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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.
| 251. |
If the speed of sound at `0^(@)C` is 330 m/s, then the speed of sound at `20^(@)C` will beA. 330 m/sB. 340 m/sC. 342 m/sD. 324 m/s |
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Answer» Correct Answer - C `v_(t) = v_(0) + 0.6 xx t` `v_(t) = 330 + 0.6 xx 20 = 330 + 12 = 342 m//s` |
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| 252. |
Compare the velocities of sound in hydrogen `(H_(2))` and carbon dioxide `(CO_(2))`. The ratio of specific heats of `H_(2) and CO_(2)` are respectively 1.4 and 1.3 (Molecular weight of `H_(2) and CO_(2)` are 2 and 44)A. 0.485B. 4.85C. 4.5D. 2.2 |
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Answer» Correct Answer - B `v_(1) = sqrt((gamma_(1)P)/(M_(1))) and v_(2) = sqrt((gamma_(2)P)/(M_(2)))` `(v_(1))/(v_(2)) = sqrt((gamma_(1))/(gamma_(2)) xx (M_(2))/(M_(1))) = sqrt((gamma_(1))/(gamma_(2)) xx (44)/(2)) = sqrt((1.4)/(1.3) xx 22)` = 4.85 |
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| 253. |
The speed of soun in a gas is v and the root mean squre speed of gas molecules is `v_(rms)`. If the ratio of the specific heats of the gas is 1.8, then the ratio `v//v_(rms)` isA. `1 : 2`B. `0.77 : 1`C. `1 : sqrt3`D. `1 : 3` |
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Answer» Correct Answer - B The time of fall of the plate is calculated by using second kinematical equation of motion, `s = (1)/(2) "gt"^(2)` `:. t = sqrt((2s)/(g)) = sqrt((2 xx 0.1)/(9.8)) = (1)/(sqrt49) = (1)/(7)` In time `t = (1)/(7)` s tuning fork complete 8 oscillations `:.` In 1s number oscillations completed are, `(1 xx 8)/((1)/(7)) = 7 xx 8 = 56 Hz` |
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| 254. |
The sound carried by air from a sitar to a listener is a wave of the following typeA. longitudinal stationaryB. transverse propressiveC. transverse stationaryD. longitudinal progressive |
| Answer» Correct Answer - D | |
| 255. |
The phase difference between particle velocity and wave velocity isA. zeroB. `(pi)/(4)`C. `(pi)/(2)`D. `(pi)/(6)` |
| Answer» Correct Answer - C | |
| 256. |
Which of the following statements is wrong?A. sound travels in a straight lineB. sound is a form of energyC. sound travels as wavesD. sound travels faster in vaccum than in air |
| Answer» Correct Answer - D | |
| 257. |
The state (or) condition of vibration of a vibrating body is known isA. amplitudeB. displacementC. phaseD. none of these |
| Answer» Correct Answer - C | |
| 258. |
Why are sound waves called mechanical waves ?A. it require material medium for its propagationB. it may not require any material medium for its propagationC. it can pass through vacuumD. it possess the property of elasticity |
| Answer» Correct Answer - A | |
| 259. |
The ratio of angular velocity to the propagation constant of the medium isA. particle velocityB. wave velocityC. group velocityD. momentum |
| Answer» Correct Answer - B | |
| 260. |
Transverse wave possesses the property ofA. Yong modulusB. modulus of rigidityC. Bulk modulusD. volume elasticity |
| Answer» Correct Answer - B | |
| 261. |
The equation of a progressive wave is given by, `y = 3 sin pi ((t)/(0.02) - (x)/(20)) m`. Then the frequency of the wave isA. 100 HzB. 25 HzC. 50 HzD. 20 Hz |
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Answer» Correct Answer - B `omega = (pi)/(0.02) :. 2pi n = (pi xx 100)/(2)` |
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| 262. |
The equation of a progressive wave is `y=8 sin [ pi ((t)/(10)-(x)/(4))+(pi)/(3)]`. The wavelength of the wave isA. 8 mB. 4 mC. 2 mD. 10 m |
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Answer» Correct Answer - A `x = 8 sin [pi ((x)/(10) - (x)/(4)) + (pi)/(3)]` `k = (pi)/(4)` `(2pi)/(lamda) = (pi)/(4) " " :. lamda = 8 m` |
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| 263. |
A plane progressive wave is `y=0.02 sin (400 t-0.12 x)`. Calculate the following: (i) amplitude of wave, (ii) frequency of wave, (iii) wavelength of wave, (iv) interval at which a compression follows a rerefaction, (v) velocity of the wave. [Hint: compare this equation with the standard `y=a sin (2 pi nt-2pi x//lambda)`] |
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Answer» Correct Answer - `(i) 0.02 m, (ii) 63.5 Hz, (iii) 52.4 m, (iv) 0.008 s, (v) 3333.3 m s^(-1)` |
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| 264. |
The wavelength of a progressive wave moving with a velocity 200 m/s is 1m. The time lag between two particles separated by a distance of 10 m isA. 0.5 sB. 0.005 sC. 0.05 sD. 0.1 s |
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Answer» Correct Answer - C `t = (d)/(v) = (1)/(200) = 0.5 xx 10^(-2) = 0.005s` |
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| 265. |
For the wave `y=5 sin 30pi[t-(x//240)]`, where `x and y` are in cm and `t` is in seconds, find the (a) displacement when `t = 0` and `x = 2 cm` (b) wavelength ( c ) velocity of the wave and (d) frequency of the wave |
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Answer» Correct Answer - A::B::C::D (a) Put `t =0, x = 2cm` (b) `lambda =(2pi)/(k) = (2pi)/((30pi//240)) = 16 cm` (c) Wave velocity `= ("Coefficient of t")/("Coefficient of x")` `=(1)/(1//240)` `=240 cm//s` (d) `f =(omega)/(2pi) = (30pi)/(2pi) =15 Hz` |
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| 266. |
What is the phase difference between two successive crests in the waveA. `pi`B. `pi//2`C. `2pi`D. `4pi` |
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Answer» Correct Answer - C The phase difference between any two successive crests is `2pi` rad |
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| 267. |
What is the phase difference between two successive crests in the waveA. `pi`B. `pi//2`C. `2 pi`D. `4 pi` |
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Answer» Correct Answer - C Phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles, i.e. how far one particle leads the other or lags behind the other. Relatin of path difference and phase difference is given by `Deltaphi= (2pi)/(lamda) xx Deltax`, where `Deltax` is path difference. But path difference between two crests, `Deltax = lamda` Hence, `Delta phi = (2pi)/(lamda) xx lamda = 2pi` |
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| 268. |
The distance between two successive particles which differ in phase byA. `pi` radian is a wavelengthB. `2pi` radian is a wavelengthC. `pi//2` radian is a wavelengthD. `2pi//3` radian is wavelength |
| Answer» Correct Answer - B | |
| 269. |
Two waves of frequencies 20 Hz and 30 Hz travel out from a common point. How wil they different phase at the end of 0.75 s ?A. `pi`B. `7pi`C. `15pi`D. `2pi` |
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Answer» Correct Answer - C `delta phi = (omega_(2) - omega_(1)) t` `= 2pi (n_(2) - n_(1)) t = 2pi (30 - 20) xx 0.75` `= 2pi xx 10 xx (3)/(4)` `= 15 pi` rad |
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| 270. |
The intensity ratio of two waves is `1 : 16`. The ratio of their amplitudes isA. `(1)/(4)`B. `(1)/(2)`C. `(1)/(10)`D. `(16)/(17)` |
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Answer» Correct Answer - A `I_(1) prop A_(1)^(2) and I_(2) prop A_(2)^(2)` `(I_(1))/(I_(2)) = (A_(1)^(2))/(A_(2)^(2)) :. (A_(1))/(A_(2)) = sqrt((I_(1))/(I_(2))) = sqrt((1)/(16)) = (1)/(4)` |
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| 271. |
If two sound waves of equal intensity I produce beats, then the maximum intensity of sound produced in beats will beA. IB. 4 IC. 2 ID. `I//2` |
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Answer» Correct Answer - B `I_("max") = (sqrtI_(1) + sqrtI_(2))^(2)` `= (sqrtI + sqrtI)^(2) = (2sqrtI)^(2) = 4I` |
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| 272. |
If two waves of intensity I and 4 I are super impose, then the minimum and maximum intensities will beA. 3I, 5IB. 1, 9C. I, 9ID. I, 3I |
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Answer» Correct Answer - C `I_("max") = (sqrtI_(1) + sqrtI_(2))^(2)` `= (sqrtI + 2sqrtI)^(2) = (3 sqrtI)^(2) = 9 I` `I_("min") = (sqrtI_(1) - sqrtI_(2))^(2) = (sqrtI - 2sqrtI)^(2) = (-sqrtI)^(2)` = I |
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| 273. |
The amplitude of sound is doubled and the frequency is reduced to one fourth. The intensity of sound at the same point will beA. increases by a factor of 2B. decreases by a factor of 2C. decreases by a factor 4D. remains unchanged |
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Answer» Correct Answer - C `I_(1) prop A_(1)^(2) n_(1)^(2) and I_(1) prop A_(2)^(2) n_(2)^(2)` `:. (I_(2))/(I_(1)) = ((2A_(1))^(2))/(A_(1)^(2)) xx ((n_(1)//4)^(2))/(n_(1)^(2)) " " :. I_(1) = (I_(1))/(4)` |
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| 274. |
The production of echo is due toA. reregaction of sound wavesB. interference of sound wavesC. reflection of sound wavesD. refraction of sound waves |
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Answer» Correct Answer - C |
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| 275. |
A wave shown by the eqauction y=A `cos (omega t-phi ) ` is totally reflected by a closed end.After reflectionA. ` phi ` does not changeB. only `phi` changeC. ` omega ` changesD. both ` omega and phi ` change |
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Answer» Correct Answer - B |
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| 276. |
A wave is reflected from a rigid support. The change in phase on reflection will beA. 0 radB. `pi//4` radC. `pi//2` radD. `pi` rad |
| Answer» Correct Answer - D | |
| 277. |
A wave is reflected from a rigid support. The change in phase on reflection will beA. `pi//2`B. `pi`C. `3pi//2`D. zero |
| Answer» Correct Answer - B | |
| 278. |
Calculate the speed of sound in hydrogen at N.T.P., if density of hydrogen at N.T.P. is `1//16^(th)` of air. Given that the speed of sound in air is 332 m/s.A. 664 m/sB. 332 m/sC. 1328 m/sD. `332 xx sqrt2 m//s` |
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Answer» Correct Answer - C `v_(a) = sqrt((gamma p)/(rho_(a))) and v_(H) = sqrt((gammap)/(rho_(H)))` `:. (v_(H))/(v_(a)) = sqrt((rho_(a))/(rho_(H))) = sqrt16 = 4` `:. V_(H) = 332 xx 4 = 1328 m//s` |
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| 279. |
The velocity of sound in air at N.T.P is `331 ms^(-1)`.Calculate the velocity at `91^@ C`. |
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Answer» Correct Answer - `382.1 ms^(-1)` |
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| 280. |
Every `^(@)C` rise in temperature, the speed of sound increases byA. 0.61 m/sB. 1.22 m/sC. 0.34 m/sD. 1.19 m/s |
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Answer» Correct Answer - A When temperature rises by `1 .^(@)C`, then the speed of sound increases by 0.61 m/s |
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| 281. |
If the speed of sound in helium be `960 ms^(-1)`, when will be the speed of sound in hydrogen at the same temperature ? The value of `gamma` for He and `H_2` are 1.67 and 1.40 respectively and the ratio of the molecular masses is 2:1. |
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Answer» Correct Answer - `1243 ms^(-1)` |
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| 282. |
The velocity of sound in air at NTP is 330 m/s, What will be its value when temperature is tripled and pressure is halved ?A. 330 m/sB. 165 m/sC. `330 sqrt3m//s`D. `330//sqrt3m//s` |
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Answer» Correct Answer - C `(V_(2))/(V_(1)) = sqrt((T_(2))/(T_(1))) :. V_(2) = 1.73 xx 330 = 330 sqrt3m//s` |
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| 283. |
Every `1^(@)F` rise in temperature, the speed of sound increases byA. 0.61 m/sB. 1.22 m/sC. 1.19 m/sD. 0.34 m/s |
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Answer» Correct Answer - D The temperature rises per `.^(@)C` is 0.61 m/s `1 .^(@)F = (5)/(9) .^(@)C = (5)/(9) xx 0.61 = 0.34 m//s` |
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| 284. |
The velocity of sound in a gas is 300 m/s . The root mean square velocity of the molecules is (`gamma = 1.4`)A. 471.4 m/sB. 400 m/sC. 231 m/sD. 462 m/s |
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Answer» Correct Answer - A `C_(rms) = v sqrt((3)/(gamma)) = 330 sqrt((3)/(1.4)) = 330 sqrt((30)/(14)) = 471.4 m//s` |
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| 285. |
The speed of sound waves in air at 300 K is `248 ms^(-1)`. At what temperature will the speed be `402 ms^(-1)`. |
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Answer» Correct Answer - 788.2 K |
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| 286. |
What is the ratio of velocity of sound in hydrogen `(gamma=7//5)` to that in helium `(gamma=5//3)` at the same temperature? |
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Answer» Correct Answer - `sqrt(42)//5` |
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| 287. |
The speed of sound in a gas is v and the root mean square speed of gas molecules is `v_(rms)`. If the ratio of the specific heats of the gas is 1.5 then the ratio `v//v_(rms)` isA. `1 : 2`B. `1 : sqrt3`C. `1 : sqrt2`D. `1 : 3` |
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Answer» Correct Answer - C `(V)/(V_(rms)) = sqrt((gamma)/(3))` |
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| 288. |
If the direction of the vibration of particles is parallel to the direction of the propagation of a wave , then the wave isA. transverse waveB. stationary waveC. longitudinal waveD. electromagnetic waves |
| Answer» Correct Answer - C | |
| 289. |
The sound energy can be transferred from one place to another place through theA. bulk motion of matterB. in the form of transverse wavesC. without bulk motion of the matterD. without material medium |
| Answer» Correct Answer - C | |
| 290. |
two pipes have each of length 2 m, one is closed at on end and the other is open at both ends. The speed of sound in air is 340 m/s . The frequency at which both can resonate is ?A. 340 HzB. 510 HzC. 42.5 HzD. none of these |
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Answer» Correct Answer - D |
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| 291. |
The speed of sound in air and water is 340 m/s and 1420 m/s respectively. If sound waves have a wavelength of 2m in air, then the frequency of the same sound waves in water will beA. 100 HzB. 125 HzC. 340 HzD. 170 Hz |
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Answer» Correct Answer - D `(lamda_(m))/(lamda_(a)) = (v_(m))/(v_(a)) = (1420)/(340)` `lamda_(m) = (142)/(34) xx lamda_(a) = (142)/(34) xx 2 = (142)/(17)` `v_(m) = n lamda_(m) " " :. n = (v_(m))/(lamda_(m)) = (1420)/(142) xx 17` `n = 170 Hz` |
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| 292. |
If a sound wave of frequency 500 Hz and velocity 350 m/s. Then the distance between the two particles of a phase difference of `60^(@)` will be nearlyA. 0.7 cmB. 70 cmC. 12 cmD. 120 cm |
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Answer» Correct Answer - C `v = n lamda " " :. Lamda = (v)/(n) = (350)/(500) = (7)/(10)` `delta = (2pi x)/(lamda) " " :. x = (delta lamda)/(2pi)` `= (pi)/(3) xx (7)/(10) xx (1)/(2pi) = 12 cm` |
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| 293. |
The velocity of sound is generally greater in solids than in gases becauseA. the density of solids is high and the elasticityB. the density of solids is high but the elasticity of solids is very highC. both the density and elasticity of solids are lowD. the density of solids is low, but the elasticity is high |
| Answer» Correct Answer - B | |
| 294. |
Consider a function `y=5.0e^((-25x^(2)-9t^(2)-30xt))` (a) Does this represent a travelling wave? (b) What is direction of propagation of the wave? (c) Find wave speed. (d) Sketch the wave at t = 0 |
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Answer» Correct Answer - (a) Yes (b) Negative x direction (c) 0.6 unit |
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| 295. |
The figure represent a longitudianl waave length travelling in positive x-direction . Then A. part ABC represent combinationB. part ABC represent rearefractionC. part CDE represent compressionD. part CDE represent rerefraction |
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Answer» Correct Answer - A::D |
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| 296. |
A set of tuning forks is arranged in ascending order of frequency each tuning fork gives 5 beats s with the preceding one. If frequency of the firs tuning fork is 100 Hz and the last fork is 150 Hz then the number of tuning forks arranged will brA. 9B. 10C. 11D. 12 |
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Answer» Correct Answer - C Number of tuning forks arranged in series `= ("difference in frequency")/("beat frequency") + 1` `= (150 - 100)/(5) + 1 = (50)/(5) + 1 = 10 + 1 = 11` |
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| 297. |
Two waves of wavelengths 52.5 cm and 52 cm produces 5 beats per second.their frequencies areA. 490 HZ, 495 HzB. 500Hz, 505 HzC. 525 Hz, 520 HzD. 500 Hz, 495 Hz |
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Answer» Correct Answer - C `n_(2) - n_(1) = n` `(v)/(lamda_(2)) - (v)/(lamda_(1)) = 5` `v [(1)/(52) - (1)/(52.5)] = 5` `[(52.5 - 52)/(52 xx 52.5)] = 5` `v = (5 xx 52 xx 52.5)/(0.5)` `n_(1) = (v)/(lamda_(1)) = (5 xx 52 xx 52.5)/(0.5 xx 52.5)` `n_(1) = 520 Hz` `n_(2) = n_(1) + 5` `= 520 + 5 = 525 Hz` |
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| 298. |
Wavelengths of two sound waves in air are ` (110)/( 177) m and ( 110)/( 175)` m respectively.When they are sounded together,they produce 6 beats per second.The frequency of the two waves respectively will beA. ` 531, Hz ,525 HzB. 525 Hz ,519 HzC. 537 HZ , 531 HzD. 519 Hz ,513 Hz |
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Answer» Correct Answer - B |
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| 299. |
Two sound waves of wavelengths 40 cm and 40.5 cm produce 10 beats per second. What will be the speed of sound in air ?A. 324 m/sB. 340 m/sC. 330 m/sD. 360 m/s |
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Answer» Correct Answer - A `n_(1) - n_(2) = n` `(v)/(lamda_(1)) - (v)/(lamda_(2)) = 10` `v[(1)/(40) - (1)/(40.5)] xx 10^(2) = 10` `v = (40 xx 40.5 xx 10)/(0.5 xx 10^(2)) = (4 xx 405)/(5)` `v = 4 xx 81 = 324 m//s` |
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| 300. |
The speed of sound wave in a gas, in which two waves of wavelengths 1.0m and 1.02m produce 6 beats per second isA. 350m/sB. 306m/sC. 380m/sD. 410m/s |
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Answer» Correct Answer - B |
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