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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Two loudspeakers `L_1` and `L_2` driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is `330ms^-1` then the frequency at which the first maximum is observed isA. 165 HzB. 330 HzC. 496 HzD. 660 Hz |
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Answer» Correct Answer - B |
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| 52. |
Two identical straight wires are stretched so as to produce ` 6 beats//s` when vibrating simultaneously . On changing the tension slightly in one of them , the beats frequency remains unchanged . If `T_(1) and T_(2)` are initial tensions in strings such that `T_(1) gt T_(2)` then it may be said while making above changes in tension :A. `T_(2)` was decreasedB. `T_(1)` was increasedC. both `T_(1) and T_(2)` were increasedD. either `T_(2)` was increased or `T_(1)` was decreased |
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Answer» Correct Answer - D Eithere , frequency of first wire should decrease or frequency of second wire should increase. |
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| 53. |
A standing wave can be produced by combiningA. two longitudinal travelling wavesB. two transverse travelling wavesC. two sinusoidal travelling waves travelling in opposite directionsD. all of the above |
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Answer» Correct Answer - D Sound waves in an organ pipe (which are standing in nature) is an example of superposition of two longitudinal travelling waves. Standing waves on a string is an example of superposition of two transverse travelling waves on a string travelling in opposite directions. |
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| 54. |
Regarding an open organ pipe , which of the following is correct?A. Both the ends are pressure antinodesB. Both the ends are displacement nodesC. Both the ends are pressure nodesD. Both (a) and (b) |
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Answer» Correct Answer - C In an open organ pipe , both the ends are free ends ,hence both are displacement antinodes and hence pressure nodes. |
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| 55. |
When two sound waves with a phase difference of ` pi //2`, and each having amplitude A and frequency `omega` , are superimposed on each other, then the maximum amplitude and frequency of resultant wave is |
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Answer» Correct Answer - `sqrt(2)A: omega` |
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| 56. |
The displacement of a particle is given by ` x = 3 sin ( 5 pi t) + 4 cos ( 5 pi t)`. The amplitude of particle isA. `3`B. `4`C. `5`D. `7` |
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Answer» Correct Answer - C Standard equation : `x = a sin omega t + b cos omega t` ` x = sqrt (a^(2) + b^(2)) sin ( omega t + tan^(-1) (b//a))` Given equation ` x = 3 sin ( 5 pi t) + 4 cos ( 5 pi t)` ` x = sqrt ( 9 + 16) sin ( 5 pi t + tan^(-1) ( 4//3))` |
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| 57. |
In a resonance column experiment , the first resonance is obtained when the level of the water in the tube is at `20 cm` from the open end . Resonance will also be obtained when the water level is at a distance ofA. `40 cm` from the open endB. `60 cm` from the open endC. `80 cm` from the open endD. `100 cm` from the open end |
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Answer» Correct Answer - B For second resonance `L_(2) = ( 3lambda)/(4) = 3L_(1) = 3 xx 20 = 60 cm` |
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| 58. |
A radio transmitter at position `A` operates at a wavelength of `20 m`. A second , identical transmitter is located at a distance `x` from the first transmitter , at position `B`. The transmitters are phase locked together such that the second transmitter is lagging `pi//2` out of phase with the first . For which of the following values of `BC - CA` will the intensity at `C` be maximum . A. `BC - CA = 60 m`B. `BC - CA = 65 m`C. `BC - CA = 55 m`D. `BC - CA = 75 m` |
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Answer» Correct Answer - C::D `y_(A) = A sin [ omega t - k (AC)]` `y_(B) = sin [ omega t - (pi)/(2) - k (BC)]` For maximum intensity at `C` `k (BC - AC) + (pi)/(2) = 2 n pi` `BC - AC = ( n lambda - (lambda)/(4)) = 15 , 35 , 55 , 75 ,….` |
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| 59. |
A source of sound attached to the bob of a simple pendulum execute `SHM` . The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean frequency of the source . The velocity of the source at the mean position is ( velocity of sound in the air is `340 m//s`) [Assume velocity of sound `lt lt` velocity of sound in air ]A. `1.4 m//s`B. `3.4 m//s`C. `1.7 m//s`D. `2.1 m//s` |
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Answer» Correct Answer - B ` f_(1) = f_(0) ((V_(0))/(V_(0) - V)) f_(2) = f_(0)((V_(0))/(V_(0) + V))` `f_(1) - f_(2) = f_(0) V_(0) ((1)/( V_(0) - V) - (1)/( V_(0) + V))` ` = f_(0) V_(0) (V_(0) + V - V_(0) + V)/( V_(0)^(2) - V^(2)) = f_(0) V_(0) xx ( 2V)/( V_(0)^(2)) = f_(0) ( 2V)/( V_(0))` given `( 2 Vf_(0))/(V_(0)) = 0.02 xx f_(0) rArr V = 0.01 V_(0) = 3.4 m//s`. |
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| 60. |
Two waves having equaitons `x_1=asin(omegat+phi_1)`,`x=asin(omega+phi_2)` If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them isA. `(pi)/(6)`B. `(2pi)/(3)`C. `(pi)/(4)`D. `(pi)/(3)` |
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Answer» Correct Answer - B |
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| 61. |
In a stationary wave all the particlesA. on either side of a node vibrate in the same phaseB. in the region between two nodes vibrate in the same phaseC. in the region between two antinodes vibrate in the same phaseD. of the medium vibrate in the same phase |
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Answer» Correct Answer - B |
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| 62. |
When a stationary wave is formed, then its frequency isA. same as that of the individual wavesB. twice that of the individual wavesC. half that of the individual wavesD. none of the above |
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Answer» Correct Answer - A |
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| 63. |
If the ratio of amplitude of two waves is `4:3`, then the ratio of maximum and minimum intensity is |
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Answer» Correct Answer - `49 : 1` |
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| 64. |
An unknown frequency `x` produces 8 beats per seconds with a freuquency of `250` Hz and 12 beats with `270Hz`. Source then `x` isA. 285 HzB. 242 HzC. 262 HzD. 282 Hz |
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Answer» Correct Answer - A |
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| 65. |
The diagram below shows two pulses traveling towards each other in a uniform medium with same speed . Pulses in the figure are at the same distance from `X` and has same height & width. Which diagram best represents the medium when the pulses meet at point `X`? A. B. C. D. |
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Answer» Correct Answer - D Apply superposition . |
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| 66. |
Two pulses travel in mutually opposite directions in a string with a speed of `2.5 cm//s` as shown in the figure. Initially the pulses are `10 cm` apart. What will be the state of the string after two seconds?A. B. C. D. |
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Answer» Correct Answer - C |
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| 67. |
Two wave function in a medium along x direction are given by `y_(1)=1/(2+(2x-3t)^(2))m`, `y_(2)=-1/(2+(2x+3t-6)^(2))m` Where x is in meters and t is in secondsA. at `x=(3)/(2)m` the resultant displacement will be zero at all times.B. at t = 1 s which resultant displacement will be zero everywhere.C. Both waves travel along the same direction.D. Both waves travel in the opposite directions. |
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Answer» Correct Answer - A::B::D |
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| 68. |
An open pipe resonates with a tuning fork of frequency `500 Hz` . It is observed that two successive notes are formed at distance `16 and 46 cm` from the open end. The speed of sound in air in the pipe isA. `230 m//s`B. `300 m//s`C. `320 m//s`D. `360 m//s` |
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Answer» Correct Answer - B ` (lambda)/(2) = 46 - 16 rArr (lambda)/(2) = 30 cm` or `lambda = 60 cm` `:. v = lambda f = (60)/(100) xx 500 = 300 m//s` |
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| 69. |
An ideal organ pipe resonates at successive frequencies of `50 Hz , 150 Hz , 250 Hz` , etc. ( speed of sound ` = 340 m//s`) The pipe isA. Open at both ends and of length `3.4 m`B. Open at both ends and of length `6.8 m`C. Closed at one end , open at the other , and of length `1.7 m`D. Closed at one end , open at the other , and of length `3.4 m` |
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Answer» Correct Answer - C The frequencies given are odd multiple of fundamental . Hence close organ pipe `50 = (1)/( 4 l) xx 340 rArr l = 1.7 m` |
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| 70. |
The breaking stress of steel is `7.85 xx 10^(8) N//m^(2)` and density of steel is `7.7 xx 10^(3) kg//m^(3)`. The maximum frequency to which a string `1 m` long can be tuned isA. `15.8 Hz`B. `158 Hz`C. `47.4 Hz`D. `474 Hz` |
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Answer» Correct Answer - B Maximum frequency `v = (1)/( 2l) sqrt((T)/( mu)) = (1)/( 2l) sqrt((T)/( A rho))` ` = (1)/(2) sqrt((7.85 xx 10^(8))/(7.7 xx 10^(3))) = 158 Hz` |
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| 71. |
In the arrangement shown in Fig. 7.100 , mass can be hung from a string with a linear mass density of `2 xx 10^(-3) kg//m` that passes over a light pulley . The string is connected to a vibrator of frequency `700 Hz` and the length of the string between the vibrator and the pulley is `1 m`. The string is set into vibrations and represented by the equation ` y = 6 sin (( pi x)/( 10)) cm cos (14 xx 10^(3) pi t)` where `x` ane `y` are in `cm` , and `t` in `s`, the maximum displacement at `x = 5 m` from the vibrator isA. `6 cm`B. `3 cm`C. `5 cm`D. `2 cm` |
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Answer» Correct Answer - A `A(x) = 6 sin ( pi x)/(10)` Maximum displacement at `x = 5 cm` `A(x) = 6 sin ((pi)/(10) xx 5 ) = 6 cm` |
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| 72. |
In the arrangement shown in Fig. 7.100 , mass can be hung from a string with a linear mass density of `2 xx 10^(-3) kg//m` that passes over a light pulley . The string is connected to a vibrator of frequency `700 Hz` and the length of the string between the vibrator and the pulley is `1 m`. If the standing waves are observed , the largest mass to be hung isA. `16 kg`B. `25 kg`C. `32 kg`D. `400 kg` |
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Answer» Correct Answer - D For largest mass , `p = 1` `n = (p)/( 2L) sqrt((T)/(mu))` `700 = (p)/( 2L) sqrt((T)/(mu))` ` m = 2 xx 10^(-3) kg//m , L = 1 m` `T = [ ( 700)^(2) xx 4 xx 1 xx 2 xx 10^(-3)] = 3920 N` Largest mass to be hang , `M_(max) = 3920//9.8 = 400 kg` |
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| 73. |
In the arrangement shown in Fig. 7.100 , mass can be hung from a string with a linear mass density of `2 xx 10^(-3) kg//m` that passes over a light pulley . The string is connected to a vibrator of frequency `700 Hz` and the length of the string between the vibrator and the pulley is `1 m`. If the mass suspended is `16 kg` , then the number of loops formed in the string isA. `1`B. `3`C. `5`D. `8` |
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Answer» Correct Answer - C Number of loops formed `P = 2nL sqrt((mu)/(T)) = 2 xx 700 xx 1 sqrt((2 xx 10^(-3))/(16 xx 9.8))` ` = ( 2xx 700)/( 280) = 5` |
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| 74. |
A long tube contains air at a pressure of `1 atm` and a temperature of `107^(@) C`. The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of `500 Hz`. Resonance is produced when the piston is at distance `19 , 58.5 and 98 cm` from the open end. The speed of sound at `10^(@) C` isA. `330 m//s`B. `340 m//s`C. `395 m//s`D. `495 m//s` |
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Answer» Correct Answer - C Velocity of sound at `107^(@) C = 2 n (l_(2) - l_(1)) = 2 xx 500(58.5 - 19) xx 10^(-2)` `= 395 m//s` |
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| 75. |
The displacement `y` of a particle periodic motion is given by `y = 4 cos ((1)/(2) t) sin (1000 t)` This expression may be considered as a result of the superposition ofA. twoB. threeC. fourD. five |
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Answer» Correct Answer - B `y =(4)/(2) [ 2 cos ^(2) ((t)/(2)) ] sin (1000 t)` or ` y = 2 ( 1 + cos t) sin 1000 t` or t = 2 sin 1000 t + 2 sin 1000 t cos t` or ` y = 2 sin 1000 t + sin (1001 t) + sin (999 t)` So , the given the expression is a result of the superposition of three independent harmonic motions . |
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| 76. |
Ten tuning forks are arranged in increasing order of frequency is such a way that any two nearest tuning forks produce `4 beast//sec`. The highest freqeuncy is twice of the lowest. Possible highest and the lowest frequencies areA. 80 and 40B. 100 and 50C. 44 and 22D. 72 and 36 |
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Answer» Correct Answer - D |
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| 77. |
If two tuning fork A and B are sounded together they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will beA. 250B. 252C. 260D. 262 |
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Answer» Correct Answer - B |
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| 78. |
Two tuning forks `A and B` give ` 4 beats//s` when sounded together . The frequency of `A is 320 Hz`. When some wax is added to `B` and it is sounded with `A , 4 beats//s per second` are again heard . The frequency of `B` isA. `312 Hz`B. `316 Hz`C. `324 Hz`D. `328 Hz` |
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Answer» Correct Answer - C Since there is no change in beats . Therefore the original frequency of `B` is ` n_(2) = n_(1) + x = 320 + 4 = 324` |
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| 79. |
On sounding fork `A` with another tuning fork `B` of frequency `384 Hz , 6 beats` are produced per second .After loading the prongs of `A` with wax and then sounding it again with `B , 4 beats` are produced per second. What is the frequency of the tuning fork `A`.A. `388 Hz`B. `80 Hz`C. `378 Hz`D. `390 Hz` |
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Answer» Correct Answer - D Probable frequency of `A is 390 Hz and 378 Hz ` after loading the beats are decreasing from ` 6 "to" 4` so the original frequency of `A` will `390 Hz`. |
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| 80. |
Two tuning forks have frequencies 380 and 384 Hz respectively. When they are sounded together they produce 4 beats. After hearing the maximum sound how long will it take to hear the minimum soundA. `(1)/(2)sec`B. `(1)/(4)sec`C. `(1)/(8)sec`D. `(1)/(16)sec` |
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Answer» Correct Answer - C |
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| 81. |
A `75 cm` string fixed at both ends produces resonant frequencies `384 Hz and 288 Hz` without there being any other resonant frequency between these two . Wave speed for the string isA. `144 m//s`B. `216 m//s`C. `108 m//s`D. `72 m//s` |
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Answer» Correct Answer - A `Delta v = 384 - 288 = 96` Thus ` 288 and 384 (96 xx 3 , 96 xx 4)` are third and fourth harmonics . For fundamental mode : `(lambda)/(2) = 0.75` `lambda = 1.5` `eta = 96` `v = 96 xx 1.5 = 144 m//s` |
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| 82. |
Consider a standing wave formed on a string . It results due to the superposition of two waves travelling in opposite directions . The waves are travelling along the length of the string in the `x` - direction and displacements of elements on the string are along the `y` - direction . Individual equations of the two waves can be expressed as `Y_(1) = 6 (cm) sin [ 5 (rad//cm) x - 4 ( rad//s)t]` `Y_(2) = 6(cm) sin [ 5 (rad//cm)x + 4 (rad//s)t]` Here `x` and `y` are in `cm`. Answer the following questions. If one end of the string is at `x = 0` , positions of the nodes can be described asA. `x = n pi//5 cm , where n = 0 , 1, 2,…`B. `x = n 2pi//5 cm , where n = 0 , 1, 2,…`C. `x = n pi//5 cm , where n = 0 , 1, 3, 5,…`D. `x = n pi//10 cm , where n = 0 , 1, 3,5,…` |
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Answer» Correct Answer - A `y = y_(1) + y_(2) = ( 12 sin 5 x) cos 4 t` Maximum value of `y` - position in `SHM`of an element of the string that is located at an antinode `= +- 12 cm ( sin 5 x = +- 1)` For the position nodes amplitudeshould be zero. So , `sin 5 x = 0 rArr 5 x = n pi` `x = ( n pi)/(5)` where `n = 0 , 1 ,2 , 3 ,.....` Value of amplitude at `x = 1.8 cm` `A = 12 sin ( 5 xx 1.8) = 4.9 cm` At any instant say `t = 0` , instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero. |
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| 83. |
Consider a standing wave formed on a string . It results due to the superposition of two waves travelling in opposite directions . The waves are travelling along the length of the string in the `x` - direction and displacements of elements on the string are along the `y` - direction . Individual equations of the two waves can be expressed as `Y_(1) = 6 (cm) sin [ 5 (rad//cm) x - 4 ( rad//s)t]` `Y_(2) = 6(cm) sin [ 5 (rad//cm)x + 4 (rad//s)t]` Here `x` and `y` are in `cm`. Answer the following questions. Figure 7.104( c) shows the standing wave pattern at `t = 0` due to superposition of waves given by `y_(1)` and `y_(2)` in Figs.7.104(a) and (b) . In Fig. 7.104 (c ) , `N` is a node and `A` and antinode . At this instant say ` t = 0` , instantaneous velocity of points on the string A. is different for different pointsB. is zero for all pointsC. changes with position of the pointD. is constant but not equal to zero for all points |
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Answer» Correct Answer - B `y = y_(1) + y_(2) = ( 12 sin 5 x) cos 4 t` Maximum value of `y` - position in `SHM`of an element of the string that is located at an antinode `= +- 12 cm ( sin 5 x = +- 1)` For the position nodes amplitudeshould be zero. So , `sin 5 x = 0 rArr 5 x = n pi` `x = ( n pi)/(5)` where `n = 0 , 1 ,2 , 3 ,.....` Value of amplitude at `x = 1.8 cm` `A = 12 sin ( 5 xx 1.8) = 4.9 cm` At any instant say `t = 0` , instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero. |
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| 84. |
A standing wave pattern is formed on a string One of the waves if given by equation `y_1=acos(omegat-kx+(pi)/(3))` then the equation of the other wave such that at `x=0` a node is formed.A. `y_(2) = a sin (omegat + kx +(pi)/(3))`B. `y_(2) = a sin (omegat + kx +(pi)/(3))`C. `y_(2) = a cos (omegat + kx +(2pi)/(3))`D. `y_(2) = a cos (omegat + kx +(4pi)/(3))` |
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Answer» Correct Answer - D |
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| 85. |
Consider a standing wave formed on a string . It results due to the superposition of two waves travelling in opposite directions . The waves are travelling along the length of the string in the `x` - direction and displacements of elements on the string are along the `y` - direction . Individual equations of the two waves can be expressed as `Y_(1) = 6 (cm) sin [ 5 (rad//cm) x - 4 ( rad//s)t]` `Y_(2) = 6(cm) sin [ 5 (rad//cm)x + 4 (rad//s)t]` Here `x` and `y` are in `cm`. Answer the following questions. Maximum value of the `y` - positions coordinate in the simple harmonic motion of an element of the string that is located at an antinode will beA. `+- 6 cm`B. `+- 8 cm`C. `+- 12 cm`D. `+- 3 cm` |
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Answer» Correct Answer - C `y = y_(1) + y_(2) = ( 12 sin 5 x) cos 4 t` Maximum value of `y` - position in `SHM`of an element of the string that is located at an antinode `= +- 12 cm ( sin 5 x = +- 1)` For the position nodes amplitudeshould be zero. So , `sin 5 x = 0 rArr 5 x = n pi` `x = ( n pi)/(5)` where `n = 0 , 1 ,2 , 3 ,.....` Value of amplitude at `x = 1.8 cm` `A = 12 sin ( 5 xx 1.8) = 4.9 cm` At any instant say `t = 0` , instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero. |
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| 86. |
Consider a standing wave formed on a string . It results due to the superposition of two waves travelling in opposite directions . The waves are travelling along the length of the string in the `x` - direction and displacements of elements on the string are along the `y` - direction . Individual equations of the two waves can be expressed as `Y_(1) = 6 (cm) sin [ 5 (rad//cm) x - 4 ( rad//s)t]` `Y_(2) = 6(cm) sin [ 5 (rad//cm)x + 4 (rad//s)t]` Here `x` and `y` are in `cm`. Answer the following questions. Amplitude of simple harmonic motion of a point on the string that is located at `x = 1.8 cm` will beA. `3.3 cm`B. `6.7 cm`C. `4.9 cm`D. `2.6 cm` |
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Answer» Correct Answer - C `y = y_(1) + y_(2) = ( 12 sin 5 x) cos 4 t` Maximum value of `y` - position in `SHM`of an element of the string that is located at an antinode `= +- 12 cm ( sin 5 x = +- 1)` For the position nodes amplitudeshould be zero. So , `sin 5 x = 0 rArr 5 x = n pi` `x = ( n pi)/(5)` where `n = 0 , 1 ,2 , 3 ,.....` Value of amplitude at `x = 1.8 cm` `A = 12 sin ( 5 xx 1.8) = 4.9 cm` At any instant say `t = 0` , instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero. |
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| 87. |
Statement I : In standing waves on a string , the medium particles , i.e., different striung elements remain at rest . Statement II : In standing waves all the medium particles attain maximum velocity twice in one cycle.A. Statement I is true , Statement II is true , Statement II is a correct explanation for Statement I.B. Statement I is true , Statement II is true , Statement II is NOT a correct explanation for Statement I.C. Statement I is true , Statement II is false.D. Statement I : is false , Statement II is true. |
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Answer» Correct Answer - D In standing waves the medium particles are oscillating and hence are not at rest , though few particles present at the location of node remain at rest. |
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| 88. |
Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below `f_(0) = 1250 Hz`. The length of the pipe is `l = 85 cm`. The velocity of sound is `v = 340 m//s`. Consider two cases the pipe is closed from one end .A. `2`B. `4`C. `8`D. `6` |
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Answer» Correct Answer - D Pipe is closed from one end : An air column in a pipe closed from one end oscillates only odd harmonics [ `I st` harmonic ( fundamental mode) , `3 rd` harmonic ( `Ist` overtone) , `5 th` harmonic ( `2 nd` overtone) , `7 th` harmonic ( `3 rd` overtone) etc.] Fundamental frequency ` = ( V)/( 4l) = ( 340)/( 4 xx 85/100) = 100 Hz` Other modes of oscillation are ` 3 rd` harmonic frequency ` = 3 xx 100 = 300 Hz` `5 th` harmonic frequency `= 5 xx 100 = 500 Hz` `7 th` harmonic frequency ` = 7 xx 100 = 700 Hz` `9 th` harmonic frequency ` = 9 xx 100 = 900 Hz` `11 th` harmonic frequency ` = 11 xx 100 = 1100 Hz` `13 th` harmonic frequency ` = 13 xx 100 = 1300 Hz` Only those natural oscillations are to be counted whose frequencies lie below `f_(0) = 1250 Hz` , the harmonics till `11 th` harmonic are to be counted . Since , the number of piossible natural oscillations `= 1 ( i st "harmonic") + 1( 3 rd "harmonic") + 1( 5 th "harmonic") + 1( 7 th "harmonic") + 1( 9 th "harmonic") + 1 ( 11 th "harmonic") = 6`. Second Method All the frequencies possible are integral multiples of fundamental frequency which is `100 Hz`. Using the fact that integer which is multiplied by fundamental frequency is the number of harmonic itself you get , highest predicted ` = [ 12.50//100]` where `[ x]` represents greatest integer less than or equal to `x = [ 12.5] = 12`. Now for closed pipe , only odd harmonics are possible , highest harmonic possible ` = 11 th` . Th epossible harmonics are `1 , 3 , 5 ,5 , 7 , 9 , 11` where are six in number. |
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| 89. |
Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below `f_(0) = 1250 Hz`. The length of the pipe is `l = 85 cm`. The velocity of sound is `v = 340 m//s`. Consider two cases the pipe is opened from both ends.A. `3`B. `7`C. `6`D. `9` |
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Answer» Correct Answer - C Pipe opened from both ends Fundamental frequency `= (V)/( 2 l) = (340)/(2 xx 85) xx 100 = 200 Hz` Frequency of the other natural modes of oscillational are : `2 nd` harmonic frequency ` = 2 xx 200 = 400 Hz ` `3 rd` harmonic frequency ` = 3 xx 200 = 600 Hz` `4 th` harmonic frequency ` = 4 xx 200 = 800 Hz` `5 th` harmonic frequency ` = 6 xx 200 = 1200 Hz` `7 th` harmonic frequency `= 7 xx 200 = 1400 Hz` You have to count only those harmonics whose frequencies are below `1250 Hz`. All the harmonics till `6 th` harmonic are possible , and obviously they are six in number . Second Method Fundamental frequency ` = 200 Hz` ` = n xx` fundamental frequency `= n xx 200` `[ n is 1 , 2, ....]` Maximum value of `n = [ 12.50 //200] = 6 ([x]` represents greatest less than or equal to `x` ). Now `n` is also equal to the number of harmonic for which frequency is being calculated , highest harmonic possible ` = 6 th`. A all harmonics are possible in case of open tube , harmonics possible are `1 st , 2 nd , 3 rd , 4 th, 5 th , 6 th. ` Number of harmonics possible in this case `= 6`. |
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| 90. |
If the sound waves produced by the tuning fork can be expressed as `y = 0.2 (cm) sin (kx - omega t)` , where `K = 2 pi//lambda` and `omega = 2 pi f (f = 512 Hz)`, maximum value of amplitude in a beat will beA. `0.4 cm`B. `0.6 cm`C. `0.8 cm`D. `0.2 cm` |
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Answer» Correct Answer - A Waves expressed by tuningfork `y = 0.2 sin ( kx - omega t)` Maximum value of amplitude of beat is `2 A` `y = 2 xx 0.2 = 0.4 cm` |
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| 91. |
Following are equations of four waves : (i) `y_(1) = a sin omega ( t - (x)/(v))` (ii) `y_(2) = a cos omega ( t + (x)/(v))` (iii) `z_(1) = a sin omega ( t - (x)/(v))` (iv) `z_(1) = a cos omega ( t + (x)/(v))` Which of the following statements are correct ?A. On superposition of waves (i) and (iii) , a travelling wave having amplitude `a sqrt(2)` will be formedB. Superposition of waves (ii) and (iii) is not possibleC. On superposition of waves (i) and (ii) , a travelling wave having amplitude `a sqrt(2)` will be formedD. On superposition of (iii) and (iv) , a transverse stationary wave will be formed |
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Answer» Correct Answer - A::D New wave ` = y_(1) hat(j) + z_(1) hat(k)` ` = ( a hat(j) + a hat(k)) sin omega(( t - x)/(v))` Amplitude ` = | a hat(j) + a hat(k)| = a sqrt(2)` (i) and (ii) are travelling in opposite directions , so they will form stationary waves. Similarly (iii) and (iv) will make the stationary wave. |
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| 92. |
Consider the three waves `z_(1), z_(2) "and" z_(3)` as `z_(1) = A "sin"(kx - omega t)` `z_(2) = A "sin"(kx + omega t)` `z_(3) = A "sin"(ky - omega t)` Which of the following represents a standing wave?A. `z_(1) + z_(2)`B. `z_(2) + z_(3)`C. `z_(3) + z_(1)`D. `z_(1) + z_(2) + z_(3)` |
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Answer» Correct Answer - A |
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| 93. |
Which two of the given transverse waves will give stationary wave when get superimposed? `z_(1) = a cos (kx-omegat) " "(A)` `z_(2) = a cos (kx +omegat) " "(B)` `z_(3) = a cos (ky - omegat) " "(C )`A. A and BB. A and CC. B and CD. Any two |
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Answer» Correct Answer - A |
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| 94. |
The following equations represent progressive transverse waves `z_(1) = A cos ( omega t - kx)` `z_(2) = A cos ( omega t + kx)` `z_(3) = A cos ( omega t + ky)` `z_(4) = A cos (2 omega t - 2 ky)` A stationary wave will be formed by superposingA. `z_(1) and z_(2)`B. `z_(1) and z_(4)`C. `z_(2) and z_(3)`D. `z_(3) and z_(4)` |
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Answer» Correct Answer - A The direction of wave must be opposite and frequencies will be same then by superposition , standing wave formation takes place. |
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| 95. |
Two vibrating tuning forks produce progressive waves given by , `y_(1) = 4 sin (500 pi t) and y_(2) = 2 sin (506 pi t)`. These tuning forks are held near the ear of person . The person will hearA. ` 3 beats//s` with intensity ratio between maxima and minima equal to `2`B. ` 3 beats//s` with intensity ratio between maxima and minima equal to `9`C. ` 6 beats//s` with intensity ratio between maxima and minima equal to `2`D. ` 6 beats//s` with intensity ratio between maxima and minima equal to `9` |
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Answer» Correct Answer - B `v_(1) = 250 Hz , v_(2) = 253 Hz , v_(2) - v_(1) = 3` Now , `(( a_(1) + a_(2))^( 2))/((a_(1) - a_(2))^(2)) = (( 4 + 2)^(2))/(( 4 - 2)^(2)) = (36)/(4) = 9` |
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| 96. |
A sound increases its decibel reading from `20 "to" 40 dB`. This means that the intensity of the soundA. is doubledB. is `20 `times greaterC. is `100` times greaterD. is the old intensity `20` |
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Answer» Correct Answer - C The decibel scale is logarithmic `dB = 10 log(I//I_(0))`. Each increase in intensity by a power of ten increases the decibel reading by `10 units`. Hence , to increase the decibel reading by `20` ,there should be an increase in the intensity of `10 xx10 = 100`. |
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| 97. |
When beats are produced by two progressive waves of nearly the same frequency, which one of the following if correct?A. The particle vibrates simple harmonically , with the frequency equal to the difference in the component frequenciesB. The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two wavesC. The frequency of beats depends upon the position , where the observer isD. The frequency of beats changes at the time progresses |
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Answer» Correct Answer - B As ` y = A_(b) sin ( 2 pi n_(av) t)` where `A_(b) = 2 A cos ( 2 pi n_(A) t)` ltwhere `n_(A) = (n_(1) - n_(2))/(2)` |
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| 98. |
A string is fixed at both end transverse oscillations with amplitude `a_(0)` are excited. Which of the following statements are correct ?A. (a) Energy of oscillations in the string is directly proportional to tension in the stringB. (b) Energy of oscillations in nth overtone will be equal to `n^(2)` times of that in first overtoneC. ( c) Average kinetic energy of string (over an oscillation period) is half of the oscillation energyD. (d) None of the above |
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Answer» Correct Answer - A::C If a string of length `l` has cross - sectional area `A`, density of its material `rho` then its oscillation energy is given by `E = pi^(2) A rho a_(0)^(2) l f^(2)` where `f` is frequency of transverse stationary wave formed in the string . But `f = (v)/(lambda) = (1)/(lambda) sqrt((T)/(m))` where `lambda` is wavelength , `T` is tension in the string and `m = A rho`. Since , string has a fixed length , therefore , wavelength of a tone excited in the string is constant . Hence , energy `E prop T`. Therefore , option `(a)` is correct . If the frequency of fundamental tone is `f_(0)`, then the frequency on `n th ` overtone will be equal to `( n + 1) f_(0)`. Hence , oscillation energy of the string will be equal to : ` E_(n) = pi^(2) A rho a_(0)^(2) l f_(0)^(2) ( n + 1)^(2)` Since `E_(n)` is not directly proportional to `n^(2)` , therefore , option `(b)` is wrong . Since , every particle of the string performs `SHM`, therefore , `r.m.s.` speed of a particle ` = 1// sqrt(2) xx` its maximum speed Hence , average `KE` is half of maximum `KE`. But maximum `KE` is equal to oscillation energy of the string . therefore , option `(c )` is correct. |
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| 99. |
The minimum intensity of audibility of sound is `10^(-12) W//m^(2) s` and density of air ` = 1.293 kg//m^(3)`. If the frequency of sound in `1000 Hz` , then the corresponding amplitude the vibration of the air particles is [ Take velocity of sound `= 332 m//s`]A. `1.1 xx 10^(-7) m`B. `1.1 xx 10^(-9) m`C. `1.1 xx 10^(-11) m`D. `1.1 xx 10^(-14) m` |
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Answer» Correct Answer - C `I = 2 pi ^(2) a^(2) v^(2) rho v` `a^(2) = (I)/( 2 pi^(2) v^(2) rho v) or a = (1)/( pi v) sqrt((I)/( 2 rho v))` or ` a = (7)/( 22 xx 1000) sqrt((10^(-12))/( 2 xx 1.293 xx 332)) m` or ` a = 1.1 xx 10^(-11) m` |
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| 100. |
Two waves of nearly same amplitude , same frequency travelling with same velocity are superimposing to give phenomenon of interference . If `a_(1)` and `a_(2)` be their respectively amplitudes , `omega` be the frequency for both , `v` be the velocity for both and `Delta phi` is the phase difference between the two waves then ,A. the resultant intensity varies periodically with time and distance.B. the resulting intensity with `(I_(min))/(I_(max)) = (a_(1) - a_(2))/(a_(1) + a_(2))^(2)` is obtained.C. both the waves must have been travelling in the same direction and must be coherent.D. `I_(R) = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos ( Delta phi)`, where constructive interference is obtained for path difference that are odd multiple of `1//2 lambda` and destructive interference is obtained for path difference that are even multiple of `1//2 lambda`. |
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Answer» Correct Answer - B::C This is the case of sustained interference in which position of maxima and minima remains fixed all over the screen. `(I_(min))/(I_(max)) = (( a_(1) - a_(2))/( a_(1) + a_(2)))^(2) ` And both waves must have beeen travelling in the same direction with a constant phase difference ( condition for coherence) |
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