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JEE • PHYSICS • WAVES

Waves practice · Medium

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Practice format and chapter quiz

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 524 questions

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361 Medium
361 of 524
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.
362 Medium
362 of 524
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.
363 Medium
363 of 524
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.
364 Medium
364 of 524
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.
365 Medium
365 of 524
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.
366 Medium
366 of 524
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.
367 Medium
367 of 524
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.
368 Medium
368 of 524
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.
369 Medium
369 of 524
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.
370 Medium
370 of 524
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.
371 Medium
371 of 524
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.
372 Medium
372 of 524
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.
373 Medium
373 of 524
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.
374 Medium
374 of 524
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.
375 Medium
375 of 524
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.
376 Medium
376 of 524
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.
377 Medium
377 of 524
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.
378 Medium
378 of 524
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.
379 Medium
379 of 524
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.
380 Medium
380 of 524
Understanding +4 | -1
For which of the following do the longitudinal waves exist?
View answer and explanation
Correct option: D
Both (2) and (3)
Option D is correct. Longitudinal waves require a medium to propagate. Both air and water can support longitudinal waves (sound waves).
381 Medium
381 of 524
Remembering +4 | -1
Transverse waves can propagate in
View answer and explanation
Correct option: B
Solids
Option B (Solids) is correct because transverse waves require a medium with shear strength to propagate, which is present in solids.
382 Medium
382 of 524
Application +4 | -1
The frequency of a note emitted by a source changes by 20% as it approaches an observer. As it recedes away him, the apparent frequency will differ from the actual frequency by
View answer and explanation
Correct option: C
0.143
Option C (14.30%) is correct because the apparent frequency changes as the source approaches and recedes, and the percentage difference is calculated using the Doppler effect formula.
383 Medium
383 of 524
Understanding +4 | -1
Which of the following properties of a wave does not change with a change in medium?
View answer and explanation
Correct option: A
Frequency
Option A (Frequency) is correct because the frequency of a wave remains constant when it travels from one medium to another.
384 Medium
384 of 524
Application +4 | -1
Two sources AA and BB are sounding notes of frequency 680 Hz680 \mathrm{~Hz}. A listener moves from AA to BB with a constant velocity uu. If the speed of sound is 340 ms−1340 \mathrm{~ms}^{-1}, what must be the value of uu, so that he hears 10 beats per second?
View answer and explanation
Correct option: B
2.5 ms−12.5 \mathrm{~ms}^{-1}
Option B (2.5 ms−12.5 \mathrm{~ms}^{-1}) is correct because it's the velocity required for the listener to hear 10 beats per second, calculated using the Doppler effect formula.
385 Medium
385 of 524
Application +4 | -1
Two sound waves travel in the same direction in a medium. The amplitude of each wave is AA and the phase difference between the two waves is 120∘120^\circ. The resultant amplitude will
View answer and explanation
Correct option: D
AA
Option D (AA) is correct because the resultant amplitude is calculated using the superposition principle, considering the phase difference between the two waves.
386 Medium
386 of 524
Remembering +4 | -1
A travelling wave reflected at an open boundary undergoes a phase change of
View answer and explanation
Correct option: B
0
Option B (0) is correct because when a travelling wave is reflected at an open boundary (rarer medium), there is no phase change.
387 Medium
387 of 524
Understanding +4 | -1
Statement-I: In stationary wave, all the particles of the medium between two nodes vibrate in same phase with same frequency but with different amplitude.
Statement-II: In stationary wave, every particle of the medium vibrates with its own energy and it does not share or transmit it to any other particle.
View answer and explanation
Correct option: A
Statement-I is correct
Option A (Statement-I is correct) is correct because in a stationary wave, particles between two nodes vibrate in phase with the same frequency but different amplitudes.
388 Medium
388 of 524
Understanding +4 | -1
The equation of a spherical progressive wave is
View answer and explanation
Correct option: D
y=arsin⁡(ωt−kr)y = \frac{a}{r}\sin \left( {\omega t - kr} \right)
Option D (y=arsin⁡(ωt−kr)y = \frac{a}{r}\sin \left( {\omega t - kr} \right)) is correct because the amplitude of a spherical wave decreases with distance as 1/r.
389 Medium
389 of 524
Application +4 | -1
A source of sound vibrates according to the equation y=0.5  cos⁡(πt).y = 0.5\,\,\cos \left( {\pi t} \right). If the wave velocity is 1.5   ms−1,1.5\,\,\,{\text{m}}{{\text{s}}^{ - 1}}, the wave length is
View answer and explanation
Correct option: C
3 m
Option C (3 m) is correct because it is calculated using the formula λ=vn\lambda = \frac{v}{n}, where n is derived from the given equation.
390 Medium
390 of 524
Understanding +4 | -1
The reason for introducing Laplace correction in the expression for the velocity of sound in a gaseous medium is:
View answer and explanation
Correct option: B
No change in the heat of the medium during the propagation of the sound through it
Option B (No change in the heat of the medium during the propagation of the sound through it) is correct because Laplace's correction assumes adiabatic conditions where no heat exchange occurs during sound propagation.
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