33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

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Telegram @unacademyplusdiscounts102 NEET-AIPMT Chapterwise Topicwise Solutions PhysicsANSWER KEY1. (d) 2. (a) 3. (c) 4. (c) 5. (b) 6. (a) 7. (a) 8. (b) 9. (b) 10. (b)11. (d) 12. (c) 13. (b) 14. (b) 15. (a) 16. (d) 17. (c) 18. (d) 19. (b) 20. (b)21. (c) 22. (a) 23. (a) 24. (a) 25. (d) 26. (d) 27. (a) 28. (a,d) 29. (d) 30. (b)31. (c) 32. (d) 33. (b) 34. (a) 35. (b) 36. (b) 37. (b) 38. (c) 39. (a) 40. (b)41. (c) 42. (d) 43. (d) 44. (a) 45. (a) 46. (d) 47. (b) 48. (d) 49. (b) 50. (a)51. (a) 52. (a) 53. (a) 54. (d) 55. (a) 56. (c) 57. (*) 58. (a) 59. (b) 60. (a)61. (a) 62. (b) 63. (d) 64. (c) 65. (a) 66. (a) 67. (a) 68. (c) 69. (a) 70. (d)71. (b) 72. (c) 73. (d) 74. (d) 75. (a) 76. (*) 77. (c) 78. (d) 79. (a) 80. (a)81. (c) 82. (c) 83. (a) 84. (d) 85. (a) 86. (a) 87. (a)Hints & Explanations1. (d) : When a beam of cathode rays (or electrons) aresubjected to crossed electric (E) and magnetic (B) fields,the beam is not deflected, ifForce on electron due to magnetic field = Force onelectron due to electric fieldEBev = eE or v = ...(i)BIf V is the potential difference between the anode and thecathode, then21 2 e vmv = eV or = ...(ii)2m 2VSubstituting the value of v from equation (i) in equation(ii),we get2e E=m 22VB2e ESpecific charge of the cathode rays =m 22VB2. (a) : Collisions of the charged particles with theatoms in the gas.3. (c) 4. (c)5. (b) : Cathode rays are basically negatively chargedparticles (electrons). If the cathode rays are allowed topass between two plates kept at a difference of potential,the rays are found to be deflected from the rectilinearpath. The direction of deflection shows that the rayscarry negative charges.6. (a) 7. (a)8. (b) : Thermionic emission : When a metal is heatedto a high temperature, the free electron gain kineticenergy and escape from the surface of the metal.Secondary emission : When an electron strikes thesurface of a metallic plate, it emits other electrons fromthe surface.Photoelectric emission : Emission of electrons from themetal surface on irradiation with radiation of suitablefrequency.X-rays emission : They are due to transitions in the innerenergy levels of the atom.9. (b) : When a metal is heated, electrons are ejectedout of it, which are called thermions.10. (b) : By changing the position of source of lightfrom photo cell, there will be a change in the intensity oflight falling on photo cell.As stopping potential is independent of the intensityof the incident light, hence stopping potential remainssame i.e., V 0 .11. (d) : The photoelectic emission occurs only whenthe incident light has more than a certain minimumfrequency. This minimum frequency is called thresholdfrequency.12. (c) : The stopping potential V s is related to themaximum kinetic energy of the emitted electrons K maxthrough the relationK max = eV s0.5 eV = eV s or V s = 0.5 V13. (b) : The number of photoelectrons ejected isdirectly proportional to the intensity of incident light.Maximum kinetic energy is independent of intensity ofincident light but depends upon the frequency of light.Hence option (b) is correct.14. (b) : The number of photoelectrons decide thephotocurrent. Assuming that the number of electronsemitted depends on the number of photons incident,the number of photoelectrons depend on the intensity oflight.15. (a)16. (d) : For a light source of power P watt, the intensityPat a distance d is given by I =42πdwhere we assume light to spread out uniformly in alldirections i.e., it is a spherical source.

Telegram @unacademyplusdiscountsDual Nature of Radiation and Matter1031 I1d2 11∴ I ∝ or =h( 5υ 2d I2d1 0)= hυ0 + mv2 ⇒ 4hυ0 = mv2 ...(ii)2221 vI11I1I1or,= ⎛ 1 2 v11or, 4 or, I2.I2⎝ ⎜ ⎞Dividing (i)by(ii), = or05 . ⎠⎟ = =422 2 = .v v2I2428. (a, d) : The maximum kinetic energy is given asIn a photoelectric emission, the number of photoelectronshc hcliberated per second from a photosensitive metallic Kmax h 0 = h h 0 = λ − λsurface is proportional to the intensity of the light. When0where la intensity of source is reduced by a factor of four, the0 = threshold wavelengthnumber of photoelectrons is also reduced by a factor of 4. or 1 2 hc hcmv = −2 λ λ17. (c) : The photoelectric current is directly0proportional to the intensity of illumination. ThereforeHere, h = 4.14 × 10 –15 eV s, c = 3 × 10 8 m s –1a change in the intensity of the incident radiation will l o = 3250 × 10 –10 m = 3250 Åchange the photocurrent.l = 2536 × 10 –10 m = 2536 Å,1equation,2in the first case isE= W0+ mv2hcKmax = −φWhen frequency of incident light is 2u 0 .1 0λ11and that in the second case ish( 2υ0)= hυ0 + mv1 ⇒ hυ0 = mv1 ...(i)hc 2hc22K = − φ0 = −φWhen frequency of incident light is 5u λ0λ0m = 9.1 × 1018. (d) : Photoelectric current I ∝ intensity of light andkghc = 4.14 × 101eV s × 3 × 10 8 m s –1 = 12420 eV Åintensity ∝122 ⎡ 1 1 ⎤(distance)∴ mv = 12420 ⎢ − ⎥ eV=1.076 eV2⎣25363250⎦1∴ I ∝−1922 2. 152 eV 2. 152 × 16 . × 10(distance)v = =19. (b) 20. (b)m−3191 . × 1021. (c) : If the intensity of light of a given frequency \ v ≈ 6 × 10 5 m s –1 = 0.6 × 10 6 m s –1is increased, then the number of photons striking the Note: Options (a) and (d) are same. So both are correct.surface per second will increase in the same ratio. Thisincreased number of photons strikes more electrons ofmetals and hence number of photoelectrons emitted29. (d) : According to Einstein’s photoelectric equationmaximum kinetic energy of photoelectrons,KE max = E v – fthrough the surface increase and hence photoelectric or 2 = 5 – f \ f = 3 eVcurrent increases.When E v = 6 eV then, KE max = 6 – 3 = 3 eV22. (a) : Since the emission of photoelectrons is directly or e(V cathode – V anode ) = 3 eVproportional to the intensity of the incident light, or V cathode – V anode = 3 V = –V stoppingtherefore photocurrent increases with the intensity of \ V stopping = –3 Vlight.30. (b) : According to Einstein’s photoelectric equation,23. (a) : Photoelectric current is directly proportionalhc hceVs to the intensity of incident light.λ − λ 0hc hc24. (a) : The work function has no effect on photoelectric ∴ As perquestion, eV = − ...(i)current so long as hu > W 0 . The photoelectric current isλ λ 0proportional to the intensity of incident light. Since there eV hc hc= −is no change in the intensity of light, hence I 1 = I 2 .4 2λ λ ...(ii)025. (d) : Initially, u = 1.5 u 0From equations (i) and (ii), we getIf the frequency is halved, u′ = υ 15υ0= . hc hc hc hc< u − = −02 22λ 4λ λ0 4λ0Hence, no photoelectric emission will take place.hc 3hc⇒ = or λ0= 3λ26. (d) : Required wavelength of light,4λ4λ0λ0 = 1240 eV nm31. (c) : Let f 0 be the work function of the surface of≈ 310 nmthe material.φ 4 eVAccording to Einstein’s photoelectric equation, the27. (a) : According to the Einstein’s photoelectric maximum kinetic energy of the emitted photoelectronsmax 22

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102 NEET-AIPMT Chapterwise Topicwise Solutions Physics

ANSWER KEY

1. (d) 2. (a) 3. (c) 4. (c) 5. (b) 6. (a) 7. (a) 8. (b) 9. (b) 10. (b)

11. (d) 12. (c) 13. (b) 14. (b) 15. (a) 16. (d) 17. (c) 18. (d) 19. (b) 20. (b)

21. (c) 22. (a) 23. (a) 24. (a) 25. (d) 26. (d) 27. (a) 28. (a,d) 29. (d) 30. (b)

31. (c) 32. (d) 33. (b) 34. (a) 35. (b) 36. (b) 37. (b) 38. (c) 39. (a) 40. (b)

41. (c) 42. (d) 43. (d) 44. (a) 45. (a) 46. (d) 47. (b) 48. (d) 49. (b) 50. (a)

51. (a) 52. (a) 53. (a) 54. (d) 55. (a) 56. (c) 57. (*) 58. (a) 59. (b) 60. (a)

61. (a) 62. (b) 63. (d) 64. (c) 65. (a) 66. (a) 67. (a) 68. (c) 69. (a) 70. (d)

71. (b) 72. (c) 73. (d) 74. (d) 75. (a) 76. (*) 77. (c) 78. (d) 79. (a) 80. (a)

81. (c) 82. (c) 83. (a) 84. (d) 85. (a) 86. (a) 87. (a)

Hints & Explanations

1. (d) : When a beam of cathode rays (or electrons) are

subjected to crossed electric (E) and magnetic (B) fields,

the beam is not deflected, if

Force on electron due to magnetic field = Force on

electron due to electric field

E

Bev = eE or v = ...(i)

B

If V is the potential difference between the anode and the

cathode, then

2

1 2 e v

mv = eV or = ...(ii)

2

m 2V

Substituting the value of v from equation (i) in equation

(ii),we get

2

e E

=

m 2

2VB

2

e E

Specific charge of the cathode rays =

m 2

2VB

2. (a) : Collisions of the charged particles with the

atoms in the gas.

3. (c) 4. (c)

5. (b) : Cathode rays are basically negatively charged

particles (electrons). If the cathode rays are allowed to

pass between two plates kept at a difference of potential,

the rays are found to be deflected from the rectilinear

path. The direction of deflection shows that the rays

carry negative charges.

6. (a) 7. (a)

8. (b) : Thermionic emission : When a metal is heated

to a high temperature, the free electron gain kinetic

energy and escape from the surface of the metal.

Secondary emission : When an electron strikes the

surface of a metallic plate, it emits other electrons from

the surface.

Photoelectric emission : Emission of electrons from the

metal surface on irradiation with radiation of suitable

frequency.

X-rays emission : They are due to transitions in the inner

energy levels of the atom.

9. (b) : When a metal is heated, electrons are ejected

out of it, which are called thermions.

10. (b) : By changing the position of source of light

from photo cell, there will be a change in the intensity of

light falling on photo cell.

As stopping potential is independent of the intensity

of the incident light, hence stopping potential remains

same i.e., V 0 .

11. (d) : The photoelectic emission occurs only when

the incident light has more than a certain minimum

frequency. This minimum frequency is called threshold

frequency.

12. (c) : The stopping potential V s is related to the

maximum kinetic energy of the emitted electrons K max

through the relation

K max = eV s

0.5 eV = eV s or V s = 0.5 V

13. (b) : The number of photoelectrons ejected is

directly proportional to the intensity of incident light.

Maximum kinetic energy is independent of intensity of

incident light but depends upon the frequency of light.

Hence option (b) is correct.

14. (b) : The number of photoelectrons decide the

photocurrent. Assuming that the number of electrons

emitted depends on the number of photons incident,

the number of photoelectrons depend on the intensity of

light.

15. (a)

16. (d) : For a light source of power P watt, the intensity

P

at a distance d is given by I =

4

2

πd

where we assume light to spread out uniformly in all

directions i.e., it is a spherical source.

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