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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. |
The room temperature charges by a small amount from `T` to `T+DeltaT` (on Kelvin scale). The fundamental frequency of organ pipe changes from `f` to `f+Deltaf`, thenA. `(DeltaT)/(T)`B. `1/2(DeltaT)/(T)`C. `2(DeltaT)/(T)`D. `(-DeltaT)/(T)` |
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Answer» Correct Answer - B `v prop sqrtT " and " n prop sqrtT` `(n+Deltan)/(Deltan) = sqrt((T+DeltaT)/(T))` `(n+Deltan)/(Deltan) = (1+(DeltaT)/(T))^(1/2)` `1+(Deltan)/(n) = 1+1/2 (DeltaT)/(T)` `(Deltan)/(n) = 1/2 (DeltaT)/(T)` |
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| 52. |
Length of 0.3 m wire vibrates in fundamental mode at frequency 480Hz. Then the velocity of sound wave in air isA. 330m/sB. 320m/sC. 300m/sD. 288m/s |
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Answer» Correct Answer - D `v=2nl` `=2xx480xx0.3 = 6xx48 = 288` |
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| 53. |
An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximumA. The two endsB. Middle of pipeC. Distance l/4 inside the endsD. The distance l/8 inside the ends |
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Answer» Correct Answer - B The pressure variation is maximum at the middle of the tube at which node is formed. |
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| 54. |
Two organ pipes, each closed at one end, give 5 beats `s^(-1)` when emitting their fundamental notes. If their lengths are in the ratio `50 : 51, ` their fundamental frequencies areA. 250, 255B. 255, 260C. 260, 265D. 265, 270 |
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Answer» Correct Answer - A `n_1l_1= n_2l_2 therefore n_1/n_2 = 51/50` `therefore n_1 = 51/50n_2, n_1-n_2 =5` `therefore 51/50 n_2 -n_2 = 5` `n_2 (51-50) = 5xx50=250Hz` `n_1 = n_2 + 5 =250+5 = 255Hz` |
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| 55. |
Two organ pipes, each closed at one end, give 5 beats `s^(-1)` when emitting their fundamental notes. If their lengths are in the ratio `50 : 51, ` their fundamental frequencies areA. 250, 255B. 255, 250C. 260, 265D. 265, 270 |
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Answer» Correct Answer - b As, `n_(1)/n_(2)=l_(2)/l_(1)=51/50and n_1-n_(2)=5` On solving, we get `n_(2)=250, n_(1) = 255` |
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| 56. |
The displacement of the wave given by equation `y(x, t)=a sin (kx - omega t +phi), ` where`phi=0` at point x and t = 0 is same as that at pointA. `x+2npi`B. `x+(2npi)/k`C. `kx + 2npi`D. Both (a) and (b) |
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Answer» Correct Answer - b `y(x,0)=a sin kx = asin (kx+2npi)` `=a sin k (x+ (2npi)/k)` `rArr` The displacement at points x and `(x+ (2npi)/k)` are the same, where `n= 1, 2, 3,…` |
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| 57. |
The two individual wave functions are `y_(1)=(5" cm")sin (4x-t)and y_(2)=(5" cm") sin (4x + t)` ltbrrgt where, x and y are in centimetres. Find out the maximum displacement of the motion at ` x = 2.0` cm.A. `9.89` cmB. 8 cmC. `8.39` cmD. `7.69` cm |
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Answer» Correct Answer - a Given, `y_(1)=(5" cm")sin (4x-t)and y_(2)=(5" cm") sin (4x + t)` We know `y=(2A sin kx )cos omegat` Compare `y_(1)=(5" cm") sin (4x-t)` with `y_(1) = A sin (kx-omegat)` `A=5, k=4 and omega = 1` `Now, " y"_("max") = 10 sin 4x|_(x=2.0,` `10 xx sin (4xx2) = 10 sin (8xx(180^(@))/pi)` `rArr " y"_("max")=9.89" cm"` |
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| 58. |
The tension of a stretched string is increased by 69%. In order to keep its frequency of vibration constant, its length must be increased by :A. 0.2B. 0.3C. `sqrt(69)%`D. 0.69 |
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Answer» Correct Answer - B `l_1/l_2 = sqrt(T_2/T_1) = sqrt(169/100) = 13/10` `l_1/l_2 = 1.3` `(l_1-l_2)/(l_2) = 1.3 - 1 0.3` `(Deltal)/(l) = 30%` |
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| 59. |
The maximum or thershold wavelength of transverse stationary wave on a string of length l isA. lB. 2lC. `l/2`D. 4l |
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Answer» Correct Answer - B `lamda/2= l` therefore lamda = 2l` |
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| 60. |
A sonometer is plucked at 1/4 of its length . The most prominent would beA. EightB. FourthC. ThirdD. Second |
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Answer» Correct Answer - D When sonometer wire is plucked at `1/4l` then second mode of vibrations are observed. |
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| 61. |
The velocity of the component wave isA. 50 m/sB. 100m/sC. 500 m/sD. 250m/s |
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Answer» Correct Answer - A `k=(2pi)/(k) = 4pi " and " omega = 2pin = 200pi` `therefore lamda = 0.5 therefore n=100Hz` `v=nlamda = 100 xx0.5 = 50` m/s |
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| 62. |
In the previous Q. The wavelength of the component progressive wave isA. 6cmB. 3cmC. 12cmD. 40cm |
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Answer» Correct Answer - A `k=pi/3` `(2pi)/(lamda) = pi/3 therefore lamda = 6cm` |
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| 63. |
Equation of a plane progressive wave is given by `y=0.6 sin 2pi(t-(x)/(2)).` On reflection from a denser medium, its amplitude becomes `2//3` of the amplitude of the incident wave. The equation of the reflected wave isA. `y= 0.6 sin 2pi(t+x/2).`B. `y= -0.6 sin 2pi(t+x/2).`C. `y= 0.4 sin 2pi(t+x/2).`D. `y= -0.4 sin 2pi(t-x/2).` |
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Answer» Correct Answer - b On reflection from a denser medium, there is a phase reversal of `180^(@)` New amplitude `0.6` `therefore` Equation of reflected wave is ltbegt `y= 0.6 sin 2pi [ t +x/2 +180^(@)]=-0.6 sin 2pi (t+x//2)` |
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| 64. |
In the Q.89 the frequency of the component progressive wave isA. 20HzB. 40HzC. 1/20HzD. 1/40Hz |
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Answer» Correct Answer - A `omega = 40pi therefore 2pin=40pi` `therefore n=20Hz` |
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| 65. |
The equation of a vibrating string is `y=0.01cos4pixxsin200pit` Where all quantities are expressed in SI units. Then the amplitude of component wave isA. 0.05mB. 0.005mC. 0.1mD. 0.01m |
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Answer» Correct Answer - B `y=0.01cos4pixsin200pit` `therefore R=0.01cos4pix` . `R=2Acos4pix` `therefore 2A = 0.01 therefore A=0.005m` |
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| 66. |
The value of amplitude at an antinode , on an undamped one dimensional stationary wave of wavelength 1.2 m, is 5mm. Then the value of amplitude of particle at a distance of 15cm from an antinode isA. 3.535 mmB. 0.707 mmC. 5 mmD. 7.07 mm |
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Answer» Correct Answer - A `R=2Acos (2pix)/(lamda)` `=5xx10^(-3) cos(2pixx0.5)/(1.2)` `=5xx10^(-3)cospi/4=3.535`mm |
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| 67. |
The value of amplitude at an antinode , on an undamped one dimensional stationary wave of wavelength 1.2 m, is 5mm. Then the value of amplitude of particle at a distance of 10cm from an node isA. 5mmB. 7.07mmC. 2.5mmD. 1mm |
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Answer» Correct Answer - C `R=2Asin (2pix)/(lamda)` `=5xx10^(-3)xxsin(2pixx10^(-1))/(1.2)` `=5xx10^(-3)sin pi/6=5xx10^(-3)xx1/2` =2.5mm |
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| 68. |
A piano wire having a diameter of `0.90 mm` is replaced by another wire of the same material but with a diameter of `0.93 mm`.If the tension of the wire is kept the same , then the percentage change in the frequency of the fundamental tone isA. 0.03B. 0.003C. `+3.2%`D. `-3.26%` |
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Answer» Correct Answer - D `d_1=0.90mm, d_2 = 0.93 mm` `n_1 prop 1/d_1 " and " n_2 prop 1/d_2` `(n_2)/(n_1) - 1= (d_1)/(d_2) -1 = 90/93-1` `therefore (n_2-n_1)/(n_1) xx100 = ((90-93)/(93)) 100 = (-3)/(93) xx100` `(n_2-n_1)/(n_1)% = (-1)/(31) xx100 = -0.0326xx100` =-3.26% |
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| 69. |
A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change ?A. Nine - timesB. Three timesC. One - thirdD. One - ninth |
| Answer» Correct Answer - C | |
| 70. |
For a constant vibrating length, density of the material and tension in the string the fundamental frequency of the vibrating stringisA. Inversely proportional to radius of the vibrating stringB. Inversely proportional to the diameter of the wireC. Both a and bD. Inversely proportional to the length |
| Answer» Correct Answer - C | |
| 71. |
Sonometer is usedA. To determine the frequency of tuning forkB. To verify the laws of vibrating stringC. To determine the velocity of longitudinal wavesD. Both a and b |
| Answer» Correct Answer - D | |
| 72. |
Sonometer is based on the principle ofA. Forced vibrationB. Free vibrationsC. ResonanceD. All of the above |
| Answer» Correct Answer - C | |
| 73. |
A wire is under tension and emitting its fundamental frequency . If the wire is replaced by another wire of the same material and same tension but it is thicker , then the fundamental frequency of vibration willA. SameB. DecreaseC. IncreaseD. Can not be predicted |
| Answer» Correct Answer - B | |
| 74. |
Monocord is based on the principle ofA. ResonanceB. Forced vibrationsC. Free vibrationD. All of these |
| Answer» Correct Answer - A | |
| 75. |
If the sonometer wire is vibrating in the second overtone, then there will be formation ofA. Two nodes and two antinodeB. One node and two antinodeC. Four nodes and three antinodesD. Three nodes and three antinodes |
| Answer» Correct Answer - C | |
| 76. |
The linear density of a vibrating string is `1.3 xx 10^(-4) kg//m` A transverse wave is propagating on the string and is described by the equation `y= 0.021 sin (x + 30 t)` where x and y are measured in meter and t`t` in second the tension in the string is :-A. 0.48 NB. 1.20 NC. 0.117 ND. 4.8 N |
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Answer» Correct Answer - C `omega = 30 therefore 2pin = 30 therefore n=(15)/(pi)` `K=1 therefore (2pi)/(lamda) =1 therefore lamda = 2pi` `v=nlamda = (15)/(pi) xx2pi = 30m//s` `v=sqrt(T/m) therefore v^2=T/m` `therefore T=v^2m = 30xx30xx1.3xx10^(-4) = 0.177 N` |
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| 77. |
Particle velocity at the node point in a stationary waves isA. MaximumB. ZeroC. MinimumD. Average velocity |
| Answer» Correct Answer - B | |
| 78. |
Phase differnce between two particles of a medium lying on the opposite sides of a node isA. ZeroB. `pi/3`C. `pi`D. `2pi` |
| Answer» Correct Answer - C | |
| 79. |
Two identical waves each of frequency 10Hz, are travelling in opposite directions in a medium with a speed of 20 cm/s . Then the distance between adjacent nodes isA. 1cmB. 1.2 cmC. 1.5 cmD. 2.0 cm |
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Answer» Correct Answer - A `v=vlamda therefore lamda=v/n=20/10=2cm` The distance between two adjacent antinode is `lamda/2 = 2/2 = 1cm` |
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| 80. |
The phenomenon of setting a body into vibrations by a strong periodic force is calledA. Free vibrationB. Forced vibrationsC. Resonant vibrationsD. None of these |
| Answer» Correct Answer - B | |
| 81. |
In free vibrations, energyA. Changes continuouslyB. Remains constantC. Gradually decreases with timeD. a and c |
| Answer» Correct Answer - C | |
| 82. |
In freee vibrations, the body vibrates withA. Frequency of external forceB. Its own natural frequencyC. Exactly equal to half to the external forceD. Nothing can be said |
| Answer» Correct Answer - B | |
| 83. |
The natural frequency of the body depends uponA. Dimensions and elastic property of the bodyB. Only on the dimension of the vibrating body sC. Elastic property of the body onlyD. Can not be predicted |
| Answer» Correct Answer - A | |
| 84. |
In resonanceA. The energy released by the vibrating body is maximumB. Energy absorbed by the vibrating body is maximumC. Neither energy absorbed by the vibrating body nor energy releasedD. Can not be predicted |
| Answer» Correct Answer - B | |
| 85. |
The frequency of natural vibrations of a body is determined byA. Elasticity of bodyB. InertiaC. Both elasticity and inertiaD. None of these |
| Answer» Correct Answer - C | |
| 86. |
Resonsance is a special case ofA. Forced oscillationsB. Damped oscillationsC. Free oscillationsD. Natural oscillations |
| Answer» Correct Answer - A | |
| 87. |
The cause of damping in an oscillatory motion isA. Restoring forceB. FrictionC. BothD. None of these |
| Answer» Correct Answer - B | |
| 88. |
The frequency of vibration of due to elasticity are independent ofA. Modulus of elasticityB. Restoring forceC. AmplitudeD. None of the above |
| Answer» Correct Answer - C | |
| 89. |
Resonance takes place when the applied frequency is ……… the natural frequency of the systemA. Greater thanB. Less thanC. Equal toD. Either a or b depending on system |
| Answer» Correct Answer - C | |
| 90. |
The acceleration of body executing free damping vibration isA. ConstantB. ChangesC. IncreasingD. Decreasing |
| Answer» Correct Answer - D | |
| 91. |
`y=3cos 100pi(2t-x)`, the value of `lamda` isA. 4cmB. 6cmC. 2cmD. 1cm |
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Answer» Correct Answer - C `y=3cos100pi(2t-x)` `k=100pi` `(2pi)/(lamda) = 100pi` `lamda = (2)/(100) = 0.02 m = 2cm` |
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| 92. |
In the fundamental mode , time taken by the wave to reach the closed end of the air filled pipe is 0.01 s . The fundamental frequency isA. 25B. `12.5`C. 20D. 15 |
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Answer» Correct Answer - a In the fundamental mode, frequency, `n=v/lambda rArr n = v/(4l)` ` n=l/(txx4l)[because v=l/t]` `rArr n=1/(0.01xx4)=25` |
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| 93. |
A pipe of lengths 10 cm, closed at one edn, has frequency equal to half the `2^(nd)` overtone of another pipe open at the ends. The lengths of the open pipe isA. 10cmB. 20cmC. 35cmD. 30cm |
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Answer» Correct Answer - D `(3n_0)/(2) =n therefore n_0 = (2n)/(3)` `(v)/(2l_0) = 2/3xx(v)/(4l_c)` `l_0 = 3xx10=30cm` |
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| 94. |
`n_(1)` is the frequency of the pipe closed at one and `n_(2)` is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formedA. `(n_(1)n_(2))/(n_(2)+2n_(1))`B. `(n_(1)n_(2))/(2n_(2)+n_(1))`C. `(n_(1)+2n_(2))/(n_(2)n_(1))`D. `(2n_(1)+n_(2))/(n_(2)n_(1))` |
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Answer» Correct Answer - a Frequency of closed pipe `n_(1)=v/(4l_(1))rArr l_(1)=v/(4n_(1))` Frequency of open pipe, `n_(2)=v/(2l_(1))rArr l_(2)=v/(2n_(2))` When both pipe are joined, then length of closed pipe `l=l_(1)=l_(2) rArr v/(4n)=v/(4n_(1))+v/(2n_(2))` `rArr 1/(2n)= 1/(2n_(1)) + 1/(n_(2)) or 1/(2n) = (n_(2)+2n_(1))/(2n_(2)n_(2))` `rArr n=(n_(1)n_(2))/(n_(2)+2n_(1))` |
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| 95. |
`n_(1)` is the frequency of the pipe closed at one and `n_(2)` is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formedA. `(n_1n_2)/(2n_2+n_1)`B. `(2n_1n_2)/(2n_2+n_1)`C. `(2n_2n_1)/(n_1+n_2)`D. `(n_2+2n_1)/(n_1n_2)` |
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Answer» Correct Answer - A `n_2/n_1 = sqrt(T_2/T_1) = sqrt(4) = 2` |
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| 96. |
In a stationary waves , energy in a loop isA. Varies instantaneouslyB. Change periodicallyC. Depends upon the types of stationary waveD. Confinet |
| Answer» Correct Answer - D | |
| 97. |
The amplitude of S.H.M. at x=0.05 m isA. 0.02 mB. 0.01 mC. 0.0141mD. 1.414 m |
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Answer» Correct Answer - C `R=2A cos (2pix)/(0.4)==0.02cos((2pixx0.05)/(0.4))` `=0.02cospi/4=0.02xx(1)/(sqrt2)=0.0141m` |
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| 98. |
A brick hung a somometer wire . If the brick is immersed in oil then its frequency willA. IncreaseB. DecreaseC. Remain unchangedD. Increases due to viscosity of oil |
| Answer» Correct Answer - B | |
| 99. |
A somometer wire produces 2 beats per second with a tuning fork, when the length of the wire is either 102cm or 104 cm . The frequency of the tuning fork isA. 206 HzB. 204 HzC. 200 HzD. 198 Hz |
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Answer» Correct Answer - A `n_1 gt n gt n_2` `n_1=n=2` `(n-n_2=2)/(n_1-n_2=4)` `n_1/n_2 = l_2/l_1= 104/102` `(n_1-n_2)/(n_2) = (104-102)/(102)` `(4)/(n_2) = 2/102` `n_2 = 102xx1=204` `n=n_2+2 = 204+2 = 206 Hz.` |
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| 100. |
In brass the velocity of longitudinal wave is 100 times the velocity of transverse wave if `Y=1xx10^11N//m^2`, then stress in the wire isA. `1xx10^13 Nm^(-2)`B. `1xx10^9Nm^(-2)`C. `1xx10^11 Nm^(-2)`D. `1xx10^7 Nm^(-2)` |
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Answer» Correct Answer - D `V_omega = 100 V_T` `sqrt(y/rho) = 100 sqrt(T/m)` `y/rho = 10^4 T/m = 10^4 (T)/(rhoA)` `T/A = (Y)/(10^4) = (10^11)/(10^4) = 10^7` Stress `=10^7N//m^2` |
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