33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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WavesTelegram @unacademyplusdiscounts12179. A whistle revolves in a circle with angular speedw = 20 rad/s using a string of length 50 cm. If thefrequency of sound from the whistle is 385 Hz,then what is the minimum frequency heard by anobserver which is far away from the centre (velocityof sound = 340 m/s)(a) 385 Hz (b) 374 Hz(c) 394 Hz (d) 333 Hz (2002)80. Two stationary sources each emitting waves ofwavelength l, an observer moves from one source toanother with velocity u. Then number of beats heardby him2u u(a) (b) (c) uλ (d)u(2000)λ λ2λ 81. A vehicle, with a horn of frequency n is moving witha velocity of 30 m/s in a direction perpendicularto the straight line joining the observer and thevehicle. The observer perceives the sound to have afrequency n + n 1 . Then (if the sound velocity in air is300 m/s)(a) n 1 = 0.1n (b) n 1 = 0(c) n 1 = 10n (d) n 1 = –0.1n (1998)82. Two trains move towards each other withthe same speed. The speed of sound is340 m/s. If the height of the tone of the whistle ofone of them heard on the other changes to 9/8 times,then the speed of each train should be(a) 20 m/s(b) 2 m/s(c) 200 m/s (d) 2000 m/s (1991)ANSWER KEY1. (c) 2. (b) 3. (c) 4. (a) 5. (c) 6. (a) 7. (a) 8. (c) 9. (c) 10. (c)11. (a) 12. (a, b) 13. (d) 14. (b) 15. (d) 16. (c) 17. (b) 18. (c) 19. (c) 20. (a)21. (c) 22. (c) 23. (d) 24. (c) 25. (c) 26. (c) 27. (a) 28. (c) 29. (b) 30. (b)31. (b) 32. (b) 33. (c) 34. (b) 35. (a) 36. (a) 37. (b) 38. (a) 39. (b) 40. (a)41. (d) 42. (b) 43. (d) 44. (c) 45. (b) 46. (c) 47. (a) 48. (c) 49. (a) 50. (d)51. (a) 52. (a) 53. (a) 54. (c) 55. (d) 56 (b) 57. (d) 58. (c) 59. (d) 60. (b)61. (d) 62. (a) 63. (b) 64. (a) 65. (c) 66. (a) 67. (a) 68. (c) 69. (b) 70. (c)71. (a) 72. (d) 73. (c) 74. (a) 75. (c) 76. (b) 77. (b) 78. (b) 79. (b) 80. (a)81. (b) 82. (a)Hints & Explanations1. (c) : Light waves are electromagnetic waves. Lightwaves are transverse in nature and do not require amedium to travel, hence they can travel in vacuum. Soundwaves are longitudinal waves and require a medium totravel. They do not travel in vacuum.2. (b) : With the propagation of a longitudinal wave,energy alone is propagated.3. (c) : The standard equation of a wave travellingalong +ve x-direction is given byy = Asin(kx – wt)where A = Amplitude of the wavek = angular wave numberw = angular frequency of the wave1Given: A = 1m, λ= 2πm,υ = Hz π2π2π1As k = = = 1 ; ω= 2πυ= 2π× = 2λ 2ππ\ The equation of the given wave isy = 1 sin (1x – 2t) = sin(x – 2t)4. (a) : y 1 = asin(wt + kx + 0.57)\ Phase, f 1 = wt + kx + 0.57⎛ π ⎞y2= acos( ωt+ kx) = asint + kx +⎝⎜ω2 ⎠⎟π∴ Phase, φ2= ωt+ kx +2Phase difference, Df = f 2 – f 1⎛ π ⎞π= + +⎝⎜ωt kx⎠⎟ − ( ωt + kx + 057 . ) = − 057 .22= (1.57 – 0.57) radian = 1 radian5. (c) : Amplitude = 2 cm = 0.02 m, v = 128 m/s4λ= = 08 . m; υ= 128 = 160 Hz50.82π2πω= 2πυ= 2π× 160 = 1005 ; k = = = 785 .λ 08 .\ y = (0.02)m sin(7.85x – 1005t)6. (a) : y = 0.25sin(10px – 2pt)2πy max = 0.25, k = = 10π ⇒ λ = 02 . mλw = 2pf = 2p ⇒ f = 1 Hz
Telegram @unacademyplusdiscounts122 NEET-AIPMT Chapterwise Topicwise Solutions PhysicsThe sign is negative inside the bracket. Therefore thiswave travels in the positive x-direction.7. (a) : y 1 = 10 –6 sin[100t + (x/50) + 0.5]y 2 = 10 –6 cos[100t + (x/50)]= 10 –6 sin[100t + (x/50) + p/2]= 10 –6 sin[100t + (x/50) + 1.57]The phase difference = 1.57 – 0.5 = 1.07 radians8. (c) : The equation of progressive wave travelling inpositive x-direction is given by2πy= asin ( vt −x)λHere a = 0.2 m, v = 360 m/s, l = 60 m∴2π⎡ ⎛ x ⎞ ⎤y= 02 . sin ( t− x = ⎢ −⎝⎜ t⎠⎟60 360 ) 02 . sin⎥⎣π 6 60 ⎦9. (c) : Displacement, y max = a, y min = 0Time taken = T/4\ T/4 = 0.170 \ T = 0.68The frequency of wave = 1/T = 1.47 Hz10. (c) : Sound wave equation isy = 0.0015 sin (62.4 x + 316 t)Comparing it with the general equation of motion⎡ x t ⎤ 2πy= Asin 2π ⎢ +⎣ T⎥ , we get = 62.4λ ⎦ λ2πor λ = = 01 . unit62.4π11. (a) : Phasedifferenceθ= 60°=radPhasedifference() θ32π= × Pathdifferenceλπ λ λThereforePathdifference = × =3 2π6 .12. (a, b) : Option (a) represents a harmonic progressivewave in the standard form whereas (b) also represents aharmonic progressive wave, both travelling in the positivex- direction. In (b), a is the angular velocity, (w) and b isk; c is the initial phase. (d) represents only S.H.M.13. (d) : Compare with the equation,y = acos(2put + f)This give 2pu = 2000; υ = 1000 Hzπ14. (b) : The standard equation of a progressive wave is⎡ ⎛ t x ⎞y= a ⎢ −⎝⎜T ⎠⎟⎣+ ⎤sin 2π φλ⎥⎦The given equation can be written as⎡ ⎛ t x ⎞y = ⎢ −⎝⎜⎠⎟⎣+ π⎤4sin2π10 18 6⎥⎦\ a = 4 cm, T = 10 s, l = 18 cm and f = p/6.15. (d) : Wavelength of pulse at the lower end,λ1 ∝ velocity( v1)=Similarly, λ2 ∝ v2=λ2T2∴ = =λ1T1T1µT2µ( m1+m2)g=mg 2m1+m2m216. (c) : Since 4.0 g of a gas occupies 22.4 litres at NTP,so the molecular mass of the gas is M = 4.0 g mol –1RTAs the speed of the sound in the gas is v = γMwhere g is the ratio of two specific heats, R is the universalgas constant and T is the temperature of the gas.2Mv∴ γ =RTHere, M = 4.0 g mol –1 = 4.0 × 10 –3 kg mol –1 , v = 952 m s –1 ,R = 8.3 J K –1 mol –1 andT = 273 K (at NTP)−3 −1 −12(. 40×10 kg mol )( 952 ms )∴ γ == 16 .−1 −1(. 83JK mol )( 273 K)CpBy definition, γ= or Cp= γCvCvBut g = 1.6 and C v = 5.0 J K –1 mol –1\ C p = (1.6)(5.0 J K –1 mol –1 ) = 8.0 J K –1 mol –117. (b) : The given wave equation isy= 3 sin π ( t−x)2 50⎛ π ⎞y= 3sint−x⎝⎜25π2 ⎠⎟...(i)The standard wave equation isy = Asin(wt – kx) ...(ii)Comparing (i) and (ii), we getπω= 25π, k =2ω 25π−Wave velocity, v = = = 50 ms 1k ( π / 2)dy dParticle velocity, vp = sin t xdt= ⎛ ⎛dt ⎝⎜ − π ⎞⎞⎠⎟⎝⎜3 25π2 ⎠⎟⎛ π ⎞= 75πcos −⎝⎜25πt x2 ⎠⎟Maximum particle velocity, (v p ) max = 75p m s –1( vp)max 75π3∴ = = πv 50 218. (c) : Here, v air = 350 m/s, v brass = 3500 m/sWhen a sound wave travels from one medium to anothermedium its frequency remains the same
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Waves
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79. A whistle revolves in a circle with angular speed
w = 20 rad/s using a string of length 50 cm. If the
frequency of sound from the whistle is 385 Hz,
then what is the minimum frequency heard by an
observer which is far away from the centre (velocity
of sound = 340 m/s)
(a) 385 Hz (b) 374 Hz
(c) 394 Hz (d) 333 Hz (2002)
80. Two stationary sources each emitting waves of
wavelength l, an observer moves from one source to
another with velocity u. Then number of beats heard
by him
2u u
(a) (b) (c) uλ (d)
u
(2000)
λ λ
2λ
81. A vehicle, with a horn of frequency n is moving with
a velocity of 30 m/s in a direction perpendicular
to the straight line joining the observer and the
vehicle. The observer perceives the sound to have a
frequency n + n 1 . Then (if the sound velocity in air is
300 m/s)
(a) n 1 = 0.1n (b) n 1 = 0
(c) n 1 = 10n (d) n 1 = –0.1n (1998)
82. Two trains move towards each other with
the same speed. The speed of sound is
340 m/s. If the height of the tone of the whistle of
one of them heard on the other changes to 9/8 times,
then the speed of each train should be
(a) 20 m/s
(b) 2 m/s
(c) 200 m/s (d) 2000 m/s (1991)
ANSWER KEY
1. (c) 2. (b) 3. (c) 4. (a) 5. (c) 6. (a) 7. (a) 8. (c) 9. (c) 10. (c)
11. (a) 12. (a, b) 13. (d) 14. (b) 15. (d) 16. (c) 17. (b) 18. (c) 19. (c) 20. (a)
21. (c) 22. (c) 23. (d) 24. (c) 25. (c) 26. (c) 27. (a) 28. (c) 29. (b) 30. (b)
31. (b) 32. (b) 33. (c) 34. (b) 35. (a) 36. (a) 37. (b) 38. (a) 39. (b) 40. (a)
41. (d) 42. (b) 43. (d) 44. (c) 45. (b) 46. (c) 47. (a) 48. (c) 49. (a) 50. (d)
51. (a) 52. (a) 53. (a) 54. (c) 55. (d) 56 (b) 57. (d) 58. (c) 59. (d) 60. (b)
61. (d) 62. (a) 63. (b) 64. (a) 65. (c) 66. (a) 67. (a) 68. (c) 69. (b) 70. (c)
71. (a) 72. (d) 73. (c) 74. (a) 75. (c) 76. (b) 77. (b) 78. (b) 79. (b) 80. (a)
81. (b) 82. (a)
Hints & Explanations
1. (c) : Light waves are electromagnetic waves. Light
waves are transverse in nature and do not require a
medium to travel, hence they can travel in vacuum. Sound
waves are longitudinal waves and require a medium to
travel. They do not travel in vacuum.
2. (b) : With the propagation of a longitudinal wave,
energy alone is propagated.
3. (c) : The standard equation of a wave travelling
along +ve x-direction is given by
y = Asin(kx – wt)
where A = Amplitude of the wave
k = angular wave number
w = angular frequency of the wave
1
Given: A = 1m, λ= 2πm,
υ = Hz π
2π
2π
1
As k = = = 1 ; ω= 2πυ= 2π× = 2
λ 2π
π
\ The equation of the given wave is
y = 1 sin (1x – 2t) = sin(x – 2t)
4. (a) : y 1 = asin(wt + kx + 0.57)
\ Phase, f 1 = wt + kx + 0.57
⎛ π ⎞
y2
= acos( ωt+ kx) = asin
t + kx +
⎝
⎜ω
2 ⎠
⎟
π
∴ Phase, φ2
= ωt
+ kx +
2
Phase difference, Df = f 2 – f 1
⎛ π ⎞
π
= + +
⎝
⎜ωt kx
⎠
⎟ − ( ωt + kx + 057 . ) = − 057 .
2
2
= (1.57 – 0.57) radian = 1 radian
5. (c) : Amplitude = 2 cm = 0.02 m, v = 128 m/s
4
λ= = 08 . m; υ= 128 = 160 Hz
5
0.8
2π
2π
ω= 2πυ= 2π× 160 = 1005 ; k = = = 785 .
λ 08 .
\ y = (0.02)m sin(7.85x – 1005t)
6. (a) : y = 0.25sin(10px – 2pt)
2π
y max = 0.25, k = = 10π ⇒ λ = 02 . m
λ
w = 2pf = 2p ⇒ f = 1 Hz
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