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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.
| 301. |
The velocity of sound in a gas ,in which two waves of length 1.00 m and 1.01 m produce 10 beats in 3 second isA. 336. 67 m/sB. 326.67 m/sC. 346.67m/sD. 356.67m/s |
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Answer» Correct Answer - B |
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| 302. |
In the above problem, when will be the fourth echo heardA. 4sB. 1.5 sC. 5.5 sD. 3.5 s |
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Answer» Correct Answer - C `t_(4) = 2t_(1) + t_(2) = 3 + 2.5 = 5.5s` |
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| 303. |
In the above problem, when will be the third echo heard.A. 4sB. 1sC. 5sD. 3s |
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Answer» Correct Answer - A `t_(3) = t_(1) + t_(2) = 1.5 + 2.5 = 4s` |
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| 304. |
Fifty-six tuning forks are arranged in order of increasing frequencies so that each fork gives 4 beats per second with the next one. The last fork gives the octave of the first. Find the frequency of the first. |
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Answer» Correct Answer - `220` |
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| 305. |
The frequency and amplitude of the problem number 130 isA. 5m, 100 HzB. 5 cm, 100 HzC. 3m, 150 HzD. 3m, `100pi Hz` |
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Answer» Correct Answer - B `omega = 200 pi` `2pin = 200pi` `n = 10 Hz, A = 5 m` |
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| 306. |
A wave along a string has the equation `y = 0.02 sin (30 t - 4x)`, where x and y are in m and t in second the amplitude of the wave isA. 0.02 cmB. 0.02 mC. 4mD. 0.4 cm |
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Answer» Correct Answer - B `y = 0.02 sin (30 t - 4x)` `y = A sin (omega t - kx)` `:. A = 0.02 m` |
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| 307. |
A fork of unknown frequency gives 4 beats per second when sounded with another of frequency `256`. The fork is now loaded with a piece of wax and again 4 beats per second are heard. Calculate the frequency of the unknown fork. |
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Answer» Correct Answer - `260` |
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| 308. |
Two waves are represented by the equations `y_(1)=a sin (omega t+kx+0.785)` and `y_(2)=a cos (omega t+kx)` where, x is in meter and t in second The phase difference between them and resultant amplitude due to their superposition areA. `45^(@)` and 1.84 aB. `30^(@)` and aC. `30^(@)` and 2aD. `45^(@)` and a |
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Answer» Correct Answer - A `y_(1)= a sin (omega t+kx+0.57)` `:.` Phase `phi_(1)=omegat+kx+0.57` `y_(2)=a cos (omega t+kx)` `=a sin (omega t+kx +pi/2)` `:.` Phase, `phi_(2)=omegat+kx+pi/2` Phase diferent, `Delta phi=phi_(2)-phi_(1)` `=[omega t+kx+pi/2]-(omegat+kx+0.57)` `=pi/2 -0.785` `=(1.57-0.785) rad =0.785` rad `=pi/4 rad = 45^(@)` Resultant amplitude, `a=2 a "cos" phi/2=2a " cos"pi/8=1.84 a` |
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| 309. |
Calculate the velocity of sound in a gas in which two waves of wavelength `50 cm` and `50.5 cm` produce 6 beats per second. |
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Answer» Correct Answer - `303 m s^(-1), |
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| 310. |
In problem No. 133, the phase difference beteen two points septated by 0.785 m isA. `pi`B. `2pi`C. `pi//2`D. `3pi//2` |
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Answer» Correct Answer - A `delta = (2pi x)/(lamda) = (2pi xx 0.785)/(pi//2) = 4 xx 0.785` `delta = 3.140 = pi` rad |
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| 311. |
In the problem No.133, the velocity of the wave isA. 30 m/sB. 7.5 m/sC. `15pi`D. `3pi//2 m//s` |
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Answer» Correct Answer - B `v = (omega)/(k) = (0)/20)/(4) = 7.5 m//s` |
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| 312. |
In the problem No. 133, the wavelength of the wave isA. `pi//2 m`B. 5mC. `4pi m`D. 4m |
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Answer» Correct Answer - A `k = 4` `(2pi)/(lamda) = 4 " " :. lamda = (2pi)/(4) = (pi)/(2)m` |
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| 313. |
Two sound waves of wavelength `1 m` and `1.01 m` in a gas produce `10` beats in `3`s. The velocity of sound in the gas isA. 332 m/sB. 336.7 m/sC. 83 m/sD. 166 m/s |
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Answer» Correct Answer - B `n_(1) - n_(2) = (10)/(3) :. (v)/(lamda_(1)) - (v)/(lamda_(2)) = (10)/(3)` `v[(1)/(1) - (1)/(1.01)] = (10)/(3)` `v = (10 xx 3.01)/( 3 xx 0.01) = (1010)/(3) = 336.7 m//s` |
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| 314. |
WHICH OF THE FOLLOWING FUNCTIONS OF X AND T REPRESENTS A PROGRESSIVE WAVE ?A. Y=SIN (4T-3X)B. `y = (1)/(4+(4T-3X)^(2)`C. Y = `1/(4T+3X)`D. 1/(E4T+3X)` |
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Answer» Correct Answer - A::B |
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| 315. |
In problem No. 133, the phase difference between two points separated by time interval 0.2098 isA. `2pi`B. `pi`C. `pi//2`D. `pi//4` |
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Answer» Correct Answer - A `delta = (2pi x)/(lamda) = (2pi Delta t)/(T) = 2pi` |
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| 316. |
A wave has SHM (simple harmonic motion) whose period is 4s while another periods 3 s. If both are combined, then the resultant wave will have the period equal toA. 4sB. 1 sC. 12sD. 3s |
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Answer» Correct Answer - C `T = (T_(1)T_(2))/(T_(1) - T_(2))` |
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| 317. |
In problem No.133, the distance moved by the wave in 4s is,A. 30 mB. 7.5 mC. 15 mD. 60 m |
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Answer» Correct Answer - A `x = Vt = 7.5 xx 4 = 30` |
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| 318. |
The equation of a wave travelling on a string is given by Y(mn) = 8 sin[`(5m^(-1)x-(4s^(-1)t`]. ThenA. velocity of wave is 0.8 m/sB. the displacement of a particle of the sting at t= 0 and `x = (pi)/(30)` m from the mean position is 4 mnC. the displacement of th mean position at t = 0, `x = (pi)/(30)`m is 8 m/sD. velocity of the wave is 8 m/s |
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Answer» Correct Answer - A::B |
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| 319. |
A steel wire is rigidly fixed at both ends. Its length mass and cross-sectionl area are 1m, 0.1kg and `10^(-8)m^(2)` respectively. Tension in the wire is produced by lowering the temperature by `20^(@)C`. If the transerverse waves are some up by plucking the wire at 0.25m from one end and assuming that the wire viberates with minimum number of loops possble for such a case. The frequency of viberation (in Hz) is found to be. 1.11. Find the value of K. Given `alpha=1.21xx10^(-5).^(@)C^(-1). Y-2xx10^(11)N//m^(2)` |
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Answer» Correct Answer - 2 |
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| 320. |
A 400 gm block B is suspended with uniform string S of mass 100gm and length 20cm as shown. Variation of tension T with distance x from the end block is hanging is T=4+4x, where T is (mN) anad x in meters. Find the value of K `(in N//m).(g=10m//s^(2)` |
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Answer» Correct Answer - 5 |
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| 321. |
in a plane progaressive harmonic wave particle speed is always less than the wave speed if.A. amplitude of wave is less than `lambda/(2 pi)`B. amplitude of wave is greater than `lambda /(2 pi ) `C. amplitude to wave is less than ` lambda`D. amplitude of wave is greater than ` lambda / pi)` |
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Answer» Correct Answer - A |
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| 322. |
The velocity of sound in air is `332 ms^(-1)` at `0^@ C` .At what temperature will the velocity become 1/2 time that at `0^@C`? |
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Answer» Correct Answer - `-204.75^@ C` |
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| 323. |
If the velocity of sound in air at `0^(@)C " is " 332 ms^(-1)`, its velocity at `30^(@)C` isA. `200 ms^(-1)`B. `300 ms^(-1)`C. `350 ms^(-1)`D. `996 ms^(-1)` |
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Answer» Correct Answer - C When temperature of medium rises by `1^(@)C` then speed of sound increases by 0.60 m/s. If the temperature rises by `30^(@)C`, then the speed of sound increases by `0.6 xx 30 = 10 m//s` Speed = 332 Hz = 350 m/s `v_(t) = v_(0) + v_(t)` `:. 332 + 0.60 xx 30 = 332 + 18 = 350` |
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| 324. |
The relation between wave velocity and maximum particle velocity is (Where `V_(p)` = Particle velocity, V = Wave velocity)A. `v_(p) = v`B. `v_(p) = (lamda)/(2pi) v`C. `v_(p) = (2piA)/(lamda)v`D. `v = (lamda)/(2pi) v_(p)` |
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Answer» Correct Answer - C `(v_("max"))/(v) = (A omega)/(n lamda) = (A xx 2pi n)/(n lamda)` `v_("max") = (2pi A)/(lamda) v` |
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| 325. |
The equation of a wave is, `y = 1.5 sin (314 t - 12.56 x)m`. The phase difference between two points 7.5 m apart isA. `100pi` radB. `4pi` radC. `10pi` radD. `30pi` rad |
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Answer» Correct Answer - D `y = 1.5 sin (314 t - 12.56 x) m` Comparing it with standard equation, `y = A sin (omega t - kx)` `k = 12.5, x = 7.5` delta = (2pi x)/(lamda)` `kx = 12.56 xx 7.5` `= 4pi xx 7.5 = 30 pi` |
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| 326. |
Calculate the velocity of transverse wave in a copper wire 1 `mm^(2)` in cross-section, under the tension produced by 1 kg wt. The density of copper =`8.93 kg m^(-3)` |
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Answer» Correct Answer - `33.12 ms^(-1)` |
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| 327. |
A string, of length L, clamped at both ends is vibrating in its first overtone mode. Answer the following questions for the moment the string looks flat (a) Find the distance between two nearest particles each of which have half the speed of the particle having maximum speed. (b) How many particles in the string have one eighth the speed of the particle travelling at highest speed? |
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Answer» Correct Answer - (a) `L/6` (b) `4` |
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| 328. |
If two waves ` x_1 =A sin (omegat -0.1 x ) and x_2 A sin (omegat -0.1x -phi //2 ) ` are combined with each other ,then resultant amplitude of the combined waves isA. ` 2A cos (phi )/(4) `B. ` A sqrt( 2cos (phi )/( 2))`C. ` 2A cos (phi )/(2)`D. ` A sqrt( ( 1+cos (phi )/( 4))` |
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Answer» Correct Answer - A |
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| 329. |
Let `n_(1) and n_(2)` be the two slightly different frequencies of two sound waves. The time interval between waxing and immediate next waning isA. `(1)/(n_(1) - n_(2))`B. `(2)/(n_(1) - n_(2))`C. `(n_(1) - n_(2))/(2)`D. `(1)/(2(n_(1) - n_(2)))` |
| Answer» Correct Answer - D | |
| 330. |
An open pipe of sufficient length is dipping in water with a speed v vertically. If at any instant l is lengths of tube avoce water. Then the rate at which fundamental frequency of pipe changes , is ( speed of sound = c) A. `cv//2l^(2)`B. `cv//4l^(2)`C. `c//2 v^(2)l^(2)`D. `c//4] v^(2)l^(2)` |
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Answer» Correct Answer - B |
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| 331. |
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. 72, 144B. 36, 72C. 18, 36D. 9, 18 |
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Answer» Correct Answer - B `N = (n_(L) - n_(F))/(x) +1` `10 = (2n_(F) - n_(F))/(4) + 1` `9 xx 4 = n_(F)` `n_(F) = 36` |
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| 332. |
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 32D. 72 and 36 |
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Answer» Correct Answer - D In n is frequency of first fork, then frequency of the last (10th fork) `= n + 4(10 -1) = 2n` `:. N = 36 and 2n = 72` |
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| 333. |
Given frequency of source f=100Hz,`v_(s)=20m//s `and `v=330 m//s. The beat frequencies `f_(1)` and `f_(2) are A. `f_(1)=20Hz`B. `f_(2)=10Hz`C. both(a) and (b) are wrongD. both (a) and (b) are correct |
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Answer» Correct Answer - C |
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| 334. |
Two sound waves with wavelengths `5.0 m` and `5.5 m` respectively, each propagates in a gas with velocity `30m//s` We expect the following number of beats per second:A. 1B. 6C. 12D. 0 |
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Answer» Correct Answer - D |
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| 335. |
Beats are produced due toA. diffraction nB. superpositionC. polarizationD. refraction |
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Answer» Correct Answer - B |
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| 336. |
When two sound waves are superimposed, beats are produced when they haveA. different amplitudes and phases.B. different velositiesC. different phasesD. different frequencies |
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Answer» Correct Answer - B |
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| 337. |
The figure shows a snap photograph of a vibrating string at `t = 0`. The particle `P` is observed moving up with velocity `20sqrt(3) cm//s`. The tangent at `P` makes an angle `60^(@)` with x-axis. (a) Find the direction in which the wave is moving. (b) Write the equation of the wave. (c) The total energy carries by the wave per cycle of the string. Assuming that the mass per unit length of the string is `50g//m`. |
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Answer» Correct Answer - A::B::C::D (a) `v_(P) = - v((dy)/(dx))` As `v_(P) `and `("slope")_(P)` are both positive, `v` must be negative. Hence the wave is moving in negative x-axis. (b) `y = A sin (omegat - kx + phi)` …(i) `k=(2pi)/(lambda) = (pi)/(2) cm^(-1)` `A = 4 xx 10^(-3)m =0.4 cm` At `t =0,x = 0 "slope" (dy)/(dx) = +ve` `:. v_(P) = - v("slope")= +ve` `:. v_( p) = -v("slope")=+ve` Further at `t = 0, x=0, y =+ve` `:. phi=(pi)/(4)` Further, `20sqrt(3) = - v tan60^(@)` `:. v =-20 cm//s` `f = (v)/(lambda) =5 Hz` `:. omega = 2pif = 10pi` `:. y = (0.4 cm) sin (10pit +(pi)/(2)x + (pi)/(4))` ( c ) `P =2pi^(2) A^(2)f^(2)muv` `:.` Energy carried per cycle `E = PT = (P)/(f) = 2pi^(2)A^(2)fmuv` Subsitituting he values, we have `E =1.6 xx10^(-5)J` |
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| 338. |
The production of echo is due toA. rarefaction of sound wavesB. interference of sound wavesC. reflection of sound wavesD. reflection and refraction of sound waves |
| Answer» Correct Answer - C | |
| 339. |
The echo and original sound will have sameA. frequencyB. amplitudeC. intensityD. all of above |
| Answer» Correct Answer - A | |
| 340. |
When a transvers wave pulse is reflected from free end or yielding support, the phase change produced isA. `pi//2`B. `pi`C. `3pi//4`D. zero |
| Answer» Correct Answer - D | |
| 341. |
A wave travelling along positive x -axis is given by `y = A sin (omega t- kx)`. If it is reflected from rigid boundary such that 80% amplitude is reflected, then equation of reflected wave isA. `y=A sin (omega t-kx)`B. `y=-0.8 A sin (omega t+kx)`C. `y=0.8 A sin (omega t+kx)`D. `y=A sin (omega t+0.8 kx)` |
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Answer» Correct Answer - B On getting reflected form a rigid boundary the wave sufferes an additional phase change of `pi`. `implies y-0.8 A sin (omega t+kx+pi)` `=-0.8 A sin (omega t+kx)` |
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| 342. |
The equation of a wave is `y=A sin (2pint) `When it is reflected at a free end its amplitude becomes 90% The equation of the reflected wave isA. ` y= 0.9 A sin ( 2pi nt + pi ) `B. ` y=(A)/( 9)sin ( 2pint ) `C. ` y= 0.9 A sin (2pint ) `D. ` y= 9 A sin (2pint ) ` |
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Answer» Correct Answer - C |
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| 343. |
A bat flying above lake emits ultrasonic sound of 100 kHz ,When this wave falls on the water surface, it is partly reflected and partly transmitted .The wavelengths of the reflected and the transmitted waves are (The speed of sound in air is 340 m/s and in water is 1450 m/s)A. ` 6.8 ` mm and 2.9 mmB. 3.4 mm and 1.45 cmC. 3.4 mm and 7.8 mmD. 6.8mm and 1.45 cm |
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Answer» Correct Answer - B |
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| 344. |
At t=0 , observer and source are at same place. Now the source is projected with velocity `60 sqrt2 `m/s at `45 ^(@)` . Natural frequency of source is 1000 Hz.find the frequency heard by the observer at t=2s. Take speed of sound = 340 m/sA. 930 HzB. 860 HzC. 826 HzD. 970Hz |
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Answer» Correct Answer - C |
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| 345. |
there are three strings RP, Pqand QS as shown. Their mass and length are RP = (0.1 kg , 2 m), PQ = ( 0.2 kg , 3 m) ,QS= ( 0.15 kg , 4 m) respectively. All the strings are under the same tension. Wave -1 is incident at P. it is partly reflected (wave -2 ) adn partly transmitted (wave -3) . now wave - 3 is incident at Q . it is again partly transmitted (wave -5)[ and partly reflected (wave -4) . phase difference between wave -1 and wave A. 2 is `pi`B. 4 is zeroC. both (a) and (b) are correctD. |
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Answer» Correct Answer - C |
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| 346. |
Two waves are represented by the equations `y_(1)=asin(omegat+kx+0.57)m` and `y_(2)=acos(omegat+kx)`m, where x is in metres and t is in seconds. The phase difference between them isA. 1.25 radB. 1.57 radC. 0.57 radD. 1.0 rad |
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Answer» Correct Answer - D `y_(1)=a sin (omega t+kx+0.57) m` and `y_(2)=a cos (omega t+kx) m` `= a sin (pi/2 +omega t+kx) m` Phase difference, `Delta phi=phi_(2)-phi_(1)` `=pi/2-0.57=1.57-0.57=1` rad |
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| 347. |
A string of length 0.4 m and mass `10 ^(-2)` kg is tightly clamped at its ends. The tension in the string is 1. 6 N. identical wave pulses are produced at one end at equal intervals of time `Delta t`. The value of `Delta t` which allows construction tnterference betwenn successive pulses is |
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Answer» Correct Answer - 1 |
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| 348. |
A wave travelling in the `+ve` x-direction having displacement along y-direction as 1`m`, wavelength `2pi` m and frequency of `1//pi` Hz is represented byA. `y= sin (x-2t)`B. `y = sin (2pi x-2pi t)`C. `y = sin (10 pi x-20 pi t)`D. `y = sin (2 pi x+2pi t)` |
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Answer» Correct Answer - A Given, `a=1 m` As `y=a sin (kx-omega t)=sin ((2pi)/(2pi)x-2pi xx1/2 t)` `=sin (x-2t)` |
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| 349. |
A hypothetical pulse is travelling along positive x direction on a taut string. The speed of the pulse is `10 cm s^(–1)`. The shape of the pulse at t = 0 is given as `{:(y=x/6+1,"for",-6 lt x le 0),(=-x+1,"for",0 le x lt1),(=0,"for all other values of x",):}` x and y are in cm. (a) Find the vertical displacement of the particle at x = 1 cm at t = 0.2 s (b) Find the transverse velocity of the particle at x = 1 cm at t = 0.2 s. |
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Answer» Correct Answer - (a) `5/6 cm` (b) `-5/3 cm s^(-1)` |
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| 350. |
A sinusoidal wave travels along a taut string of linear mass density `0.1 g//cm`. The particles oscillate along y-direction and the disturbance moves in the positive x-direction. The amplitude and frequency of oscillation are `2` mm and `50` Hz respectively. The minimum distance between two particles oscillating in the same phase is 4 m. (a) Find the tension in the string. (b) Find the amount of energy transferred through any point of the string in one second. (c) If it is observed that the particle at `x = 2 m` is at `y = 1` mm at `t = 2 s`, and its velocity is in positive y-direction, then write the equation of this travelling wave. |
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Answer» Correct Answer - (a) `400 N` (b) `pi^(2)/25 J` (c) `y=2 sin ((pix)/2-100 pit-30^(@))` |
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