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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.
| 451. |
minimum frequency of audible sound is 20 Hz and maximum frequency is 20 , 000 Hz. If we compare the sound levels of these two frequencies of same amplitudes, their difference will beA. 30 dBB. 60 dBC. 90 dBD. 120 dB |
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Answer» Correct Answer - B |
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| 452. |
Two waves ` y_(1) =A_(1) sin (omega t - beta _(1)), y_(2)=A_(2) sin (omega t - beta_(2)` Superimpose to form a resultant wave whose amplitude isA. ` sqrt (A_1 +A_2^(2) + 2A_1A_2 cos (beta _1 -beta __2))`B. ` sqrt ( A_1^(2) +A_2^(2) +2A _1A_2 sin (beta _1-beta _2))`C. ` |A_1-A_2|`D. ` |A_1 +A_2|` |
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Answer» Correct Answer - A |
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| 453. |
Two sources of intensity I and 4 are used in an interference experiment. Find the intensity at point where the waves from two sources superimpose with a phase difference (i) zero (ii) `pi//2` and (iii) `pi`.A. 9 IB. 5 IC. ID. 0 |
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Answer» Correct Answer - B `I = I_(1) + I_(2) + 2 sqrt(I_(1)I_(2)) "cos" (pi)/(2)` `= I + (4I) + 2 sqrt(I_(1)I_(2))xx 0 = 5I` |
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| 454. |
Two waves of wavelength 2 m and 2.02 m , with the same speed, superimpose to produce 2 beats per second ,The speed of each wave isA. 400 m/sB. 404 m/sC. 402 m/sD. 406 m/s |
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Answer» Correct Answer - B As `lamda_(2) gt lamda_(1) " " :. n_(1) - n_(2) = 2` `(v)/(lamda_(1)) - (v)/(lamda_(2)) = 2 " " :. v [(1)/(2) - (1)/(2.02)] =2` `:. v [(2.02 -2)/(2 xx 2.02)] = 2` `:. v = (2 xx 2 xx 2.02)/(0.02) = 404 m//s` |
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| 455. |
The intensity of sound at a point A is `10^(-4) Wm^(-2)` and that at point B is `10^(-8) Wm^(-2)`.What is the difference between the intensity levels at A and B ? |
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Answer» Correct Answer - 40 dB |
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| 456. |
Two waves of wavelength 2 m and 2.02 m , with the same speed, superimpose to produce 2 beats per second ,The speed of each wave isA. ` 400 ms ^(-1)`B. ` 404 ms ^(-1)`C. ` 402 ms ^(-1)`D. ` 406 ms ^(-1)` |
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Answer» Correct Answer - A |
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| 457. |
Find the intensity level of a sound wave that has an intensity of `10^(-5) Wm^(-2)`. |
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Answer» Correct Answer - 70 dB |
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| 458. |
Three sound waves of equal amplitudes have frequencies (n-1) ,n (n+1) .They superimpose to give beats.The number of beats produced per second will beA. 2B. 1C. 4D. 3 |
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Answer» Correct Answer - A |
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| 459. |
A line source emits a cylindrical expanding wave. Assuming the medium absorbs no energy find how the ampitude and intensity of wave depend on the distance from the source? |
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Answer» Correct Answer - A Suppose power of linre source is P. Then, at distance r, surface area is `2pirl`. `:. l = (P)/(S) =(P)/(2pirl)` or `I prop (l)/( r )` Further, `I prop A^(2)` `:. A prop (1)/(sqrt(r))` |
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| 460. |
A tuning fork A produces 4 beats/ s with tuning fork, B of fequency 256 Hz. When the fork A is filled beats are found to occurs at shorter intervals, then the original frequency will beA. 252 HzB. 260 HzC. 256 HzD. 262 Hz |
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Answer» Correct Answer - B After filling the fork, the frequency of the fork increases. |
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| 461. |
For energy density, power and intensity of any wave choose the correct options.A. `u = "energy density" = (1)/(2)rhoomega^(2)A^(2)`B. `P = "power" = (1)/(2)rhoomega^(2)A^(2)Sv`C. `I = "intensity"= (1)/(2)rhoomega^(2)A^(2)Sv`D. `I = (P)/(S)` |
| Answer» Correct Answer - A::D | |
| 462. |
A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibreting string of a piano.The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased.The frequency of the piano string before increasing the tension wasA. 256+5HzB. 256+2HzC. 256-2HzD. 256-5Hz |
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Answer» Correct Answer - A |
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| 463. |
A tuning fork whose frequency is given by the manufacturer as 512 Hz is being tested using an accurate oscillator.It is found that they produce 2 beats per second,when the oscillator reads 514 Hz and 6 beats per second,when it reads 510 Hz.The actual frequency of the fork isA. 508 HzB. 512 HzC. 516 HzD. 518 Hz |
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Answer» Correct Answer - D |
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| 464. |
The string of a musical instrument was being tuned using a tuning fork of known frequency, `f_(0) = 1024 Hz`. The tuning fork and the string were set to vibrate together. Both vibrated together for `10 s` and no beat was heard. What prediction can be made regarding the frequency of the string? |
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Answer» Correct Answer - `1023.9 Hz lt f_("string") lt 1024.1 Hz` |
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| 465. |
At any instant a wave travelling along the string shown in figure. Here, point A is moving upward. Which of the following statement is true? A. The wave is travelling to the rightB. The displacement amplitude of wave is equal to displacement of B at this instantC. At this instant C also directed upward.D. None of the above |
| Answer» Correct Answer - B | |
| 466. |
(i) A harmonic wave in a stationary medium is represented by `y = a sin (kx – omegat)`. Write the equation of this wave for an observer who is moving in negative `x` direction with constant speed `v_(0)`. (ii) The Doppler flow meter is a device that measures the speed of blood flow, using transmitting and receiving elements that are placed directly on the skin. The transmitter emits a continuous sound wave whose frequency is `5 M Hz`. When the sound is reflected from the red blood cells, its frequency is changed in a kind of Doppler effect. The cells are moving with the same velocity as the blood. The receiving element detects the reflected sound, and an electronic counter measures its frequency, which is Doppler- shifted relative to the transmitter frequency. From the change in frequency the speed of the blood flow can be determined. Typically, the change in frequency is around `600 Hz` for flow speeds of about `0.1 m//s`. Assume that the red blood cell is directly moving away from the source and the receiver. (a) Estimate the speed of the sound wave in the blood? (b) A segment of artery is narrowed down by plaque to half the normal cross-sectional area. What will be the Doppler change in frequency due to reflection from the red blood cell in that region? |
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Answer» Correct Answer - (i) `y=a sin [(omega_kv_(0))t-kx]` (ii) (a) 1700 m/s (b) 1200 Hz |
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| 467. |
A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y_(2)=-A sin(kx-omegat)`B. `y_(2)=-A cos(kx-omegat)`C. `y_(2)=A sin(kx-omegat)`D. `y_(2)=A cos(kx-omegat)` |
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Answer» Correct Answer - B |
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| 468. |
A block of mass `M = 2kg` is suspended from a string AB of mass `6 kg` as shown in figure. A transverse wave pulse of wavelength `lambda_(0)` is produced at point B. find its wavelength while reaching at point A. |
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Answer» Correct Answer - A::B::D `lambda =(v)/(f) =sqrt(T//mu)/(f)` or `lambda prop sqrt(T)` (as `mu` and if are constants) `:. (lambda_(B))/(lambda_(A))= sqrt((T_(B))/(T_(A))` `:. lambda_(B) = (sqrt(T_(B)/(T_(A)))) lambda_(A)` `= [sqrt(((2+ 6)g)/(2g))]` `=2 lambda` |
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| 469. |
In a wave motion `y= a sin(kx-omegat)`, y can represent: (a) electric field , (b) magnetic field , (c) displacement , (d) pressure |
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Answer» `(a,b,c,d)` In case of sound wave,y can represent pressure and displacement , which in the case of electromagnetic wave, it represent electric and magnetic fields. |
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| 470. |
If wave y = A cos `(omegat + kx)` is moving along x-axis The shape of pulse at t = 0 and t = 2 sA. are differentB. are sameC. may not be sameD. None of these |
| Answer» Correct Answer - B | |
| 471. |
At t=0, a transverse wave pulse in a wire is described by the function `y=6//(x^(2)-3)` where x and y are in metres. The function y(x,t) that describes this wave equation if it is travelling in the positive x direction with a speed of `4.5 m//s` isA. `y=(6)/((x+4.5t)^(3)-3)`B. `y = (6)/((x-4.5 t^(2))+3)`C. `y = (6)/((x+4.5 t)^(2)-3)`D. `y = (6)/((x-4.5 t)^(2)-3)` |
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Answer» Correct Answer - C `y (x, t)=(a)/((x pm vt)^(2)+b)` is another form of progressive wave equation propagating with a speed v. Negative sign to be taken for propagation along + x-axis and positive sign to be taken for propagation along - x-axis. |
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| 472. |
The equation of a wave travelling on a stretched string along the x-axis is `y = ae^(-(bx+ct))`. The direction of propagation of wave isA. along negative y-axisB. along positive y-axisC. along negative x-axisD. along positive x-axis |
| Answer» Correct Answer - C | |
| 473. |
Prove that the equation `y = a sin omegat` does not satisfy the wave equation and hence it does not represent a wave. |
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Answer» `(del^(2)y)/(delt^(2)) = -omega^(2) a sin omegat` and `(del^(2)y)/(delx^(2)) = 0` Since, `(del^(2)y)/(delt^(2)) != (del^(2)y)/(delx^(2))` (constant) Hence, the given equation does not represent a wave equation. |
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| 474. |
A wave pulse is described by `y(x, t) = ae^-(bx - ct)^(2)`, where a,b,and c are positive constants. What is speed of this wave? |
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Answer» Correct Answer - B::C Speed of wave `= ("Coefficient of `t`")/("Coefficient of `x`" ) = (c)/(b)` |
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| 475. |
There are three sources of sound of equal intensity with frequencies 400, 401 and 402 vib/sec . The number of beats heard per second isA. zeroB. 1C. 2D. 4 |
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Answer» Correct Answer - C `n_(1)=401 Hz, n_(2) =402` Hz and `n_(3)=403` Hz `:.` Beats `=n_("max")-n_("min")` `=403-401=2 sec^(-1)` |
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| 476. |
Two tunning forks A and B produce notes of frequencies 256 Hz & 262 Hz respectively. An unknown note sounded at the same time with A produce beats. When the note is sounded with B, beats frequency is twice as large. The unknown frequency could beA. 256 HzB. 254 HzC. 300 HzD. 280 Hz |
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Answer» Correct Answer - B `n_(A)=258` Hz and `n_(B)=262` Hz Let n is the frequency of unknown tuning fork. It produces x beats with 258 and 2x with 262, `implies 262-(258-x) =2x` `implies 262-268+x=2x` `implies x=4` `implies n=254 Hz` |
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| 477. |
A and B are two wires whose fundamental frequencies are 256 and 382 Hz respectively. How many beats in 2 seconds will be heard by the third harmonic of A and second harmonic of B? |
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Answer» Correct Answer - 8 |
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| 478. |
A person is talking in a small room and the sound intensity level is 60dB everywhere within the room. If there ar eight people talking simultaneously in the room, what is the sound intensity level?A. 96dBB. 69dBC. 74dBD. 81dB |
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Answer» Correct Answer - B |
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| 479. |
The equation of a transverse wave is given by `y=10 sin pi (0.01 x -2t )` where x and y are in cm and t is in second. Its frequency is |
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Answer» Correct Answer - 10 cm, 1 Hz , 200 cm `s^(-1)` , 200 cm |
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| 480. |
Equation of a plane wave is given by `4 sin .(pi)/(4)[2t+(x)/(8)]`. The phase difference at any given instant of two particles 16 cm apart isA. `60^(@)`B. `90^(@)`C. `30^(@)`D. `120^(@)` |
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Answer» Correct Answer - B `y = 4 sin. (pi)/(4) (2t+(x)/(8)) " "….(i)` or `y = 4 sin((pi)/(2)t + (pix)/(32))" "….(ii)` The standard equation is `y = 4 sin (omega t pm kx)` Comparing the Eq. (i) with Eq. (ii), we get ` k = (pi)/(2)` `therefore " "(2pi)/(lambda)=(pi)/(32)` `therefore " "(lambda)/(2)=32` `therefore " "lambda = 64` `therefore " "Delta phi = (2pi)/(lambda)Deltax` `= (2pi)/(64) xx 16 = (pi)/(2) = 90^(@)` |
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| 481. |
Two very long string are tial together at the point x = 0 In region x lt 0, the wave speed is `v_(1)`,while in the regon x gt0, the speed is `v_(2)`. A sinusoidal wave is incident on the knot from the left(x lt 0). Part of the wave is redlected and part is transmitted. For X lt 0 the the displacement of the wave is described by y(x,t) = A sin(`K_(1)x-wt`)+B sin (`k_(1)x+wt)`, while for x gt 0, y(x,t)=Csin (`K_(1))x-wt`, where`w//k_(1) =v_(1)"and "w//k_(2) = v_(2)`. Which of the following is /are correct .A. `C/A=(2v_(2))/(v_(1)+v_(2))`B. `B/A=(v_(2)-v_(1))/(v_(1)+v_(2)`C. `B^(2)+(v_(1))/(v_(2))C^(2)=A^(2)`D. A^(2)+(v^(1))/(v_(2))C^(2)=B^(2)` |
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Answer» Correct Answer - A::B::C |
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| 482. |
Shape of a string transmitting wave along x-axis some instant is shown. Velocity of point P v - (`4pi`)cm /s (`theta`) = `tan^(-1) (0.004`pi`) A. Amplitube of wave is 2 mmB. Velocity of wave is 10 m/sC. Maximum acceleration of particle is `80pi^(2)cm//sec^(2)`D. Wave is travelling in negative x-direction |
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Answer» Correct Answer - A::B::C::D |
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| 483. |
A waves undergoes reflection at a rigid wall The parameter changed during this reflection isA. frequencyB. wavelengthC. amplitudeD. velocity |
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Answer» Correct Answer - D |
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| 484. |
A sonometer wire of length 114 cm is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio `1 : 3 : 4`?A. At 36 cm and 84 cm from one endB. At 24 cm and 72 cm from one endC. At 48 cm and 96 cm from one endD. At 72 cm and 96 cm from one end |
| Answer» Correct Answer - C | |
| 485. |
the fundamental frequency of a sonometer wire of length is `f_(0)`.A bridge is now introduced at a distance of `Deltal`from the centre of the wire `(Deltal lt lt l )`. The number of beats heard if their fundamental mode areA. `(8f_(0)Delta|)/|`B. `(f_(0)Delta|)/|`C. `(2f_(0)Delta|)/|`D. `(4f_(0)Delta|)/|` |
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Answer» Correct Answer - A |
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| 486. |
Two speakeer connected to the same source of fixed frequency are placed 2m apart in a box. A sensitive microphone placed at a distance of 4m from the midpoint alon the perpendicular bisector shown maximum response. The box is slowly rotated till the speaker are in line with the microphone, The distance between the midpoint of the speakers and the microphone remains unchanged. Exactly 5 maximum responses (inculuding the initial and last one) and observed in the microphone in doing this. The wavelength of the sound wave is (o.x) meter. Find the value of x. |
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Answer» Correct Answer - 5 |
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| 487. |
A sensitive microphone with its receiving surface turned towards a long vertical wall is placed at a distance of 2 m from the wall. A strong source of sound of `500 Hz` is placed between the wall and microphone on the line perpendicular to the wall and passing through the position of microphone. Find the position of the source where no sound will be heard in the microphone. (Velocity of sound in air `=350 m s^(-1)`) |
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Answer» Correct Answer - at distance `0.175 m, 0.525 m, 0.875m, 1.225m, 1.575 m` and `1.925 m` from the wall. |
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| 488. |
A wave train of a plane wave with wavelength 1.8 cm, travels from deep water into a shall water. Then the velocity of waves on surface shallow water, if its wavelength in shallow wave is 1 cm, is (The velocity of waves in surface deep water is 36 cm/s)A. 35 cm/sB. 20 cm/sC. 64.8 cm/sD. 37.4 cm/s |
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Answer» Correct Answer - B `(v_(s))/(v_(d)) = (lamda_(s))/(lamda_(d)) = (1)/(1.8)` `v_(s) = (v_(d))/(1.8) = (36)/(1.8) = (360)/(1.8) = 20 cm//s` |
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| 489. |
The distance between two consecutive crests in a wave train produced in string is 5 m. If two complete waves pass through any point per second, the velocity of wave is :-A. 10 cm/sB. 2.5 cm/sC. 5 cm/sD. 15 cm/s |
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Answer» Correct Answer - A `lamda = 5 cm, n =2` `v = n lamda = 2 xx 5 = 10 cm//s` |
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| 490. |
A travelling wave passes a point of observation. At this point, the time interval between successive crests is 0.2 seconds andA. wavelength is 5 m.B. frequency is 5HzC. velocity of propagation is 5m/sD. wavelength is 0.2 m |
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Answer» Correct Answer - B |
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| 491. |
The quality of a note changes when change occurs inA. pitchB. loudnessC. nature of overtoneD. waveforms |
| Answer» Correct Answer - D | |
| 492. |
The speed of a periodic wave is the product of itsA. wavelength and periodB. wavelength and frequencyC. period and frequencyD. amplitude and frequency |
| Answer» Correct Answer - B | |
| 493. |
Wave motion is periodic inA. spaceB. space and timeC. timeD. direction |
| Answer» Correct Answer - B | |
| 494. |
Oscillatory disturbance travelling through the medium isA. energyB. momentumC. waveD. wave motion |
| Answer» Correct Answer - C | |
| 495. |
Dimensions of wavelength of progressive wave isA. ` [M^(0) L^(0) T^(-1)]`B. ` [M^(0) L^(-1) T^(0)]`C. ` [M^(-1) L^(0) T^(0)]`D. ` [M^(0) L^(0) T^(0)]` |
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Answer» Correct Answer - D |
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| 496. |
A stone is dropped on the surface of water in pond. Name the type of waves produced.A. transverseB. stationaryC. longitudinalD. electromagnetic waves |
| Answer» Correct Answer - A | |
| 497. |
Which of the following in NOT the characteristic of the progressive wave?A. All the vibrating particles of the medium have different amplitudes and frequencyB. State of oscillation change from particles to particleC. For is propagation,medium has elasticity and inertiaD. The form of wave repeats itself at equal intervals |
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Answer» Correct Answer - A |
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| 498. |
The dimensions of the propagation constant of the wave isA. `[L^(1) M^(0) T^(0)]`B. `[L^(0) M^(1) T^(1)]`C. `[L^(1) M^(1) T^(0)]`D. `[L^(-1) M^(0) T^(0)]` |
| Answer» Correct Answer - D | |
| 499. |
The propagation of wave through the medium is possible only when the medium hasA. properly of elasticityB. inertial propertyC. low frictional resistanceD. all of the above |
| Answer» Correct Answer - D | |
| 500. |
A medium can carry a longitudinal wave because it has the propertyA. Yong modulusB. modulus of elasticityC. Bulk modulusD. all the modulus |
| Answer» Correct Answer - C | |