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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. |
Doppler shift in frequency does not depend uponA. the actual frequency does not depend uponB. distance of the source from the listenerC. the velocity of the sourceD. velocity of the observe |
| Answer» Correct Answer - B | |
| 52. |
Doppler effect is not applicableA. when the source and observer both are at restB. when there is relative motion between source and observerC. when source is at rest and observer is movingD. when source is moving and observer is at rest |
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Answer» Correct Answer - A |
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| 53. |
Two notes A and B sounded together produce 2 beats per second. When notes B and C are sounded together 3 beats with per second are produced. The notes A and C separately produce the same number of beats with a standard tuning fork of frequnecy 456 Hz. the possible frequency of note B isA. 458.8 HzB. 456.5 HzC. both are correctD. both are wrong |
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Answer» Correct Answer - B |
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| 54. |
Two notes of wavelength 2.04 m and 2.08 m produce 200 beats per minute in a gas. Find the velocity of sound in the gas. |
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Answer» Correct Answer - `356.6 ms^(-1)` |
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| 55. |
The equation of the progressive wave is `Y=3sin [pi ((1)/(3)- (x)/(5) )+( pi)/(4)]` where x and y are in metre and time in second.Which of the following in correct?A. velocity V = 1.5 m/sB. amplitude A =3cmC. frequency F= 0.2 HzD. wavelength `lambda = 10m` |
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Answer» Correct Answer - B |
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| 56. |
The equation of the progressive wave is y =a `sin pi (nt -(x)/(5))` the ratio maximum paritcle velocity to wave velocity isA. `(pi a)/(5)`B. `(2pi a)/(5)`C. `(3pi a)/(5)`D. `(4pi a)/(5)` |
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Answer» Correct Answer - B `y = a sin 2pi [nt - (x)/(5)], (V_(p))/(V) =` ? `(V_(p))/(V) = (A omega)/(n lamda) = (A xx 2pi n)/(n lamda)` `(V_(p))/(V) = AK` Here, `A = a and k = (2pi)/(5)` `:. (V_(p))/(V) = a xx (2p)/(5) = (2pi a)/(5)` |
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| 57. |
For the transverse wave equation `y = A sin(pix + pit)`, choose the correct option at `t =0`A. points at `x = 0 and x = 1` are at mean positionsB. points at `x = 0.5 and x = 1.5` have maximum accelerationsC. points at `x = 0.5 and x = 1.5` are at restD. the given wave is travelling in negative x-direction |
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Answer» Correct Answer - A::B::C::D `y = A sin (pix + pit)` `v_(p) = (dely)/(delt) = piA cos(pix + pit)` `a_(p) = (del^(2)y)/(delt^(2)) = pi^(2) A sin (pix + pit)` Now, subsitute `t=0` and given value of `x`. Since, `omegat and kx` are of same sign, hence the wave is travelling in negative x-direction. |
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| 58. |
The equation of a wave distrubance is a given as `y= 0.02 sin ((pi)/(2)+50pi t) cos (10 pix)`, where x and y are in metre and t is in second . Choose the correct statement (s).A. The wavelength of wave is 0.2mB. Displacement node occurs at x=0.15xmC. Displacement antinode occurs at x=0.3mD. The speed of constitutent wave is 0.2m/s |
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Answer» Correct Answer - A::B::C |
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| 59. |
The equation of a transverse travelling on a rope is given by `y=10sinpi (0.01x-2.00t)` where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is aboutA. ` 63cm //s`B. ` 75 cm//s`C. ` 100 cm //s`D. ` 121 cm //s` |
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Answer» Correct Answer - A |
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| 60. |
The wave described by y `= 0.25 sin ( 10 pix -2pi nt )` where x and y are in meters and t in seconds , is a wave travelling along theA. Positive x direction with frequency 1 Hz and wavelength `lambda =0.2 m`B. negative x direction with amplitude 0.25 and wavelength `lambda =0.2` mC. negative x direction with frequency 1 HzD. Positive x direction with frequency ` pi ` Hz and wavelength `lambda =0.2 m ` |
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Answer» Correct Answer - B |
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| 61. |
The equation of a transverse wave travelling in a rope is given by `y = 5 sin (4t - 0.02 x)`, where y and x are in cm and time t is in second. Then the maximum transverse speed of wave in the rope isA. 125 cm/sB. 200 cm/sC. 250 cm/sD. 100 cm/s |
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Answer» Correct Answer - B `v = (omega)/(k) = (4)/(0.02) = (400)/(2) = 200` |
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| 62. |
Following is given the equation of a travelling wave (all is SI unit) y = (0.02) sin `2pi`(10t - 5x) Match of following. |
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Answer» Correct Answer - (a)R, (b)P, (c )Q,(d)S |
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| 63. |
The equation of a travelling wave is `y(x, t) = 0.02 sin ((x)/(0.05) + (t)/(0.01)) m` Find (a) the wave velocity and (b) the particle velocity at `x = 0.2 m and t = 0.3 s`. |
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Answer» Correct Answer - A::B (a) Wave velocity `= ("Coefficient of t")/("Coefficient of x")` `=((1//0.01))/((1//0.05)) = 5 m//s` Since, coefficient of `t` and coefficient of `x` are of same sign. Wave is travelling in negative x-direction. or `v= - 5m//s` (b) Particle velocity, `v_(P) = (dely)/(delt)` `=2 cos ((x)/(0.05) + (t)/(0.01))` Substituting `x = 0.2` and `t = 0.3`, we have `= (2)(-0.85)` `= - 1.7 m//s` |
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| 64. |
Transverse waves on a srting have speed `12.0 m//s`, amplitude `0.05 m` and wavelength `0.4 m`. The waves travel in the `+ x-direction` and at `t = 0` the `x = 0` end of the string has zero displacement and is moving upwards. (a) Write a wave function describing the wave. (b) Find the transverse displacement of a point at `x = 0.25 m` at time `t = 0.15 s`. ( c ) How much time must elapse from the instant in part (b) until the point at `x = 0.25 m` has zero dispacement? |
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Answer» Correct Answer - A::B::C::D (a) `omega = 2pif = 2pi((v)/(lambda)) = 2pi((12)/(0.4))` `=(60 pi) rad//s` `k = (2pi)/(lambda) = (2pi)/(0.4) = (5pi) m^(-1)` Since, wave is travelling along `+ve` x-direction, `omegat and kx` should have opposite sign. Further at `t = 0 , x = 0` the string has zero displacement and moving upward (in positive cirection). Hence at `x = 0`, we should have `Asin omegat` not `- Asin omegat`. Therefore, the correct expression is `y A sin (omegat - kx)` or `y = 0.05 sin (60pit - 5pix)` (b) Putting `x = 0.2 m` and `t = 0.15 s` in above equation we have, `y = -0.035 4m` `= -3.54 cm` ( c ) In part (b), `y =A// sqrt(2)` From `y =(A)/(sqrt(2))` to `y = 0`, time taken is `t = (T)/(8) = (2pi)/(omega xx 8)` `=(2pi)/((60pi)(8))` `= 4.2 xx 10^(-3) s` `=4.2 ms` |
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| 65. |
A given element in a star is giving out radiations of certain wavelength.The wavelength , when sighted on earth, has a shift of 0.025 % towards the longer wavelength side.What is the velocity of the star in the line the sight ? Given the velocity of light, `c=3xx10^(8) ms^(-1)`. |
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Answer» Correct Answer - `7.5xx10^(4) ms^(-1)` |
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| 66. |
The displacement of a particle in a medium can be expressed as `y=10^(-6)sin(100t+20x+pi//4),` where x is in metre and t is in second. What is the speed of the wave?A. `2000 ms^(-1)`B. `5 ms^(-1)`C. `20 ms^(-1)`D. `5 pi ms^(-1)` |
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Answer» Correct Answer - B The displacement is given by `y=10^(-6) sin (100t+20x+pi/4)` Comparing with standard equation `y=a sin (omega t+kx+phi)` `omega=100, k=20` where, `omega` is angular frequency and k is angular wave number. Now, `k=(2pi)/lambda=(2 pi v)/(lambda v)=omega/v=20` `implies v=omega/20=100/20=5 ms^(-1)` So, the speed of the wave is `5 ms^(-1)`. |
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| 67. |
A string 1 has the length, twice the radius,two the tension and twice the density of another string The relation between the fundamental frequecy 1 and 2 isA. `f_(1)=2f_(2)`B. `f_(1)=4f_(2)`C. `f_(2)=4f_(1)`D. `f_(1)=f_(2)` |
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Answer» Correct Answer - C |
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| 68. |
For the wave shown in figure, write the equation of this wave if its position is shown at `t= 0`. Speed of wave is `v = 300m//s`. |
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Answer» Correct Answer - A::B::C The amplitude `A= 0.06 m` `:. (5)/(2)lambda = 0.2 m` `lambda= 0.08 m` `f = (v)/(lambda)=(300)/(0.08) = 3750 Hz` `k = (pi)/(lambda) = 78.5 m^(-1)` and `omega =2pif = 23562 rad//s` At `t = 0`,`x = 0 , (dely)/(delx) = "positive"` and the given curve is a sine curve. Hence, equation of wave travelling in positive x-direction should have the form, `y(x, t) = A sin(kx - omegat)` Substituting the values, we have `y(x, t) = (0.06 m) sin [(78.5 m^(-1) x = (23562 s^(-1)) t] m` |
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| 69. |
A string is under tension sot that its length uncreased by `1/n` times its original length . The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will beA. 1:nB. `n^(2):1`C. `sqrtn:1`D. n:1 |
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Answer» Correct Answer - C |
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| 70. |
A source of sound of frequency 600Hz is placed inside of water. The speed of sound is water is `1500m//s` and air it is `300m//s`. The frequency of sound recorded by an observer who is standing in air isA. 200HzB. 3000HzC. 120HzD. 600Hz |
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Answer» Correct Answer - D |
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| 71. |
source and observer both start moving simultaneously from origion one along y - axis with speed of source = 2 (speed of observer ). The graph between the apparent frequency observed by observer (f) and time (t) would beA. B. C. D. |
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Answer» Correct Answer - B |
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| 72. |
The second overtone of an open pipe A and a closed pipe B have the same frequencies at a given temperature. Both pipes contain air. The raito of fundamental frequency of A to the fundamental frequency of B is:A. `3:5`B. `5:3`C. `5:6`D. `6:5` |
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Answer» Correct Answer - B |
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| 73. |
Assertion : On moon you cannot hear your friend standing at some distance from you. Reason : There is a vacuum on moon.A. If both Assertion and Reason are true and the Reason is correct expanation of the Assertion.B. If both Assertion and Reason are true but the Reason is not correct expanation of the Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A Longitudinal or sound wave cannot travel in vacuum. |
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| 74. |
The transverse waves can propagate throughA. gases but not through metailsB. metals but not through gases 0C. both gases and metailsD. neither gases nor metals |
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Answer» Correct Answer - B |
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| 75. |
Vibrations of period `0.25 s` propagate along a straight line at a velocity of `48 cm//s`. One second after the the emergence of vibrations at the intial point, displacement of the point, `47 cm` from it is found to be `3 cm`. [Assume that at intial point particle is in its mean position at `t=0` and moving upwards]. Then,A. amplitude of vibrations is `6cm`B. amplitude of vibrations is `3sqrt2cm`C. amplitude of vibrations is `3cm`D. None of these |
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Answer» Correct Answer - A `because` `omega =(2pi)/(T) =(2pi)/(0.25)` `= 8pi rad//s` `v=(omega)/(k)` `:. k=(omega)/(v) =(8pi)/(0.48) = ((50)/(3))pi m^(-1)` `y = A sin(omegat - kx)` `=A sin (8pit-(50)/(3) pix)` Put, `y = 3cm`, t= 1s,x=0.47m` showing we get `A = 6 cm` |
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| 76. |
If the equation of transverse wave is `y=5 sin 2 pi [(t)/(0.04)-(x)/(40)]`, where distance is in cm and time in second, then the wavelength of the wave isA. 10 cmB. 25 cmC. 40 cmD. 60 cm |
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Answer» Correct Answer - A |
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| 77. |
If the equation of transverse wave is `y=5 sin 2 pi [(t)/(0.04)-(x)/(40)]`, where distance is in cm and time in second, then the wavelength of the wave isA. 60 cmB. 40 cmC. 35 cmD. 25 cm |
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Answer» Correct Answer - B Standard transverse equation of wave is `y=a sin 2 pi (t/T-x/lambda)` ...(i) Given equation is, `y=5 sin 2 pi (t/0.04-x/40)` ...(ii) Comparing the given Eqs. (i) and (ii), we get `x/lambda=x/40` `implies lambda =40 cm` |
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| 78. |
The displacement y (in cm ) produced by a simple harmonic wave is `y=(10)/(pi) sin (2000 pi t -(pi x)/(17))` . The periodic time and maximum velocity of the particles in the medium will respectively beA. `10^(-3) s and 330 m//s`B. `10^(-3)s` and 200 m/sC. `10^(-4)s` and 20 m/sD. `10^(-2) s` and 2000 m/s |
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Answer» Correct Answer - B `y = (10)/(pi) sin (2000 pi t - (pi)/(17) x)` `omega = 2000 pi` `2pi n = 2000 pi` `n = 10000` `T = (1)/(n) = (1)/(1000) = 10^(-3)s` `v = (omega)/(k) = (2000pi)/((pi)/(17)) = 34000 cm//s` `v = 340 m//s` |
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| 79. |
The displacement of a particle executing S.H.M. is given by `y = 10 sin [6t + (pi)/(3)]` where y is in metres and t is in seconds. Then the initial displacement and velocity of the particle isA. `5sqrt3 m and 30 ms^(-1)`B. `15 m and 5 sqrt3 ms^(-1)`C. `15 sqrt3 and 30 ms^(-1)`D. `20 sqrt3 and 30 ms^(-1)` |
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Answer» Correct Answer - A `y = 10 sin [6t + (pi)/(3)]` `y = 10 sin [6 xx 0 + (pi)/(3)]` `= 10 xx (sqrt3)/(2) = 5 sqrt3` `v = (dv)/(dt) = 10 xx 6 xx cos (6t + (pi)/(3))` `v = 60 xx cos (0 + (pi)/(3))` `= 60 xx (1)/(2) = 30 m//s` |
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| 80. |
The equation of a simple harmonic motion is given by `y = 3 sin.(pi)/(2) (50 t-x)`, where x and y are in metres and t is in seconds, the ratio of maximum particle velocity to the wave velocity isA. 25 m/sB. 30 m/sC. 50 m/sD. None of these |
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Answer» Correct Answer - C The given wave equation is `y = 3 sin .(pi)/(2)(50 t - x)` `implies " "y = 3 sin (25 pi t - (pi)/(2)x) " "...(i)` Comparing with standard equation `y = a sin (omegat - kx) " "...(ii)` `omega=25pi, k = pi//2` Wave velocity, `v = (omega)/(k)=(25pi)/(pi//2)=50m//s` |
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| 81. |
A transverse wave is derscried by the equation `y=y_(0) sin 2 pi (ft - (x)/(lamda))`. The maximum particle velocity is equal to four times the wave velocity if :-A. ` lambda =(pi y_0)/( 4) `B. ` lambda = ( pi y_0)/( 2) `C. ` lambda =pi y_0`D. ` lambda =2pi y _0` |
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Answer» Correct Answer - C |
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| 82. |
A progressive wave is, `y= 12 sin (5t - 4x)`. On this wave how far away are the two points having a phase difference of `45^@)` ?A. `pi//4`B. `pi//8`C. `pi//16`D. `pi//32` |
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Answer» Correct Answer - C `delta = (2pix)/(lamda)` here k = 4 `:. (2pi)/(lamda) = 4 " " :. lamda = (pi)/(4)` `:. x = (delta lamda)/(2pi) = ((pi)/(4) xx (pi)/(2))/(2pi) = (pi)/(16)` |
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| 83. |
The displacement x(in metres) of a particle performing simple harmonic motion is related to time t(in seconds) as `x=0.05cos(4pit+(pi)/4)` .the frequency of the motion will beA. 0.5 HzB. 1.0 HzC. 1.5 HzD. 2.0 Hz |
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Answer» Correct Answer - D Compare the given equation with the standard from `y=a cos [(2pi t)/T-(2pi x)/lambda]` Coefficient of `t=(2pi)/T=2 pi n=4 pi, n =2` Hz |
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| 84. |
A transverse wave ` y = 0.05 sin ( 20 pi x - 50 pi t)` meters , is propagating along + ve X - axis on a string light insect starts crawling on the string with velocity of 5 cm /s at t= 0 along the +ve X - axis from point where x = 5 cm . After 5 s the difference in phase of its position is equal toA. `150 pi `B. ` 250 pi `C. `-245 pi`D. `-5pi` |
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Answer» Correct Answer - B |
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| 85. |
A progressive wave is represented by `y = 12 sin (5t - 4x)` cm. On this wave, how far away are the two points having phase difference of `90^circ` ?A. `pi//4`B. `pi//8`C. `pi//16`D. `pi//32` |
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Answer» Correct Answer - B `y = 12 sin (5t - 4x)` `delta = (pi)/(2)` `delta = (2pi x)/(lamda)` `:. x = (delta)/((2pi)/(lamda)) = ((pi)/(2))/(4) = (pi)/(8)` |
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| 86. |
A transverse wave of amplitude `0.5 m` and wavelength 1 m and frequency 2 Hz is propagating in a string in the negative x - direction. The expression for this wave is [A. `y= 0.5sin (4pi t -2pix) `B. `y= 0.5cos (pi t -2pix ) `C. `y =0.5 sin (4pit +2pi x ) `D. ` y= 0.5sin (pit+ 2pix )` |
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Answer» Correct Answer - C |
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| 87. |
The equation of a transverse wave propagating in a string is given by `y = 0.02 sin (x + 30t)` where, `x and y` are in second. If linear density of the string is `1.3 xx 10^(-4)kg//m`, then the tension in the string isA. `0.12N`B. `1.2N`C. `12 N`D. `120 N` |
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Answer» Correct Answer - A `v=(omega)/(k) = sqrt((T)/(mu)` `:. T =mu((omega)/(k))^(2) =(1.3 xx 10^(-4))((30)/(1))^(2)` `=0.12N` |
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| 88. |
Two vibrating tuning forks produce waves given by `y_(1)=4 sin 53 pi t, y_(2)=2 sin 50 pi t` If they are held near the ear of a person, the person will hearA. 3 beats `s^(-1)` with intensity ratio of maximum to minima equal to 9B. 3 beats `s^(-1)` with intensity ratio of maxima to minima equal to 2C. 6 beats `s^(-1)` with intensity ratio of maxima to minima equal to 2D. 6 beats `s^(-1)` with intensity ratio of maxima to minima equal to 9 |
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Answer» Correct Answer - B `y_(1) = 4 sin 53pi t and y_(2) = 2 sin 50pi t` `:. A_(1) = 4 f_(1) = 53` `A_(2) = 2 f_(2) = 50` `:.` Beat `= 53 - 50 = 3 "sec"^(-1)` `(l_("max"))/(l_("min")) = ((A_(1) + A_(2))^(2))/((A_(1) - A_(2))^(2)) = ((4 + 2)^(2))/((4 -2)^(2)) = (36)/(4) = 9 : 1` |
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| 89. |
When two harmonic sound waves of close but not equal) frequencies are heard at the same time, we hearA. a sound of sililar frequencyB. a sound of frequency which is the average of two close frequency which is the average of two close frequenciesC. audibly distinct waxing and waning of the intensity of the sound with a frequency equal to the difference in the two close frequenciesD. All of the above |
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Answer» Correct Answer - D When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (average of two close frequencies) The intensity of the sound heard is waxing and waning (varying) with a frequency `= |v_(1) - v_(2)|`, where `v_(1) and v_(2)` are the individual frequencies of the waves. |
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| 90. |
Beats are produced when two progressive waves of frequency 256 Hz and 260 Hz superpose. Then the resultant amplitude change periodically with frequency ofA. 256 HzB. 260 HzC. `(256-260)/2 Hz`D. 4 Hz |
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Answer» Correct Answer - D Frequency of change of resultant amplitude = number of beats `s^(-1) = 260 - 256 = 4 Hz` |
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| 91. |
Two waves of same frequency have amplitudes 5 cm and 3 cm, These waves are made to superpose in the same direction. The ratio of maximum intensity to minimum intensity at various places will beA. ` 3: 5`B. ` 9: 25`C. ` 9: 4`D. ` 16: 1` |
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Answer» Correct Answer - D |
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| 92. |
Longitudinal waves do not exhibitA. reflectedB. refractedC. interferenceD. polarised |
| Answer» Correct Answer - D | |
| 93. |
When a simple harmonic progressive wave is propogating the medium, all the particles of the medium vibrate withA. different amplitude and frequencyB. the same amplitude and same frequencyC. the same amplitude and different frequencyD. the different amplitude and same frequency |
| Answer» Correct Answer - B | |
| 94. |
When plane progressive waves travelling in the same direction superpose over each other then the velocity of resultant waveA. decreaseB. increaseC. becomes zeroD. remains unchanged |
| Answer» Correct Answer - D | |
| 95. |
When simple harmonic progressive waves, travelling through a medium each succeeding particleA. leading in phase than preceding particleB. lagging behind in phase than the preceding particleC. lagging in phase by `180^(@)`D. leading in phase by `180^(@)` |
| Answer» Correct Answer - B | |
| 96. |
A set of 25 tuning forks is arranged in order of decreasing frequency. Each fork gives 3 beats with succeeding one. The first fork is octave of the last. Calculate the frequency of the first and 16th fork.A. 144 HzB. 99 HzC. 95 HzD. 85 Hz |
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Answer» Correct Answer - B `N = (n_(f) - n_(L))/(x) + 1` `25 -1 = (2n_(L) - n_(L))/(3)` `24 xx 3 = n_(L)` `72 = n_(L)` `n_(f) = 2n_(2) = 2 xx 72 = 144` `n_(1) - n_(2) = 3` `n_(1) = n_(2) + 3` `n_(1) = n_(3) + 2 xx 3` `n_(1) = n_(16) + 15 xx 3` `n_(16) = 144 - 45 = 99Hz` |
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| 97. |
A set 65 tuning forks is arranged so that each gives 3 beats per second with the previous one and the frequency of last fork is an octave of first. Then the frequencies of first and last tuning forks areA. 192 Hz, 384 HzB. 64 Hz, 26 HzC. 384 Hz, 576 HzD. 64 Hz, 64 Hz |
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Answer» Correct Answer - A `N = (n_(L) - n_(f))/(x) + 1` `65 -1 = (2n - nf)/(3)` `64 xx 3 = n_(f)` `192 = n_(f)` `n_(L) = 2n_(f) = 2 xx 192 = 384 Hz` |
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| 98. |
11 tuning forks are arranged in increasing order of frequency.Each gives 8 beat/s with previous one.If `11 ^(th)` fork is octave of ` 1^(st) ` fork then the frequency of `10^(th)` fork isA. 96 HzB. 80 HzC. 152 HzD. 64 Hz |
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Answer» Correct Answer - D |
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| 99. |
16 tuning forks are arranged in increasing order of frequency. Any two consecutive tuning forks when sounded together produce 8 beats per second. If the frequency of last tuning fork is twice that of first, the frequency of first tuning fork is :- |
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Answer» Correct Answer - 120 Hz |
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| 100. |
A set of 25 tuning forks is arranged in order of decreasing frequency. Each fork gives 3 beats with succeeding one. The first fork is octave of the last. Calculate the frequency of the first and 16th fork. |
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Answer» Correct Answer - 144 Hz, 99 Hz |
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