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JEE • PHYSICS • WAVES

Waves practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-390 of 543 questions

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361 Medium
361 of 543
Analyzing +4 | -1
Two stationary sources of sound, S1S_1 and S2S_2 having an equal frequency are fixed some distance apart. The position A is left of S1S_1, position B in the middle of the two sources and position C is to the right of S2S_2. An observer starts moving with velocity V0V_0 from position A towards S1S_1, then
View answer and explanation
Correct option: C
beats will be not heard for A and C but will be heard for B
Option C (beats will be not heard for A and C but will be heard for B) is correct because at A and C, the observer is moving directly towards or away from both sources, while at B, the motion is towards one source and away from the other.
362 Medium
362 of 543
Understanding +4 | -1
The intensity of harmonic wave:
View answer and explanation
Correct option: C
Depends upon both, its frequency and amplitude
Option C is correct because the intensity of a harmonic wave depends on both its frequency and amplitude.
363 Medium
363 of 543
Application +4 | -1
A pipe 17 cm17 \mathrm{~cm} long is closed at one end. Which harmonic mode of the pipe resonates a 1.5kHz1.5 \mathrm{kHz} source ? (\left(\right. Speed of sound in air =340 m s−1)\left.=340 \mathrm{~m} \mathrm{~s}^{-1}\right) :-
View answer and explanation
Correct option: B
Third
Option B (Third) is correct because the calculation of the harmonic mode using the given frequency, speed of sound, and pipe length results in n=3.
364 Medium
364 of 543
Understanding +4 | -1
Distance between a compression and adjoining rarefaction in a pressure wave is
View answer and explanation
Correct option: B
λ2\frac{\lambda}{2}
Option B (λ2\frac{\lambda}{2}) is correct because the distance between a compression and an adjacent rarefaction is half the wavelength of the pressure wave.
365 Medium
365 of 543
Understanding +4 | -1
When an air column of length ‘I’ open at both ends produces its fundamental note, the wavelength is given by
View answer and explanation
Correct option: B
λ=2l\lambda = 2l
Option B (λ=2l\lambda = 2l) is correct because the fundamental frequency in an open pipe has a wavelength twice the length of the air column.
366 Hard
366 of 543
Application +4 | -1
A sound has an intensity of 2×10−8 W/m22 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2. Its intensity level (in decibel) is :

(log⁡102=0.3) :- \left(\log _{10} 2=0.3\right) \text { :- }
View answer and explanation
Correct option: A
43
Option A (43) is correct because it is the intensity level calculated using the formula L = 10 log₁₀(I/I₀), where I₀ is the threshold of hearing (10⁻¹² W/m²).
367 Medium
367 of 543
Understanding +4 | -1
Assertion: In standing wave pattern particle of medium between two consecutive nodes vibrates in same phase but with different amplitude.
Reason: In stationary wave, the amplitude of vibration does not depends on the position of the particle.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because the assertion about standing waves is true, but the reason stating the amplitude doesn't depend on position is false; the amplitude varies with position in a standing wave.
368 Medium
368 of 543
Application +4 | -1
A transverse sinusoidal wave of amplitude AA, wavelength λ\lambda and frequency nn is travelling on a stretched string. The maximum speed of particle is (1/10)(1 / 10) th the speed of propagation of the wave. If A=10−3 mA=10^{-3} \mathrm{~m} and v=10 ms−1v=10 \mathrm{~ms}^{-1}, then λ\lambda and nn are given by
View answer and explanation
Correct option: A
λ=2π×10−2 m\lambda=2 \pi \times 10^{-2} \mathrm{~m}
Option A (λ=2π×10−2 m\lambda=2 \pi \times 10^{-2} \mathrm{~m}) is correct because it is the wavelength calculated using the given amplitude, frequency, and wave speed.
369 Medium
369 of 543
Understanding +4 | -1
An observer is moving towards the stationary source of sound, then
View answer and explanation
Correct option: B
Apparent frequency will be greater than the real frequency
Option B (Apparent frequency will be greater than the real frequency) is correct because the Doppler effect causes an increase in frequency when the observer moves towards the source.
370 Medium
370 of 543
Understanding +4 | -1
When a longitudinal wave propagates through a medium, the particles of the medium execute simple harmonic oscillations about their mean positions. These oscillations of a particle are characterised by an invariant
View answer and explanation
Correct option: C
Sum of kinetic energy and potential energy
Option C is correct because the total energy (sum of kinetic and potential energy) remains constant during simple harmonic oscillations.
371 Medium
371 of 543
Application +4 | -1
A whistle giving out 450  Hz450\,\,Hz approaches a stationary observer at a speed of 33  ms−133\,\,\mathrm{m}{\mathrm{s}^{ - 1}}. The frequency heard by the observer in HzHz is [velocity of sound in air =333  ms−1333\,\,\mathrm{m}{\mathrm{s}^{ - 1}}
View answer and explanation
Correct option: D
500
Option D (500) is correct because it is the frequency heard by the observer calculated using the Doppler effect formula when the source is moving towards the observer.
372 Medium
372 of 543
Understanding +4 | -1
When a sound source is moving towards a stationary observer :
View answer and explanation
Correct option: D
apparent frequency is more than actual frequency and wavelength is less than actual wavelength
Option D is correct because when a sound source moves towards a stationary observer, the apparent frequency increases (higher pitch) and the apparent wavelength decreases.
373 Medium
373 of 543
Understanding +4 | -1
Assertion (A): A driver under water is unable to hear the sound produced in air.
Reason (R): The sound produced in air undergoes reflection at the water surface.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A is correct because sound waves are reflected at the water surface, preventing the diver from hearing sounds produced in the air.
374 Medium
374 of 543
Analyzing +4 | -1
Two identical strings AA and BB, have nearly the same tension. When they both vibrate in their fundamental resonant modes, there is a beat frequency of 3 Hz3 {~Hz}. When string BB is tightened slightly, to increase the tension, the beat frequency becomes 6 Hz6 {~Hz}. This means
View answer and explanation
Correct option: D
That before and after tightening BB has higher frequency than AA
Option D is correct because increasing the tension in string B increases its frequency, and since the beat frequency increased, B must have initially had a higher frequency than A.
375 Medium
375 of 543
Understanding +4 | -1
Learned Indian classical vocalists do not like the accompaniment of a harmonium because
View answer and explanation
Correct option: D
Tempered scale is used in the harmonium
Option D is correct because harmoniums use a tempered scale, which has fixed intervals that do not perfectly align with the precise intervals preferred in Indian classical music.
376 Medium
376 of 543
Understanding +4 | -1
Assertion: Description of sound as pressure wave is preferred over displacement wave.
Reason: Sound sensors (ear or mike) detect pressure changes.
View answer and explanation
Correct option: A
If both Assertion & reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because sound is better described as a pressure wave, and sound sensors indeed detect pressure changes, making the reason a valid explanation.
377 Medium
377 of 543
Application +4 | -1
The linear density of a vibrating string is 1.3×10−4 kg m−1.1.3 \times 10^{-4} \mathrm{~kg ~m}^{-1}. A transverse wave is propagating on the string and its described by the equation y=0.021sin⁡(x+30t)y=0.021 \sin (x+30 t), where xx and yy are measured in mm and tt in ss. The tension is the string is
View answer and explanation
Correct option: A
0.12 N0.12 \mathrm{~N}
Option A (0.12 N) is correct because the tension in the string is calculated using the wave speed formula and the given linear density.
378 Medium
378 of 543
Analyzing +4 | -1
Assertion: Doppler formula for sound wave is symmetric with respect to the speed of source and speed of observer.
Reason: Motion of source with respect to stationary observer is equivalent to the motion of an observer with respect to stationary source.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D (If both assertion and reason are false) is correct. The Doppler formula is asymmetric, and the motion of source and observer are not equivalent.
379 Medium
379 of 543
Understanding +4 | -1
The characteristics of a sound with the help of which we can distinguish between shrill note and a grave note is
View answer and explanation
Correct option: B
Pitch
Option A (All the resistances are equal) is correct because a Wheatstone bridge is most sensitive when all its resistances are of the same order of magnitude, ideally equal.
380 Medium
380 of 543
Application +4 | -1
Two sound waves of wavelengths 2 m2 \mathrm{~m} and 2.02 m2.02 \mathrm{~m} produce 15 beats in 3 seconds in a medium. The velocity of sound in that medium is
View answer and explanation
Correct option: B
1010 m/s
The correct answer is B (1010 m/s) because the beat frequency is the difference between the two frequencies, which can be used to calculate the velocity of sound.
381 Medium
381 of 543
Understanding +4 | -1
The phase change between incident and reflected sound wave from a free end is
View answer and explanation
Correct option: A
0
Option A is correct because the phase change between incident and reflected sound wave from a free end is 0.
382 Medium
382 of 543
Understanding +4 | -1
An open pipe of length ll vibrates in fundamental mode. The pressure variation is maximum at
View answer and explanation
Correct option: B
The middle of pipe
Option B (The middle of pipe) is correct because in an open pipe vibrating in the fundamental mode, the pressure variation is maximum at the middle where a node is formed.
383 Medium
383 of 543
Understanding +4 | -1
The principle of superposition in wave motion tells that in a motion in which two or more waves are simultaneously producing their displacement in a particle along the same line, then the resultant
View answer and explanation
Correct option: C
Displacement is the sum of the individual displacements
Option C is correct because the principle of superposition states that the resultant displacement is the sum of individual displacements.
384 Medium
384 of 543
Understanding +4 | -1
Statement-I: A transverse wave is travelling in a string. Equation of the wave is equal of shape of the string at an instant t{t}.
Statement-II: At a point beats frequency of nHzn{Hz} is observed. It means, at that point, zero intensity is observed 2n2 {n} times per second.
View answer and explanation
Correct option: A
Statement-I is correct
Option A (Statement-I is correct) is correct because the equation of a transverse wave represents the shape of the string at a given instant.
385 Medium
385 of 543
Application +4 | -1
An aluminium rod having a length of 90.0 cm90.0 \mathrm{~cm} is clamped at its middle point and is set into longitudinal vibrations by stroking it with a rosined cloth. Assume that the rod vibrates in its fundamental mode of vibration. The density of aluminium is 2600 kg m−32600 \mathrm{~kg} \mathrm{~m}^{-3} and its Young's modulus is 7.80×1010 N m−27.80 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}. Find (a) the speed of sound in aluminium, (b) the wavelength of sound waves produced in the rod, (c) the frequency of the sound produced and (d) the wavelength of the sound produced in air. Take the speed of sound in air to be 340 ms−1340 \mathrm{~ms}^{-1}.
View answer and explanation
Correct option: A
5480 ms−1; 180 cm; 3050 Hz; 11.1 cm5480 \mathrm{~ms}^{-1};\,180\mathrm{~cm};\,3050 \mathrm{~Hz};\,11.1 \mathrm{~cm}
Option A (5480 ms−1; 180 cm; 3050 Hz; 11.1 cm5480 \mathrm{~ms}^{-1};\,180\mathrm{~cm};\,3050 \mathrm{~Hz};\,11.1 \mathrm{~cm}) is correct because it accurately calculates the speed of sound, wavelength, and frequency in the aluminum rod and the wavelength in air.
386 Medium
386 of 543
Application +4 | -1
The intensity ratio of two waves is 1:9. The ratio of their amplitudes, is
View answer and explanation
Correct option: B
0.04375
Option A (+ 3.0 D) is correct because it is the power of the lens required to correct the person's near point.
387 Medium
387 of 543
Application +4 | -1
Two sitar strings "A" and "B" playing the note "Dha" are slightly out of tune and produce beats of frequency 5 Hz5 \mathrm{~Hz}. The tension of the string " B\mathrm{B} " slightly increased and the beat frequency is found to decrease to 3 Hz3 \mathrm{~Hz}. What is the original frequency of " B\mathrm{B} " in hertz, if the frequency of " A\mathrm{A} " is 427 Hz427 \mathrm{~Hz} ?
View answer and explanation
Correct option: B
422
Option B (422) is correct because the original frequency of string B can be either 432 Hz or 422 Hz, but the condition that increasing the tension decreases the beat frequency implies that the original frequency was lower than that of string A.
388 Medium
388 of 543
Application +4 | -1
In Quincke's experiment, the sound intensity has a minimum value II at a particular position. As the sliding tube is pulled out by a distance of 16.5 mm16.5 \mathrm{~mm}, the intensity increases to a maximum of 9I9 I. Take the speed of sound in air to be 330 ms−1330 \mathrm{~ms}^{-1}. (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.
View answer and explanation
Correct option: A
5.0 kHz; 2
Option A (5.0 kHz; 2) is correct because the frequency is calculated from the wavelength derived from the path difference, and the amplitude ratio is derived from the intensity ratio.
389 Medium
389 of 543
Analyzing +4 | -1
A source of sound gives 55 beats s−1{\mathrm{s}^{ - 1}} when sounded with another source of frequency 100  Hz100\,\,Hz. The second harmonic of the source together with a source of frequency 205  Hz205\,\,Hz gives 5  beats  s(−1)5\,\,beats\,\,{s^{( - 1)}}. What is the frequency of the source?
View answer and explanation
Correct option: A
105Hz105\mathrm{ }Hz
Option A (105Hz105\mathrm{ }Hz) is correct because it satisfies both beat frequency conditions given in the problem.
390 Medium
390 of 543
Analyzing +4 | -1
A travelling wave y=Asin⁡(kx−ωt+θ)y=A \sin (k x-\omega t+\theta) passes from a heavier string to a lighter string. The reflected wave has amplitude 0.5 A0.5 \mathrm{~A}. The junction of the strings is at x=0x=0. The equation of the reflected wave is:
View answer and explanation
Correct option: D
y′=−0.5Asin⁡(kx+ωt−θ)y^{\prime}=-0.5 A \sin (k x+\omega t-\theta)
Option D (y′=−0.5Asin⁡(kx+ωt−θ)y^{\prime}=-0.5 A \sin (k x+\omega t-\theta)) is correct because the wave is reflected from a rarer medium, so there is no phase change, and the direction of propagation is reversed.
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