 3 ç ÷ = è3ø 3 ø

C æM M ö L2 ç + ÷ 2 ø è 3 As frequency reduces by 80% f2 \ f2 = 0.8 f1 Þ f = 0.8 1 Solving equations (i), (ii) & (iii) f2 =

Þ Tp = 2 3 s

45.

m M

GM (d) Acceleration due to gravity g = 2 R gp

L/2

m

1 2 ´12 ´10 –2 Dw = ´ = 10–3 rad/sec. 2 10

Also T µ

at equilibriu B0 = mg

y = 5 éësin(3pt) + 3 cos(3pt) ùû

(a) Given : æ è

44.

a

x0

Þ y = 10sin ç 3pt + 3 ÷

43.

3 4

...(i) ...(ii)

1 2p

Ratio

m = 0.37 M

...(i)

...(ii)

...(iii)

P-214

Physics

As frequency reduces by 80% f2 \ f2 = 0.8 f1 Þ f = 0.8 1 Solving equations (i), (ii) & (iii)

F=

...(iii)

52. (c) Time lost/gained per day =

m = 0.37 M 48. (b) As we know, frequency in SHM Ratio

1 k = 1012 2p m where m = mass of one atom

f=

108 ´10-3 kg Mass of one atom of silver, = 6.02 ´ 1023

(

49.

)

1 k ´ 6.02 ´ 1023 = 1012 2p 108 ´ 10-3 Solving we get, spring constant, K = 7.1N/m (d) Using y = A sin wt a = A sin wt0 b = A sin 2wt0 c = A sin 3wt0 a + c = A[sin wt0 + A sin 3wt0] = 2A sin 2wt0 cos wt0

f
54cm 47. (a) It is given that 315 Hz and 420 Hz are two resonant frequencies, let these be n th and (n + 1)th harmonies, then we have nv = 315 2l v = 420 and (n + 1) 2l n + 1 420 Þ = n 315 Þn=3 v v = 105 Hz Hence 3 ´ = 315 Þ 2 l 2l The lowest resonant frequency is when n=1 Therefore lowest resonant frequency = 105 Hz. 48. (c) The fundamental frequency for tube B closed at one end is given by lù v é Ql = ú uB = ê 4û 4l ë Where l = length of the tube and v is the velocity of sound in air. The fundamental frequency for tube A open with both ends is given by lù é v Ql = ú uA = ê 2û 2l ë

50. 51. 52. 53.

54.

55.

uA v 4l 2 \ u = 2l ´ v = 1 B (b) To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of p. The equation of the reflected wave will be y = a sin (wt + kx + p) Þ y = – a sin (wt + kx) (a) (c) Beat frequency = difference in frequencies of two waves = 11 – 9 = 2 Hz (d) (d) According to question, tuning fork gives 1 beat/second with (N) 3rd normal mode. Therefore, organ pipe will have frequency (256 ± 1) Hz. In open organ pipe, frequency NV n= 2l 3 ´ 340 Þ l = 2 m = 200 cm or, 255 = 2´ l (a) Probable frequencies of tuning fork be n ± 5 1 Frequency of sonometer wire, n µ l n + 5 100 \ = Þ 95(n + 5) = 100( n - 5) n - 5 95 or, 95 n + 475 = 100 n – 500 or, 5 n = 975 975 or, n = = 195 Hz 5 æ 5p ö (a) Given, y (x, t) = 0.5 sin ç x ÷ cos (200 pt), è 4 ø

comparing with equation – y (x, t) = 2 a sin kx cos wt 5p w = 200 p, k = 4 w 200p speed of travelling wave v = = = 160 m/s k 5p 4 56. (b) Since the point x = 0 is a node and reflection is taking place from point x = 0. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of p or a path l change of . 2 So, if yincident = a cos ( kx – wt) Þ yincident = a cos (– kx – wt + p) = – a cos (wt + kx) Hence equation for the other wave y = a cos(kx + wt + p) 57. (d) In case of destructive interference Phase difference f = 180° or p So wave pair (i) and (ii) will produce destructive interference. Stationary or standing waves will produce by equations (iii) & (iv) as two waves travelling along the same line but in opposite direction. n¢ = n + x 58. (d) y = A sin (wt – kx) + A sin (wt + kx) y = 2A sin wt cos kx

P-232

Physics

This is an equation of standing wave. For position of nodes cos kx = 0 2p p .x = (2n + 1) Þ l 2 2 n + 1) l ( Þ x= , n = 0,1, 2,3,........... 4 59. (a) Intensity of a wave 1 I = pw2 A2 v 2 Since, I µ A2w 2 \ I1 µ (2a)2 w 2

60. 61.

62.

63.

and I 2 µ a 2 (2w ) 2 I1 = I 2 In the same medium, p and v are same. Intensity depends on amplitude and frequency. (b) Maximum number of beats = Maximum frequency – Minimum frequency = ( n + 1) – ( n – 1) = 2 Beats per second (d) Frequency of fork 1, no = 200 Hz No. of beats heard when fork 2 is sounded with fork 1 = Dn = 4 Now on loading (attaching tape) on unknown fork, the mass of tuning fork increases, So the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by, n = n 0 – Dn = 200 – 4 = 196 Hz (c) It is given that tuning fork of frequency 256 Hz makes 5 beats/second with the vibrating string of a piano. Therefore, possible frequency of the piano are (256 ± 5) Hz. i.e., either 261Hz or 251 Hz. When the tension in the piano string increases, its frequency will increases. As the original frequency was 261Hz, the beat frequency should decreases, we can conclude that the frequency of piano string is 251Hz (b) Frequency of unknown fork = known frequency ± Beat frequency = 288 + 4 cps or 288 – 4 cps i.e. 292 cps or 284 cps. When a little wax is placed on the unknown fork, it produces 2 beats/sec. When a little wax is placed on the unknown fork, its frequency decreases and simultaneously the beat frequency decreases confirming that the frequency of the unknown fork is 292 cps. Note : Had the frequency of unknown fork been 284 cps, then on placing wax its frequency would have decreased thereby increasing the gap between its frequency and the frequency of known fork. This would produce high beat frequency.

64. (b) Frequency heard by the observer æ ö vsound vobserved = ç ÷ v0 è vsound - v cos q ø Observer O D q Source V

Initially q will be less so cos q more. \ vobserved more, then it will decrease.

65. (a) Let f 1 be the frequency heard by wall, æ v ö f1 = ç f0 è v - vc ÷ø Here, v = Velocity of sound, vc = Velocity of Car, f0 = actual frequency of car horn Let f2 be the frequency heard by driver after reflection from wall. æ v + vc ö æ v + vc ö f2 = ç f = f è v ÷ø 1 çè v - v ÷ø 0 c

é 345 + vc ù 12 345 + vc Þ 480 = ê = ú 440 Þ 11 345 - vc ë 345 - vc û Þ vc = 54 km/hr 66. (a) From the Doppler's effect of sound, frequency appeared at wall 330 fw = ×f ...(i) 330 - v Here, v = speed of bus, f = actual frequency of source Frequency heard after reflection from wall (f') is 330 + v 330 + v f '= × fw = ×f 330 330 - v 330 + v Þ 490 = × 420 330 - v 330 ´ 7 Þv= » 25.38 m/s = 91 km/s 91 67. (d) Permanent magnets (P) are made of materials with large retentivity and large coercivity. Transformer cores (T) are made of materials with low retentivity and low coercivity. 68. (c) From Doppler’s effect, frequency of sound heard (f1) when source is approaching c f1 = f 0 c–v Here, c = velocity of sound v = velocity of source Frequency of sound heard (f2) when source is receding

c c+v Beat frequency = f1 – f2 1 ù é 1 Þ 2 = f1 – f 2 = f 0c ê – – c v c + v úû ë 2v = f0c é v2 ù c 2 ê1 – 2 ú êë c úû For c>> v c 2c 350 1 Þ v= = = = m/s f 0 1400 4 2 f0 f2 = f0

æ v - vo ö æ 1500 - 5 ö 69. (d) f1 = f ç v - v ÷ = f çè 1500 - 7.5 ÷ø è ø s No reflected signal,

P-233

Waves

æ v + vo ö æ 1500 + 7.5 ö f2 = f1 ç v + v ÷ = f1 çè 1500 + 5 ÷ø è ø s æ 1500 - 5 ö æ 1500 + 7.5 ö f2 = 500 çè ÷ç ÷ 1500 - 7.5 ø è 1500 + 5 ø 502 Hz

70. (c) f1 = f S1

é V ù é 340 ù fapp = f0 ê ú Þ fl = f0 ê ú ë V – Vs û ë 340 – 34 û

v - v0 v + v0 and f2 = f v v

L

u

S2

But frequency, 2v0 f2 – f1 = f ´ v 2u 330 \ u = 2.5 m/s. 71. (b) Frequency of sound source (f0) = 500 Hz When observer is moving away from the source æ n - n'0 ö Apparent frequency f1 = 480 = f0 ç n ÷ ....(i) è ø And when observer is moving towards the source

or 10 = 660 ´

æ n - n"0 ö f 2 = 530 = f0 ç ÷ è n ø From equation (i) æ 300 - v '0 480 = 500 ç ç 300 è v’0 = 12 m/s From equation (ii)

....(ii)

ö ÷ ÷ ø

æ v'' ö 530 = 500 ç1 + 0 ÷ vø è \ V”0 = 18 m/s 72. (a) When source is moving towards a stationary observer, æ V -0 ö fapp = f source ç è V - 50 ÷ø æ 350 ö 1000 = fsource ç è 300 ÷ø When source is moving away from observer æ 350 ö f ' = fsource çè ÷ 350 + 50 ø

1000 ´ 300 350 ´ 350 400 f ' » 750 Hz v , v0 73. (a) f ' = f v ∗ vs

f'=

340 , 20 340 ∗ 20 \ f = 2250 Hz. 74. (b) According to Doppler’s effect, when source is moving but observer at rest

or 2000 < f

é 340 ù and, f 2 = f 0 ê ú ë 340 –17 û f 340 –17 323 f 19 \ l = = or, 1 = f 2 340 – 34 306 f 2 18 75. (a) Frequency of the sound produced by open flute. æ v ö 2 ´ 330 = 660 Hz f = 2ç ÷ = è 2l ø 2 ´ 0.5 5 25 Velocity of observer, v0 = 10 ´ = m / s 18 9 As the source is moving towards the observer therefore, according to Doppler's effect. \ Frequency detected by observer, é 25 ù ê 9 + 330 ú é v + v0 ù ê ú 660 f'= ê ú f = ê 330 ú ë v û ë û 2995 ´ 660 or, f ' = 665.55 ; 666 Hz = 9 ´ 330 76. (d) nA = 425 Hz, nB = ? Beat frequency x = 5 Hz which is decreasing (5 ® 3) after increasing the tension of the string B. Also tension of string B increasing so nB­ (Q n µ T) Hence nA – nB­= x¯ ¾¾ ® correct nB­ – nA= x¯ ¾¾ ® incorrect \ nB = nA – x = 425 – 5 = 420 Hz 77. (d) From Doppler's effect æ 340 ö f (direct) = f ç ÷ = f1 è 340 - 5 ø æ 340 ö f (by wall)=f ç ÷ =f è 340+5 ø 2 Beats = (f1 – f2) 340 ö æ 340 5=fç ÷ è 340 - 5 340 + 5 ø Þ f < 170 Hz. 78. (d) We know that the apperent frequency æ v - v0 ö f'=ç ÷ f from Doppler's effect è v - vs ø where v0 = vs = 30 m/s, velocity of observer and source Speed of sound v = 330 m/s 330 ∗ 30 \ f '< ≥ 540 = 648 Hz. 330 , 30 Q Frequency of whistle (f) = 540 Hz. é

79.

v ù 320 Hz ú= f´ v v 300 sû ë

(d ) f1 = f ê

é v ù 320 f2 = f ê Hz ú= f ´ 340 ë v + vs û

P-234

Physics

æ f2 ö æ 300 ö çè f - 1÷ø ´ 100 = çè 340 - 1÷ø ´ 100 ; 12% 1 80. (c) According to Doppler’s effect, æ V +V ö

0 Apparent, frequency f = ç V – V ÷ f 0 S ø è

æ

f

ö

V f

0 0 Now, f = ç V – V ÷ V0 + V – V è S ø s f0 So, slope = V – VS Hence, option (c) is the correct answer. 81. (a) Reflected frequency of sound reaching bat

é V - (-V0 ) ù é V + V0 ù V + 10 = ê f úf= ê úf= V - 10 ë V - Vs û ëV - Vs û æ 320 + 10 ö = ç ÷ ´ 8000 = 8516 Hz è 320 - 10 ø 82. (b) Given fA = 1800 Hz vt = v fB = 2150 Hz Reflected wave frequency received by A, fA¢ = ? Applying doppler’s effect of sound, vs f f¢ = vs - v t

æ f ö here, v t = vs ç1 - A ÷ è fB ø æ 1800 ö = 343 ç 1 ÷ è 2150 ø vt = 55.8372 m/s Now, for the reflected wave, æ vs + v t ö ¢ \ fA = ç v - v ÷ fA t ø è s

æ 343 + 55.83 ö =ç ÷ ´ 1800 è 343 - 55.83 ø = 2499.44 » 2500Hz 83. (d) Given: Frequency of sound produced by siren, f = 800 Hz Speed of observer, u = 2 m/s Velocity of sound, v = 320 m/s No. of beats heard per second = ? No. of extra waves received by the observer per second = + 4l \ No. of beats/ sec 2 æ 2ö 4 = - ç- ÷ = l è lø l æ 2´ 2 Vö = 320 çQ l = ÷ f ø è 800 2 ´ 2 ´ 800 = 10 = 320 84. (c) f = 500 Hz A 4 m/s C

B

Listener Case 1 : When source is moving towards stationary listener

æ v ö æ 340 ö = 506 apparent frequency h ' = h ç v - v ÷ = 500 ç ÷ s ø è è 336 ø Hz Case 2 : When source is moving away from the stationary listener æ v ö æ 340 ö h" = h ç ÷ = 500 ç ÷ = 494 Hz è 344 ø è v + vs ø In case 1 number of beats heard is 6 and in case 2 number of beats heard is 18 therefore frequency of the source at B = 512 Hz

85. (d)

ENGINE

A

0.9 km

H B I L L

C Let after 5 sec engine at point C AB BC 0.9 ´ 1000 BC t= + + 5= 330 330 330 330 \ BC = 750 m Distance travelled by engine in 5 sec = 900 m – 750 m = 150 m Therefore velocity of engine 150 m = = 30 m/s 5sec 86. (c) Bats catch the prey by hearing reflected ultrasonic waves. When the source and the detector (observer) are moving, frequency of reflected waves change. This is according to Doppler’s effect. 2 u=0 vm a = 2m/s 87. (a) Electric s Motor siren cycle Let the motorcycle has travelled a distances, its velocity at that point 2 vm - u 2 = 2as \ v2m = 2 ´ 2 ´ s \ vm = 2 s The observed frequency will be é v - vm ù n'= nê ú ë v û

é 330 - 2 s ù 0.94n = n ê ú Þ s = 98.01 m ë 330 û é v ù 88. (c) Apparent frequency n' = n ê ú ë v - vs û é 300 ù Þ 10000 = 9500 ê ú Þ 300 - v = 300 ´ 0.95 ë 300 - v û

Þ v = 300 - 285 = 15 ms -1 89. (c) Apparent frequency vù é v+ ú ê + v v é 5 = n é6ù n' = 6 0ù = n n' = nê ê ú ê5ú n 5 ú ë û ë v û ê v ú ë û The percentage increase in apparent n'- n 6 - 5 = ´ 100 = 20% frequency n 5

15

P-235

Electric Charges and Fields

Electric Charges and Fields TOPIC 1 1.

2.

Electric Charges and Coulomb's Law

+Q

Three charges + Q, q, + Q are placed respectively, at distance, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is: [9 Jan. 2019 I] (a) – Q/4 (b) + Q/2 (c) + Q/4 (d) – Q/2 Charge is distributed within a sphere of radius R with a

A -2r a where A and a e r2 are constants. If Q is the total charge of this charge distribution, the radius R is: [9 Jan. 2019, II]

–Q

volume charge density p(r) =

Q ö æ (a) a log ç 1 ÷ 2paA ø è

3.

4.

æ ö ç ÷ a 1 log ç (b) ÷ 2 çç 1 - Q ÷÷ 2paA ø è

æ ö ç ÷ 1 a Q ö æ (c) a log ç log ç 1 ÷ (d) ÷ 2 2paA ø è çç 1 - Q ÷÷ 2paA ø è Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to [Online April 16, 2018] 3F F 3F (b) (c) F (d) 4 2 8 Shown in the figure are two point charges +Q and –Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If s1 is the surface charge on the inner surface and Q1 net charge on it and s2 the surface charge on the outer surface and Q2 net charge on it then : [Online April 10, 2015] (a)

5.

(a) s1 ¹ 0, Q1 = 0 (b) s1 ¹ 0, Q1 = 0 s2 = 0, Q2 = 0 s2 ¹ 0, Q2 = 0 (c) s1 = 0, Q1 = 0 (d) s1 ¹ 0, Q1 ¹ 0 s2 = 0, Q2 = 0 s2 ¹ 0, Q2 ¹ 0 Two charges, each equal to q, are kept at x = – a and x = a q on the x-axis. A particle of mass m and charge q 0 = is 2 placed at the origin. If charge q0 is given a small displacement (y R (d) F = for all r 4pe0 R 2 4pe0 R 2 15. Two charged thin infinite plane sheets of uniform surface

(c) F =

charge density s + and s – , where | s + | > | s – |, intersect at right angle. Which of the following best represents the electric field lines for this system ? [Sep. 04, 2020 (I)] s–

s+

(a)

s–

s+

(b)

s–

s+

(c)

s–

Electric Field and Electric Field Lines

Charges Q1 and Q2 are at points A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then Q1/Q2 is proportional to : [Sep. 06, 2020 (I)] A Q1

(a)

(a) F =

s+

(d)

16. A particle of charge q and mass m is subjected to an electric field E = E0 (1 – ax2) in the x-direction, where a and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is : [Sep. 04, 2020 (II)] 3 1 2 (c) (d) a a a 17. A charged particle (mass m and charge q) moves along X axis with velocity V0. When it passes through the origin it r enters a region having uniform electric field E = - Ejˆ which

(a) a

(b)

extends upto x = d. Equation of path of electron in the region x > d is : [Sep. 02, 2020 (I)]

P-237 ur 20. An electric dipole of moment p = (iˆ - 3 ˆj + 2kˆ) ´ 10-29 C.m is at the origin (0, 0, 0). The electric field due to this dipole r at r = +iˆ + 3 ˆj + 5kˆ r ur (note that r . p = 0) is parallel to: [9 Jan. 2020, I]

Electric Charges and Fields

Y E O

(a) y = (c) y = 18.

qEd mV02 qEd

V0

X d

(x - d )

(b) y =

x

(d) y =

qEd æ d ö - x÷ 2 ç mV0 è 2 ø

qEd 2

x mV02 A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale). [Sep. 02, 2020 (II)] E x mV02

(a) (+iˆ - 3 ˆj - 2kˆ)

(c) (+iˆ + 3 ˆj - 2kˆ) (d) (-iˆ - 3 ˆj + 2kˆ) 21. A charged particle of mass ‘m’ and charge ‘q’ moving under the influence of uniform electric field Eiˆ and a uniform r magnetic field Bk follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, r r vi and -2vj . Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale) [9 Jan. 2020, I] Y

E P

y y

a

y

(a)

O

(b)

x

x

y

y

(c) 19.

(d)

x x Consider a sphere of radius R which carries a uniform R charge density r. If a sphere of radius is carved out of 2 ur EA it, as shown, the ratio ur of magnitude of electric EB ur ur field E A and E B , respectively, at points A and B due to the remaining portion is: [9 Jan. 2020, I]

(b) (-iˆ + 3 ˆj - 2kˆ)

B

v

Q 2v

2a

3 æ mv 2 ö (A) E = 4 çç qa ÷÷ è ø

X

3 æ mv 2 ö (B) Rate of work done by the electric field at P is 4 çç a ÷÷ è ø (C) Rate of work done by both the fields at Q is zero (D) The difference between the magnitude of angular momentum of the particle at P and Q is 2 mav. (a) (A), (C), (D) (b) (B), (C), (D) (c) (A), (B), (C) (d) (A), (B), (C) , (D) 22. Three charged particles y

2q B d

150° O

d 30° 30° d

–4q A x C –2q

A, B and C with charges – 4q, 2q and –2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is: [8 Jan. 2020, I] (a)

21 34

(b)

18 34

(c)

17 54

(d)

18 54

(a)

3q p Î0 d

2

(b)

2 3q p Î0 d

2

(c)

3q 4p Î0 d

2

(d)

3 3q 4p Î0 d 2

P-238

23.

Physics

A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale) [8 Jan. 2020, II] v

v

(a)

x

v

v

(c)

(d) x

x

Two infinite planes each with uniform surface charge density +s are kept in such a way that the angle between them is 30°. The electric field in the region shown between them is given by: [7 Jan. 2020, I]

30°

x é æ s 3ö xˆ ù s é xˆ ù yˆ + ú (a) (1 + 3) yˆ - ú (b) êç1 + ÷ ê Î0 ëêè 2 ø 2 ûú 2 Î0 ë 2û s éæ 3ö xˆ ù xˆ ù s é yˆ - ú 1 + 3 yˆ + ú (d) êç1 ÷ ê 2 Î0 ëêè 2 ø 2 ûú 2 Î0 ë 2û A particle of mass m and charge q has an initial velocity r r r v = v0 $j . If an electric field E = E0 i and magnetic field r B = B0iˆ act on the particle, its speed will double after a

(

25.

)

time:

26.

[7 Jan 2020, II] 2mv0 2mv0 3mv0 3mv0 (a) qE (b) qE (c) (d) qE0 qE0 0 0 A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by : [10 April 2019, II]

L 2p qE ö æ (a) çg+ ÷ m ø è 2p

(c)

L qE ö æ çg÷ m ø è

2p

(b)

(d)

1

R

(b) E µ

R (c) R (d) R 2 5 2 30. Two point charges q1 ( 10 mC) and q2 (– 25 mC) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, [9 Jan 2019, II] é ù 1 = 9 ´ 109 Nm 2 C -2 ú ê take 4p Î0 ë û (b)

(a) (63 ˆi – 27 ˆj ) × 102

(b) (– 63 ˆi + 27 ˆj ) × 102

(c) (81 ˆi – 81 ˆj ) × 102 (d) (–81 ˆi + 81 ˆj ) × 102 31. A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct? [Online April 15, 2018]

1 1 q2 E2 mw2 A2 + 2 2 k 2qE (b) The new equilibrium position is at a distance: k from x = 0 qE (c) The new equilibrium position is at a distance: 2k from x = 0 2 2 1 2 2 1q E (d) The total energy of the system is mw A – 2 2 k 32. A solid ball of radius R has a charge density r given by (a) The total energy of the system is

q2 E2 m2

the ball is:

L

(a)

æ qE ö g2 + ç ÷ è m ø

3

rö æ r = r0 ç1 - ÷ for 0 £ r £ R. The electric field outside è Rø

L g2 -

2p

29.

(a)

y

(c)

1 1 1 (c) E µ 4 (d) E µ 2 D D D D The bob of a simple pendulum has mass 2 g and a charge of 5.0 ¼C. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : [8 April 2019 I] (take g = 10 m/s2) (a) tan–1 (2.0) (b) tan –1 (0.2) (c) tan–1 (5.0) (d) tan –1 (0.5) For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is: [9 Jan. 2019 I]

(a) E µ 28.

(b) x

24.

27. Four point charges –q, +q, + q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D>> d, will behave as: [9 April 2019, II]

2

r0 R 3 e0 r 2

[Online April 15, 2018] (b)

4r0 R 3 3e 0 r 2

(c)

3r 0 R 3 4e 0 r 2

(d)

r0 R 3 12e0 r 2

P-239

Electric Charges and Fields

33.

34.

A long cylindrical shell carries positive surface charge s in the upper half and negative surface charge - s in the lower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic and not drawn to scale) [2015] (a)

(b)

(c)

(d)

C ield Lines

A wire of length L (=20 cm), is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges ± Q, [|Q| = 103e0 Coulomb where e0 is the permittivity (in SI units) of free space] the net electric field at the centre O of the semicircular arc would be : [Online April 11, 2015] Y

X

O

(a) (50 × 10 N/C) $j 3

35.

36.

O

(b) (50 × 103 N/C) $i

(c) (25 × 103 N/C) $j (d) (25 × 103 N/C) $i A thin disc of radius b = 2a has a concentric hole of radius ‘a’ in it (see figure). It carries uniform surface charge ‘s’ on it. If the electric field on its axis at height ‘h’ (h R , where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by [2010] r0 r æ 5 r ö 4pr0 r æ 5 r ö (a) ç - ÷ (b) 3e çè 3 - R ÷ø 4e 0 è 3 R ø 0

r0 r æ 5 r ö r0 r æ 5 r ö ç - ÷ (d) 3ε çè 4 - ÷ø R 4ε0 è 4 R ø 0 This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement-2 : The net work done by a conservative force on an object moving along a closed loop is zero. [2009] (a) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (b) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is false. Q Let r (r ) = r be the charge density distribution for p R4 a solid sphere of radius R and total charge Q. For a point ‘P’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is : [2009] 2 Q Qr1 (a) (b) 2 4p Î0 r1 4p Î0 R4

(c)

45.

(c)

46.

Qr12

r

E(r)

(d) 0 3p Î0 R4 A thin spherical shell of radus R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 £ r < ¥, where r is the distance from the centre of the shell? [2008]

O

r

R

(d) O

R

r

47. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is [2006] (a) 4 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4 48. Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is [2005]

(c)

44.

R

R

q

ˆj 4p e 0 r 4p e 0 r 2 q q ˆj ˆ (c) - 2 2 j (d) 2 2p e 0 r 2p e 0 r 2 Let there be a spherically symmetric charge distribution (a)

O

r

E(r)

i

O

E(r)

E(r)

L (b) 2 L (c) 4 L (d) 8 L 4 A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface charge density s of the sheet is proportional to [2005]

(a) 49.

P

q S

B (a) cot q (b) cos q (c) tan q (d) sin q 50. Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [2004] (a) -

(b)

Q (1 + 2 2) 4

Q Q (1 + 2 2) (1 + 2 2) (d) 4 2 A charged oil drop is suspended in a uniform field of 3×104 v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9×10–15 kg and g = 10 m/s2) [2004] (a) 1.6×10–18 C (b) 3.2×10–18 C (c) 3.3×10–18 C (d) 4.8×10–18 C

(c) -

51.

Q (1 + 2 2) 2

P-241

Electric Charges and Fields

(a) surface change density on the inner surface is uniform

Electric Dipole, Electric Flux TOPIC 3 and Gauss's Law 52.

Two identical electric point dipoles have dipole moments ®

®

P1 = P$i and P2 = - P$i and are held on the x axis at distance ‘a’ from each other. When released, they move along xaxis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is ‘m’, their speed when they are infinitely far apart is : [Sep. 06, 2020 (II)]

P 1 (a) a pe 0 ma

(c) 53.

P 1 (b) a 2pe 0 ma

P 2 a pe 0 ma

(d)

P 2 a 2 pe 0 ma

ur An electric field E = 4 xiˆ - ( y 2 + 1) ˆj N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as f1 and f11 respectively. The difference between (f1 – f11) is (in Nm2/C) _______. [9 Jan 2020, II] z A (0, 0, 2) (0, 2, 2)

54.

y

D

B

(3, 0, 2)

C

(3, 2, 2) E F x (0, 0, 0) (3, 0, 0) H G (0, 2, 0) (3, 2, 0)

In finding the electric field using Gauss law the formula r q | E | = enc is applicable. In the formula Î is 0 Î | A| 0

55.

and equal to

permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation? [8 Jan 2020, I] (a) Only when the Gaussian surface is an equipotential surface. Only when the Gaussian surface is an r (b) equipotential surface and | E | is constant on the surface. r (c) Only when | E | = constant on the surface. (d) For any choice of Gaussian surface. Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At ur its centre is a dipole p as shown. In this case : [12 April 2019, I]

56.

Q/2

4 pa 2 (b) electric field outside the shell is the same as that of a point charge at the centre of the shell. (c) surface charge density on the outer surface depends r on P (d) surface charge density on the inner surface of the shell is zero everywhere.Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of – Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then. [12 April 2019, II]

(a) a = 8–1/4 R

(b) a =

(c) a = 2–1/4 R

(d)

3R 21/ 4

a = R/ 3 57. An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency w is : [8 April 2019, II] qE qE qE 2qE (b) (c) 2 (d) md md 2md md 58. An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole? [11 Jan 2019, II] (a) –20 × 10–18 J (b) –7 × 10 –27 J (c) –10 × 10–29 J (d) – 9 × 10–20 J 59. Charges – q and + q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P¢ such that OP¢

(a)

æ yö æy ö = ç ÷ , the force on Q will be close to: ç >> 2a ÷ 3 3 è ø è ø [10 Jan 2019, II] P

Q P¢ A

(a) 3 F

O –q (b)

+q

F 3

(c) 9 F

B

(d) 27 F

P-242

60.

Physics

A charge Q is placed at a distance a/2 above the centre of the square surface of edge a as shown in the figure. The electric flux through the square surface is: [Online April 15, 2018]

Q (a) 3e 0

P a/2

Q (b) 6e 0 Q (c) 2e 0

61.

62.

a Q (d) e 0 ur An electric dipole has a fixed dipole moment p , which makes angle q with respect to x-axis. When subjected to uur ur an electric field E1 = Eiˆ , it experiences a torque T1 = t iˆ . uur When subjected to another electric field E2 = 3E1 ˆj it uur ur experiences torque T2 = -T1 . The angle q is : [2017] (a) 60° (b) 90° (c) 30° (d) 45° Four closed surfaces and corresponding charge distributions are shown below. [Online April 9, 2017]

5q

q

2q

q

S1

8q –2q –4q

–q q

q

q

S3

S2

Q b

(c) 64.

(

2

p a -b

Q 2 pa 2

2

)

(b)

(d)

(a) pa2 E

S4

a

2Q

45°

2Q pa 2

(

® E

3q

Let the respective electric fluxes through the surfaces be F 1, F 2, F 3, and F 4. Then : (a) F 1< F2 = F 3 > F4 (b) F 1> F2 > F 3 > F4 (c) F 1= F2 = F 3 = F4 (d) F 1> F 3 ; F 2 < F4 63. The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), have volume charge density A r= , where A is a constant and r is the distance from r the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is: [2016]

(a)

through a circular surface of radius 0.02 m parallel to the YZ plane is nearly: [Online April 19, 2014] (a) 0.125 Nm2/C (b) 0.02 Nm2/C 2 (c) 0.005 Nm /C (d) 3.14 Nm2/C ur ur 65. Two point dipoles of dipole moment p1 and p 2 are at a ur ur distance x from each other and p1 || p 2 . The force between the dipoles is : [Online April 9, 2013] 1 4 p1 p2 1 3 p1 p2 (a) (b) 4pe0 x 4 4pe0 x3 1 8 p1 p2 1 6 p1 p2 (c) (d) 4pe0 x 4 4pe0 x 4 66. The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field ® p E is applied at an angle with the vertical direction. The 4 electric flux through the curved surface of the hemisphere is [Online May 19, 2012]

(

)

The electric field in a region of space is given by, r E = Eoˆi + 2Eoˆj where Eo = 100 N/C. The flux of the field

)

2 (d) 2 2 2 2 67. An electric dipole is placed at an angle of 30° to a nonuniform electric field. The dipole will experience [2006] (a) a translational force only in the direction of the field (b) a translational force only in a direction normal to the direction of the field (c) a torque as well as a translational force (d) a torque only 68. If the electric flux entering and leaving an enclosed surface respectively is f1 and f2, the electric charge inside the surface will be [2003] (a) (f2 – f1)eo (b) (f1 – f2)/eo (c) (f2 – f1)/eo (d) (f1 – f2)eo 69. A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is [2002] E F

(c)

c

O

2 p b2 - a 2

2

( p + 2) pa 2 E

pa2 E

D

Q

pa2 E

(b)

H A

(a) q /4 p Î0 L (c) q/2 p Î0 L

q

q G B

L

(b) zero (d) q/3 p Î0 L

P-243

Electric Charges and Fields

y

d

1.

(a)

Fa

Fb +Q

q

d/2

5.

Force due to charge + Q, KQQ Fa = 2 d Force due to charge q, KQq Fb = 2 ædö ç ÷ è2ø For equilibrium, r r Fa + Fb = 0 Þ

kQQ d2

+

kQq

( d / 2)

2

F

+Q

d/2

(b)

Fnet =

= 0 \q = -

Q 4

A –2r/a e ( 4pr 2dr ) r2 0

æ ö R ç e –2r/a ÷ –2r/a = 4 pA ò e dr = 4 p A ç ÷ 0 ç –2 ÷ è a ø0 æ aö = 4pA ç – ÷ e –2R/a –1 è 2ø dr Q = 2paA(1–e–2R/a) æ ö r ç ÷ a 1 R = log ç ÷ 2 ç 1– Q ÷ 2paA ø è (d) Spheres A and B carry equal charge say 'q' kqq \ Force between them, F = 2 r q When A and C are touched, charge on both q A = q C = 2 Then when B and C are touched, charge on B q +q 3q qB = 2 = 2 4 Now, the force between charge qA and qB

4.

Þ

k´

F sin q

æqö 2kq ç ÷ è2ø æ y2 + a2 ö ç ÷ è ø

6.

kq2 y

2

×

y 2

y + a2

(Q y r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is : [Sep. 02, 2020 (II)]

[Sep. 03, 2020 (I)] (a) 4.5 mC on both (b) +4.5 mC and –4.5 mC (c) 3 mC and 6 mC (d) 6 mC and 3 mC Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is : [Sep. 03, 2020 (II)] (a)

3Q1 16pe 0 R

(b)

3Q2 4pe0 R

(c)

Q2 4pe0 R

(d)

3Q1 4pe0 R

6.

1/2

(a)

2 æ 4 q2 ö ç ÷ m è 15 4pe0a ø

1/2

7.

1/2

(b)

2 æ 1 q2 ö ç ÷ m è 5 4pe0a ø

1/2

2 æ 2 q2 ö 2 æ 1 q2 ö (c) (d) ç ÷ ç ÷ m è 15 4pe0a ø m è 15 4pe0a ø A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 4 Q, the new potential difference between the same two surfaces is : [8 April 2019 I] (a) – 2V (b) 2 V (c) 4 V (d) V

P-255

Electrostatic Potential and Capacitance

8.

9.

r The electric field in a region is given by E = ( Ax + B ) iˆ , where E is in NC–1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = –5 is V2, then V1 – V2 is : [8 Jan. 2019 II] (a) 320 V (b) – 48V (c) 180 V (d) – 520 V The given graph shows variation (with distance r from centre ) of : [11 Jan. 2019 I]

rO

r rO (a) Electric field of a uniformly charged sphere (b) Potential of a uniformly charged spherical shell (c) Potential of a uniformly charged sphere (d) Electric field of a uniformly charged spherical shell 10. A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be: [10 Jan. 2019 I] Q ab + bc + ca (a) 12pÎ abc 0

(b)

Q (a 2 + b 2 + c 2 ) 4pÎ0 (a 3 + b 3 + c3 )

Q (a + b + c) Q (c) 4pÎ (a + b + c) (d) 4pÎ (a 2 + b 2 + c2 ) 0 0 11. Two electric dipoles, A, B with respective dipole r r moments d A = – 4 qa iˆ and d B = – 2 qa iˆ are placed on the x-axis with a separation R, as shown in the figure

The distance from A at which both of them produce the same potential is: [10 Jan. 2019 I] R 2R (a) 2 + 1 (b) 2 +1 (c)

R

(d)

2R

2 -1 2 -1 12. Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2 = R1/R2. Then the ratio V1(on S1)/V2(on S2) of the electrostatic potentials on each sphere is: [8 Jan. 2019 II] (a) R1/R 2 (b) (R1/R2)2 3

æR ö (c) (R2/R1) (d) ç 1 ÷ è R2 ø 13. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +s, –s and +s respectively. The potential of shell B is: [2018]

s é a2 - b2 ù (a) Î ê a +c ú 0 êë ûú

s é a 2 - b2 ù (b) Î ê b +cú 0 êë ûú

s é b2 - c2 ù (c) Î ê b +a ú 0 ëê ûú

s é b 2 - c2 ù (d) Î ê c +a ú 0 ëê ûú

14. There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between in the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field ? [Online April 8, 2017] (a) 589.5 V (b) 589.2 V (c) 589.4 V (d) 589.6 V 15. Within a spherical charge distribution of charge density r(r), N equipotential surfaces of potential V0, V0 + DV, V0 + 2DV, .........V0 + NDV (DV > 0), are drawn and have increasing radii r0, r1, r2,......... rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and DV then : [Online April 10, 2016] 1 (a) r(r) = constant (b) r(r) µ 2 r 1 (c) r(r) µ (d) r(r) µ r r 16. The potential (in volts) of a charge distribution is given by V(z) = 30 – 5z2 for |z| £ 1m V(z) = 35 – 10 |z| for |z| ³ 1 m. V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume r0 (in units of e0) which is spread over a certain region, then choose the correct statement. [Online April 9, 2016] (a) r0 = 20 e0 in the entire region (b) r0 = 10 e0 for |z| £ 1 m and p0 = 0 elsewhere (c) r0 = 20 e0 for |z| £ 1 m and p0 = 0 elsewhere (d) r0 = 40 e0 in the entire region 17. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ¥) on its surface. For this sphere the equipotential surfaces with potentials 3V0 5V0 3V0 V0 , , and have radius R1, R2, R3 and R4 2 4 4 4 respectively. Then [2015] (a) R1 = 0 and R2 < (R4 – R3) (b) 2R = R4 (c) R1 = 0 and R2 > (R4 – R3) (d) R1 ¹ 0 and (R2 – R1) > (R4 – R3) r -1 exists in a region of $ 18. An electric field E = (25i$ + 30j)NC space. If the potential at the origin is taken to be zero then the potential at x = 2 m, y = 2 m is : [Online April 11, 2015] (a) –110 J (b) –140 J (c) –120 J (d) –130 J r 19. Assume that an electric field E = 30x 2 ˆi exists in space. Then the potential difference VA - VO , where VO is the potential at the origin and VA the potential at x = 2 m is: (a) 120 J/C (b) –120 J/C [2014] (c) –80 J/C (d) 80 J/C

P-256

Physics

20. Consider a finite insulated, uncharged conductor placed near a finite positively charged conductor. The uncharged body must have a potential : [Online April 23, 2013] (a) less than the charged conductor and more than at infinity. (b) more than the charged conductor and less than at infinity. (c) more than the charged conductor and more than at infinity. (d) less than the charged conductor and less than at infinity. 21. Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle q from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining them will æ 1 ö be ç = k÷. è 4p Î0 ø

[Online April 22, 2013]

(a) 2 k mg tan q

(b)

k mg tan q

(c) 4 k mg / tan q

(d)

k mg / tan q

(

)

1 ( R + r) Q (c) 4pe 0 R2 + r 2

(

)

( R + r) Q 1 (b) 4pe 0 2 R 2 + r 2

(

( R - r) Q 1 (d) 4pe 0 2 R 2 + r 2

(

q A

q B

D -q

C -q

ur E ur changes, V remains unchanged unchanged, V changes E remains ur both E and V change ur (d) E and V remain unchanged 28. The potential at a point x (measured in m m) due to some charges situated on the x-axis is given by V(x) = 20/(x2 – 4) volt. The electric field E at x = 4 m m is given by [2007] (a) (10/9) volt/ m m and in the +ve x direction (b) (5/3) volt/ m m and in the –ve x direction (c) (5/3) volt/ m m and in the +ve x direction (d) (10/9) volt/ m m and in the –ve x direction 29. Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and -q. The potential difference between the centres of the two rings is [2005] (a) (b) (c)

22. A point charge of magnitude + 1 mC is fixed at (0, 0, 0). An isolated uncharged spherical conductor, is fixed with its center at (4, 0, 0). The potential and the induced electric field at the centre of the sphere is :[Online April 22, 2013] (a) 1.8 × 105 V and – 5.625 × 106 V/m (b) 0 V and 0 V/m (c) 2.25 × 105 V and – 5.625 × 106 V/m (d) 2.25 × 105 V and 0 V/m 23. A charge of total amount Q is distributed over two concentric hollow spheres of radii r and R (R > r) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is [Online May 19, 2012]

1 ( R - r) Q (a) 4pe 0 R2 + r 2

26. An electric charge 10–3 m C is placed at the origin (0, 0) of X – Y co-ordinate system. Two points A and B are situated at ( 2, 2) and (2, 0) respectively. The potential difference between the points A and B will be [2007] (a) 4.5 volts (b) 9 volts (c) Zero (d) 2 volt 27. Charges are placed on the vertices of a square as shown. r Let E be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then [2007]

24. The electric potential V(x) in a region around the origin is given by V(x) = 4x2 volts. The electric charge enclosed in a cube of 1 m side with its centre at the origin is (in coulomb) [Online May 7, 2012] (a) 8e0 (b) – 4e0 (c) 0 (d) – 8e0 25. The electrostatic potential inside a charged spherical ball is given by f = ar2 + b where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is: [2011] (a) –6ae0r (b) –24pae0 (c) –6ae0 (d) –24pae0r

ù ú (b) R 2 + d 2 úû

q 2 p Î0

(c)

q é1 1 ê 2 4 p Î0 ê R R + d2 ë

) )

é1 ê êë R

(a)

1

ù ú (d) úû

qR 4p Î0 d 2

zero

30. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. R 2 [2003]

The electrostatic potential at a point P , a distance from the centre of the shell is 2Q (a) 4pe R o

2Q 2q (b) 4pe R - 4pe R o o

2Q q (c) 4pe R + 4pe R o o

(q + Q)2 (d) 4pe R o

P-257

Electrostatic Potential and Capacitance

Electric Potential Energy and TOPIC 2 Work Done in Carrying a Charge 31. A solid sphere of radius R carries a charge Q + q distributed uniformaly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed v when it has fallen through a vertical height y (see figure), then : (assume the remaining portion to be spherical). [Sep. 05, 2020 (I)]

Q

R

33. Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to : [Sep. 03, 2020 (II)] (a) 1 : 2 (b) 10 : 7 (c) 2 : 1 (d) 5 : 7 34. In free space, a particle A of charge 1 mC is held fixed at a point P. Another particle B of the same charge and mass 4 mg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is : é ù 1 = 9 ´ 109 Nm 2C -2 ú êTake [10 April 2019 II] 4pe 0 ë û (a) 1.0m/s (b) 3.0×104 m/s 3 (c) 2.0×10 m/s (d) 1.5×102 m/s 35. A system of three charges are placed as shown in the figure:

q y v

é ù qQ (a) v 2 = y ê + g ú 2 êë 4pe 0 R ym úû é ù qQ + gú (b) v 2 = y ê ë 4pe 0 R( R + y )m û é ù Qq R + gú (c) v 2 = 2 y ê 3 êë 4pe0 ( R + y ) m úû é ù qQ 2 + gú (d) v = 2 y ê 4 ( ) pe R R + y m ë 0 û 32. A two point charges 4q and –q are fixed on the x-axis at

If D >> d, the potential energy of the system is best given by [9 April 2019 I] (a)

1 é - q 2 - qQd ù 1 é - q 2 2qQd ù + ê ú ê 2 ú (b) 4p Î0 ë d 2 D û 4p Î0 ë d D2 û

(c)

1 é q 2 qQd ù 1 é q 2 qQd ù + 2 ú (d) - 2 ú ê+ ê4p Î0 ë d 4p Î0 ë d D û D û

36. A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to : [8 April 2019 II]

d d and x = , respectively. If a third point charge ge 2 2 ‘q’ is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will : [Sep. 04, 2020 (I)] x=-

4q

(a) increase by

3q 2 4pe0 d

(b) increase by

2q 2 3pe0 d

(c) decrease by

q2 4pe0 d

(d) decrease by

4q 2 3pe0 d

–q

+r /r (a) v µ e 0

(b) v µ

ærö ln ç ÷ è r0 ø

ærö ærö (c) v µ ln ç ÷ (d) v µ ç ÷ r è 0ø è r0 ø 37. There is a uniform spherically symmetric surface charge density at a distance Ro from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R(t) is: [12 Jan. 2019 I]

P-258

Physics V(R(t))

V(R(t))

(a)

(b) R (t)

Ro

Ro

R (t)

V(R(t))

V(R(t)) Vo

(c)

(d) Ro

R (t)

Ro

R (t)

38. Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is : [11 Jan. 2019 I] Q

+q

+q

- 2q (b) 2 +1

(a) + q

-q (d) –2q (c) 1+ 2 39. Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, – 2) and (0, – 2). The work required to put a fifth charge Q at the origin of the coordinate system will be: [10 Jan. 2019 II] (a)

Q2 4 pe 0

(c)

Q2 2 2 pe0

1 ö æ ç1 + ÷ 3ø è

(b)

Q2 4 pe 0

(d)

Q2 4pe 0

1 ö æ ç1 + ÷ 5ø è

40. Statement 1 : No work is required to be done to move a test charge between any two points on an equipotential surface. Statement 2 : Electric lines of force at the equipotential surfaces are mutually perpendicular to each other. [Online April 25, 2013] (a) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1. (b) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1. (c) Statement 1 is true, Statement 2 is false. (d) Statement 1 is false, Statement 2 is true.

41. An insulating solid sphere of radius R has a uniformly positive charge density r. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinite is zero. [2012] Statement -1 When a charge q is taken from the centre to the surface of the sphere its potential energy changes qr . by 3e0 Statement -2 The electric field at a distance r (r