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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.
| 351. |
A long taut string is connected to a harmonic oscillator of frequency `f` at one end. The oscillator oscillates with an amplitude `a_(0)` and delivers power `P_(0)` to the string. Due to dissipation of energy the amplitude of wave goes on decreasing with distance `x` from the oscillator given as `a = a_(0)e^( –kx)`. In what length of the string `(3/4)th` of the energy supplied by the oscillator gets dissipated? |
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Answer» Correct Answer - `(ln 2)/(k)` |
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| 352. |
A transverse harmonic wave travels along a taut string having a tension of `57.6 N` and linear mass density of `100 g//m`. Two points A and B on the string are `5` cm apart and oscillate with a phase difference of `pi/6`. How much does the phase of oscillation of point A change in a time interval of `5.0` ms? |
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Answer» Correct Answer - `(2pi)/5` |
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| 353. |
Consider a sinusoidal travelling wave shown in figure. The wave velocity is `+ 40 cm//s`. Find (a) the frequency (b) the phase difference between points `2.5 cm` apart ( c ) how long it takes for the phase at a given position to (d) the velocity of a particle at `P` at the instant shown. |
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Answer» Correct Answer - A::B::C::D (a) `(lambda)/(2) = 2 cm` rArr `lambda=4 cm` `f =(v)/(lambda) = (40)/(4) =10 Hz` (b) `Deltaphi = (2pi)/(lambda)Deltax = ((2pi)/(4))(2.5) = (5pi)/(4)` ( c ) `omegat = theta` or `(2pi//t) = theta` `:. t=(theta)/(2pif) = (pi//3)/((2pi)(10))` `=(1)/(60)s` (d) At `P`, particle is at mean position. So, `v= "maximum velocity"` `= omegaA` `=2pifA` `=(2pi)(10)(2)` `= (40pi) cm//s` `=125.7 cm//s = 1.26 m//s` Further, `v_(P) = - (v)((dely)/(delx))` ...(i) Sign of `v_(1)`, the wave velocity is given positive. Sign of `(dely)/(delx)`, slope of `y - x` graph is also positive. Hence, from Eq. (i) particle velocity is negative. `:. v_(P) = - 1.26 m//s` |
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| 354. |
A sinusoidal wave `y=a sin ((2pi)/lambda x-omegat)` is travelling on a stretched string. An observer is travelling along positive x direction with a velocity equal to that of the wave. Find the angle that the velocity of a particle on the string at `x=lambda/6` makes with `-x` direction as seen by the observer at time `t=0`. |
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Answer» Correct Answer - `theta=tan^(-1)((api)/lambda)` |
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| 355. |
A travelling wave in a stretched string is described by the equation `y = A sin (kx - omegat)` the maximum particle velocity isA. `A omega`B. `omega//k`C. `dx//dk`D. `x//1` |
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Answer» Correct Answer - A Given that, the displacement of a particle is `y=A sin (omega t=kx)` ...(i) The particle velocity, `v_(p)=(dy)/(dt)` ...(ii) Now, on differentiating Eq. (i) with respect to t, `(dy)/(dt)= A cos (omega t-kx)xx omega` `implies (dy)/(dt)= A omega cos (omega t-kx)` From Eq. (ii), we get `implies v_(p)= A omega cos (omega t-kx)` For maximum particle velocity, `cos (omega t-kx)=1` So, `v_(p)=A omega xx1 implies v_(p)=A omega` |
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| 356. |
A sinusoidal wave trsvelling in the positive direction on a stretched string has amplitude `2.0 cm`, wavelength `1.0 cm` and velocity `5.0 m//s`. At `x = 0` and ` t= 0` it is given that `y = 0` and `(dely)/(delt) lt 0`. Find the wave function `y (x, t)`. |
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Answer» Correct Answer - A::B::D We start with a general equation for a rightward moving wave, `y(x, t) = A sin (kx - omegat + Phi) ` The ampitude given is `A = 2.0 cm = 0.02 m` A wavelength is given as `lambda = 1.0 m` :. Angular wave number, `k =(2pi)/( lambda) = (2pi) m^(-1)` Angular frequency, `omega = vk = (10pi) rad//s` ` :. y(x, t) = (0.02) sin [ 2pi (x - 5.0t) + Phi]` We are given that for `x = 0`, `t = 0`, `y =0` and `(dely)/(delt) lt 0` i.e. `0.02 sin Phi = 0` (as `y = 0`) and `-0.2 pi cos Phi lt 0` From these conditions, we conclude that ` Phi = 2npi` , where `n = 0,2,4,6`...... Therefore, `y(x, t) = (0.02 m) sin [(2pi m^(-1)) x - (10pi rad s^(-1)) t] m` |
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| 357. |
A simple harmonic wave of amplitude `8` units travels along positive x-axis. At any given instant of time, for a particle at a distance of `10 cm` from the origin, the displacement is `+ 6 units`, and for a particle at a distance of `25 cm` from the origin, the displacement is `+ 4` units. Calculate the wavelength. |
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Answer» Correct Answer - A::B::C::D `y =A sin (2pi)/(lambda)(vt - x)` or `(y)/(A) = sin 2pi ((t)/(T) -(x)/(lambda))` In the first case, `y_(1) = + 6 , A = 8 , x_(1) = 10cm` `(6)/(8) = sin 2pi((t)/(T) - (x_(1))/(lambda))` Here, `y_(1) = + 6 , A = 8 , x_(1) =10 cm` `:. (6)/(8) = sin 2pi ((t)/(T) - (10)/(lambda))` ...(i) Similarly in the second case, `(4)/(8) = sin 2pi((t)/(T) - (25)/(lambda))` ...(ii) From Eq. (i), `2pi((t)/(T) - (10)/(lambda)) = sin^(-1) ((6)/(8)) = 0.85 rad` or `(t)/(T)-(10)/(lambda) = 0.14` ...(iii) Similarly from Eq. `(ii)`, `2pi((t)/(T) - (25)/(lambda)) = sin^(-1) ((4)/(8))= (pi)/(6) rad` or `(t)/(T) - (25)/(lambda) = 0.08` ...(iv) Substracting Eq. (iv) from Eq. (iii), we get `(15)/(lambda) = 0.06` `:. lambda =250 cm` |
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| 358. |
A sinusoidal wave travelling in the same direction have amplitudes of 3 cm and 4 cm and difference in phase by `pi//2`. The resultant amplitude of the superimposed wave isA. 7 cmB. 5 cmC. 2 cmD. 0.5 cm |
| Answer» Correct Answer - B | |
| 359. |
The equation of progressive wave travelling a long positive direction of x axis having a amplitude of 0.04 m, frequency 440 Hz and wave velocity 330 m/s, isA. `y = 0.04 sin 2pi (440t - (4x)/(3))`B. `y = 0.04 cos 2pi (440t - (4x)/(3))`C. `y = 0.04 sin 2pi (440t + (4x)/(3))`D. `y = 0.04 cos 2pi (440t + (4x)/(3))` |
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Answer» Correct Answer - A `y = A sin (omegat -kx)` `= 0.04 sin [2pi n - (2pi)/(4) x]` `y = 0.04 sin 2pi [440t - (4)/(3)x]` |
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| 360. |
A sinusoidal transverse wave of small amplitude is travelling on a stretched string. The wave equation is `y = a sin(k x – omegat)` and mass per unit length of the string is `mu`. Consider a small element of length `Deltax` on the string at `x = 0`. Calculate the elastic potential energy stored in the element at time `t = 0`. Also write the kinetic energy of the element at `t = 0`. |
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Answer» Correct Answer - `DeltaU=1/2 mu a^(2) omega^(2)Deltax; Deltak =1/2 mu a^(2) omega^(2) Deltax` |
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| 361. |
A transverse harmonic wave of amplitude 4 mm and wavelength `1.5` m is travelling in positive `x` direction on a stretched string. At an instant, the particle at `x = 1.0` m is at `y = + 2` mm and is travelling in positive `y` direction. Find the co- ordinate of the nearest particle `(x gt 1.0 m)` which is at its positive extreme at this instant. |
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Answer» Correct Answer - 2.25 m |
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| 362. |
A sinusoidal wave travelling in the positive x-direction has an amplitude of `15.0 cm`, a wavelength of `40.0 cm` and a frequency of `8.00 Hz. The verticle displacement of the medium at `t = 0 and x = 0` is also `15.0 cm` as shown in figure. (a) Find the angular wave number `k`, period `T`, angular frequency `omega` and speed `v` of the wave. (b) Write a general expression for wave function. |
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Answer» Correct Answer - A::B::C::D (a) `k = (2pi)/(lambda) = (2pi)/(40) = 0.157 rad//cm` `T = (1)/(f) = (1)/(8) =0.125 s` `omega =(2pi)/(T) =(2pi)/(0.215) = 50.3 rad//s` `v = (omega)/(k) = 320 cm//s` (b) Amplitude is `15 cm`. At `t = 0 , x = 0, y = + A`. Hence, equation should be a `cos` equation. Further, wave is travelling in positive x-direction. Hence, `omegat` and `kx` should have opposite signs. |
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| 363. |
The equation of a wave of amplitude 0.02 m and period 0.04 s travelling along a stretched string with a velocity of 25m/s will beA. ` y= 0.02 sin 2pi (0.04t -0.5x)`B. ` y= 0.02 sin 2pi (25t -2x) `C. ` y=0.02sin 2pi (0.04t-x) `D. ` y= 0.02sin 2pi (25t -x)` |
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Answer» Correct Answer - D |
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| 364. |
A 100 Hz sinusoidal wave is travelling in the positve x - direaction along a string with a linear mass density of `3.5 xx10^(-3) kgm ^(-1)` and a tension of 35 N. At time t =0 ,the point x=0 has zero displacment and the slope of the string is `pi//20`then select the wrong alternative.A. velcoity of wave is 100 m/sB. angular velocity is `(200 pi)` rad /sC. Amplitude of wave is 0.025 mD. none of the above |
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Answer» Correct Answer - D |
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| 365. |
The equation of a wave on a string of linear mass density `0.04 kg m^(-1)` is given by `y = 0.02 (m) sin [2pi((t)/(0.04(s))-(x)/(0.50(m)))]`. The tension in the string is :A. 4.0 NB. 12.5 NC. 0.5 ND. 6.25 N |
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Answer» Correct Answer - D Tension is the string is `T=mu v^(2)=mu omega^(2)/k^(2)` `=0.04 ((2pi//0.04)^(2))/((2pi//0.50)^(2))=6.25 N` |
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| 366. |
Two tuning forks A and B give 4 beats/s when sounded together. If the fork B is loaded with wax 6 beats/s are heard. If the frequency of fork A is 320 Hz, then the natural frequency of the tuning fork B will beA. 320B. 316C. 312D. 326 |
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Answer» Correct Answer - B `n_(B) = n_(A) +- 4` `= 320 +- 4 = 324 or 315 Hz` When `n_(B)` is loaded with wax and sounded with the fork `n_(A)` produces 6 b/s `:. n_(B) = n_(A) +-6` `= 320 +- 6 = 326 or 314 Hz` Before loading frequency of tuning fork increases by 2 `:. n_(B) = 316 Hz` |
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| 367. |
in an experiment it was found that string vibrates in n loops when a mass M is placed on the pan. What mass should be placed on the pan to make it vibrate in 2n loops with same frequency ? ( neglect the mass of pan )A. 2mB. `m/4`C. 4MD. `M/2` |
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Answer» Correct Answer - B |
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| 368. |
A string fastened at both ends has successive resonances with wavelengths of 0.1 m for mth harmonic and 0.08 m for (m+1)th harmonic. The value of m isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - B |
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| 369. |
A long rubber tube having mass 0.9 kg is fastened to a fixed support and the free end of the tube is attached to a cord which passes over a pulley and supports an object, with a mass of 5 kg as shown in figure. If the tube is struck by a transverse blow at one end, find the time required for the pulse to reach the other end. A. 5 sB. 0.47 sC. 4.7 sD. 3.2 s |
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Answer» Correct Answer - B Let `mu` = mass per unit length of rubber tube `mu = (0.9)/(12) = 0.075 kg//m` Tension in tube, T = mg `= 5xx9.8 = 49 N` Speed of wave on the tube `v = sqrt(((T)/(mu)))=sqrt((49)/(0.075))=25.56 m//s` Time required = t= `(AB)/(v) = (12)/(25.56) = 0.47 s` |
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| 370. |
Corresponding to `y - t` graph of a transverse harmonic wave shown in figure, A. B. C. D. |
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Answer» Correct Answer - A::D `y-t` graph is sine graph. Threfore, `v-t` graph is cos graph and `a-t` is -sine graph as `v_(p) =(dely)/(delt)` and `a_(P)=(del^(2)y)/(delt^(2))` |
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| 371. |
Two cars are approaching each other with same speed of ` 20 m//s `. A man in car A fires bullets at regular intervals of 10 seconds. What will be the time interval noted by a man in car B between 2 bullets? (velocity of sound =340 m/s )A. 11. 1sB. 10sC. 8.9 sD. 12 s |
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Answer» Correct Answer - B |
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| 372. |
(i) In a car race sound signals emitted by the two cars are detected by the detector on the straight track at the end point of the race. Frequency observed is 330 Hz and 360 Hz. The original frequency of horn is 300 Hz for both cars. Race ends with the separation of 1000 m between the cars. Assume both cars move with constant velocity and velocity of sound is 330 m/s. Find the time (in seconds) taken by the winning car to finish the race. (ii) A source of sound of frequency f is dropped from rest from a height h above the ground. An observer `O_(1)` is located on the ground and another observer `O_(2)` is inside water at a depth d from the ground. Both `O_(1)` and `O_(2)` are vertically below the source. The velocity of sound in water is 4V and that in air is V. Find (a) The frequency of the sound detected by `O_(1)` and `O_(2)` corresponding to the sound emitted the source initially. (b) The frequency detected by both `O_(1)` and `O_(2)` corresponding to the sound emitted by the source at height `h//2` from the ground. |
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Answer» Correct Answer - (i) 40 s (ii) (a) `f_(2)=f_(1)=(2vf^(2))/(2vf-g)` (b) `f_(1)=f_(2)=(2vf^(2))/(2(v-sqrt(gh))f-g)` |
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| 373. |
A policeman on duty detects a drop of `10%` in the pitch of the horn of a moving car as it crosses him. If the velocity of sound is `330 m//s`, the speed of the car will be |
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Answer» Correct Answer - `17.37 ms^(-1)` |
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| 374. |
Two cars are approaching each other on a straight road and moving with a velocity of `30 kmh^(-1)`.if the sound produced in a car is of frequency 500 Hz, what will be the frequency of sound as heard by the person sitting in the other car ? When the two cars have crossed each other and are moving away from each other.what will be the frequency of sound as heard by the same person ? Speed of sound =`330 ms^(-1)`. |
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Answer» Correct Answer - 526 Hz, 475 Hz |
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| 375. |
A motor car is approaching towards a crossing with a velocity of `75 kmh^(-1)`.The frequency of sound of its horn as heard by a policeman standing on the crossing is 260 Hz.What is the real frequency of the horn ? Speed of sound =`332 ms^(-1)`. |
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Answer» Correct Answer - 244 Hz |
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| 376. |
A car passing a check post gives sound of frequency 1000 c.p.s. If the velocity of the car is `72km//h` and of sound is `350m//s` , find the change is apparent frequency as it crosses the post. |
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Answer» Correct Answer - 114.66 Hz |
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| 377. |
A motor car blowing a horn of frequency `124vib//sec` moves with a velocity `72km//hr` towards a tall wall. The frequency of the reflectedf sound heard by the driver will be (velocity of sound in air is `330 m//s`)A. 109 vib/sB. 132 vib/sC. 140 vib/sD. 248 vib/s |
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Answer» Correct Answer - B |
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| 378. |
A police car with a siren of frequency `8 KHz` is moving with uniform velocity `36 Km//hr` towards a ball building which reflects the sound waves. The speed of sound in air is `320 m//s`. The frequency of the siren heard by the car driver isA. 8.25 kHzB. 8.50 kHzC. 7.75 kHzD. 7.50 kHz |
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Answer» Correct Answer - B `n_("inicident") = n_("reflected") = (320)/(320 - 10) xx 8 kHz` `n_("observed") = (320 + 10)/(320) n_("reflected")` `= 8 xx (330)/(310) = 8.51 kHz = 8.5 kHz` |
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| 379. |
An ambhulance blowing a siren of frequency 700Hz is travelling slowly towards vertical reflectoing wall with a speed 2m/s. The speed of sound is 350m/s. How many beats ar heard per sec to the driver of het ambluance? |
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Answer» Correct Answer - 8 |
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| 380. |
A siren emitting a sound of frequency 1000 vibrations per second is moving with a speed of `10 ms^(-1)`.What will be the frequency of sound, which a listener will hear when (i)the siren is moving towards him ? (ii)the siren is moving away from him ? Speed of sound in air =`340 ms^(-1)`. |
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Answer» Correct Answer - 1033.3 Hz and 971.4 Hz |
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| 381. |
Two trains A and B are approaching a platfrom from opposite directions .The siren in the station is making a sound at a frequency 4kHz.The passengers in trains A and B hear siren at frequencies 4.5 and 5kHz respectively.Then the velocities of the trains A and B are (velocity of sound in air =340 m/s)A. 42.5 m/s ,85m/sB. 75 m/s, 55 m/sC. 85 m/s, 8.5 m/sD. 42.5 m/s ,62.m/s |
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Answer» Correct Answer - B |
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| 382. |
The equation of a travelling wave is `y=60 cos(1800t-6x)` where y is in microns, t in seconds and x in metres. The ratio of maximum particle velocity to velocity of wave propagation isA. ` 3.6`B. ` 3.6xx 10^(-6)`C. ` 3.6xx 10^(-11)`D. ` 3.6xx 10^(-4)` |
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Answer» Correct Answer - C |
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| 383. |
The minimum distance between source of sound and reflection surface for the clear hearing of sound isA. 17 mB. 1.7 mC. 20 mD. 19.2 m |
| Answer» Correct Answer - A | |
| 384. |
The human ear cannot distinguish sound notes of the time intervalA. greater than `(1//10)^(th)` secondB. within `(1//100)^(th)` secondC. within `(1//10)^(th)` secondD. equal to 10 second |
| Answer» Correct Answer - C | |
| 385. |
Beats are result ofA. destructive intergerenceB. diffraction of sound wavesC. constructive and destructive interferenceD. constructive interference |
| Answer» Correct Answer - C | |
| 386. |
In a constructive intergerence of sound the resultant intensity of sound at a point of medium isA. maximumB. zeroC. minimumD. can not be predicted |
| Answer» Correct Answer - A | |
| 387. |
For constructive interference of sound waves the mathematical phase difference between the two arriving waves isA. `n lamda`B. `(2n -1) xx lamda//2`C. odd multiple of `pi`D. even multiple of `pi` |
| Answer» Correct Answer - D | |
| 388. |
For destructive interference of sound waves the mathematical phase difference between the two arriving waves isA. `n lamda`B. `(2n -1) xx lamda//2`C. odd multiple of `pi`D. even multiple of `pi` |
| Answer» Correct Answer - C | |
| 389. |
The ratio of intensities between two cohernt sound sources is 4:1. The difference of loudness is decibel (bD) between maximum and minimum intensitiesm, when they interface in space isA. 10log2B. 20log3C. 10log3D. 20log2 |
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Answer» Correct Answer - B |
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| 390. |
For destructive interference of sound waves the mathematical path difference between the two arriving waves isA. `n lamda`B. `(2n -1) xx lamda//2`C. odd multiple of `pi`D. even multiple of `pi` |
| Answer» Correct Answer - B | |
| 391. |
Decibel isA. musical instrumentB. musical noteC. measure of sould levelD. wavelength of noise |
| Answer» Correct Answer - C | |
| 392. |
For constructive intergerence of sound waves the mathematical path difference between the two arriving waves isA. `n lamda`B. `(2n -1) xx lamda//2`C. odd multiple of `pi`D. even multiple of `pi` |
| Answer» Correct Answer - A | |
| 393. |
A set of tones whose frequencies are integral multiples of the fundamental frequency are calledA. HarmonicsB. OvertonesC. Doppler frequencyD. Beat frequency |
| Answer» Correct Answer - A | |
| 394. |
A source frequency f gives t beats when sounded with a frequency 200Hz. The second harmonic of same source gives 10 beats when sounded with a source of frequency 420Hz. The value of f isA. `200 s^(-1)`B. `205 s^(-1)`C. `195 s^(-1)`D. `210 s^(-1)` |
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Answer» Correct Answer - B Here, `n = 200 +- 5 and 2n = 420 +- 10`. This is possible only when `n = 200 + 5 = 205 s^(-1)` |
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| 395. |
A source frequency f gives 5 beats when sounded with a frequency 200Hz. The second harmonic of same source gives 10 beats when sounded with a source of frequency 420Hz. The value of f isA. 200HzB. 210HzC. 205HzD. 195Hz |
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Answer» Correct Answer - C |
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| 396. |
Two progressive waves having equation `x_(1) = 3 sin omegatau` and `x_(2) = 4 sin (omegatau + 90^(@))` are superimposed. The amplitude of the resultant wave isA. 5 unitB. 1 unitC. 3 unitD. 4 unit |
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Answer» Correct Answer - D `x_(1) = 2 sin omega t " and "x_(2) = 4 sin (omegat +90^(@))` The phase difference between the two waves is `90^(@)` So, resultant amplitude `a = sqrt((3)^(2)+(4)^(2))=sqrt(9+16)=sqrt(25) = 5 " unit"` |
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| 397. |
The temperature at which the speed of sound in air becomes double of its value at `27^(@)C` isA. `273^(@)C`B. `1200^(@)C`C. `927^(@)C`D. `1027^(@)C` |
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Answer» Correct Answer - C `v_(1) prop sqrt(T_(1)) and v_(2) prop sqrtT_(2)` `(v_(2))/(v_(1)) = sqrt((T_(2))/(T_(1))) " " :. (2v_(1))/(v_(1)) = sqrt((T_(2))/(T_(1)))` `:. 4 = (T_(2))/(T_(1))` `T_(2) = 4T_(1) = 300 xx 4 = 1200 .^(@)K` `t_(2) = 1200 - 273` `= 927^(@)C` |
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| 398. |
The temperature at which the speed of sound in air becomes double of its value at `0^(@)C` isA. `546^(@)C`B. `819^(@)C`C. `273^(@)C`D. `1092^(@)C` |
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Answer» Correct Answer - B `v_(1) prop sqrtT_(1) and v_(2) prop sqrtT_(2)` `(v_(2))/(v_(1)) = sqrt((T_(2))/(T_(1))) " " :. ((v_(2))/(v_(1)))^(2) = (T_(2))/(T_(1))` `:. ((2v_(1))/(v_(1)))^(2) = (T_(2))/(T_(1)) " " :. 4 = (T_(2))/(T_(1))` `:. T_(2) = 4T_(1) = 4 xx 273 = 1092 .^(@)K` `= 1092 - 273 = 819^(@)C` |
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| 399. |
The velocity of sound in air when temperature is halved and pressure doubled will be (The velocity of sound in air at NTP is 330 m/s)A. 234.2 m/sB. 466.62 m/sC. 165 m/sD. 330 m/s |
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Answer» Correct Answer - A `v_(1) prop sqrtT_(1) and v_(2) prop sqrtT_(2)` `(v_(2))/(v_(1)) = sqrt((T_(2))/(T_(1))) " " :. (v_(2))/(v_(1)) = sqrt(((1)/(2)T_(1))/(T_(1))) = sqrt((1)/(2))` `:. V_(2) = v sqrt((1)/(2)) = (v)/(sqrt2) = 330 xx 0.707 = 234.2 m//s` |
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| 400. |
The equation of a travelling wave is given as `y=5sin10pi(t-Q.01x),` along the x-axis . Here, all quantities are in SI units. The phase difference between the points separated by a distance of 10m alond x-axis isA. `(x)/(2)`B. `pi`C. `2pi`D. `(pi)/(4)` |
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Answer» Correct Answer - B |
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