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JEE • PHYSICS • WAVES

Waves practice · Mixed difficulty

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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 271-300 of 543 questions

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271 Medium
271 of 543
Remembering +4 | -1
In the equation y=Asin⁡(kx−ωt+ϕ0),y=A \sin (k x-\omega t+\phi_0), the term phase is defined as
View answer and explanation
Correct option: D
kx−ωt+ϕ0k x-\omega t+\phi_0
Option D (kx−ωt+ϕ0k x-\omega t+\phi_0) is correct because the phase in a wave equation represents the argument of the sinusoidal function, which includes the position, time, and initial phase constant.
272 Medium
272 of 543
Understanding +4 | -1
Statement-I: Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases.
Statement-II: Solid possess elasticity of length as well as shape, but gases possess only Bulk modulus.
View answer and explanation
Correct option: C
Both statement-I &\& II are correct
Option C (Both statement-I &IIarecorrect)iscorrectbecausesolidscansupportbothtypesofwavesduetotheirelasticityinlengthandshape,whilegasesonlysupportlongitudinalwavesduetopossessingonlybulkmodulus.\& II are correct) is correct because solids can support both types of waves due to their elasticity in length and shape, while gases only support longitudinal waves due to possessing only bulk modulus.
273 Medium
273 of 543
Understanding +4 | -1
Which one of the following statements is true-
View answer and explanation
Correct option: B
The sound waves in air are longitudinal while the light waves are transverse
Option B (The sound waves in air are longitudinal while the light waves are transverse) is correct because sound waves are compressional and light waves are electromagnetic.
274 Medium
274 of 543
Application +4 | -1
If the lengths of the first and second resonating air columns are 16.5 cm and 51.5 cm with tuning fork of frequency 512 Hz. Calculate the velocity of sound in air
View answer and explanation
Correct option: C
358.4 m/s
Option C (358.4 m/s) is correct because the velocity of sound is calculated using the difference between the lengths of resonating air columns and the frequency of the tuning fork.
275 Medium
275 of 543
Application +4 | -1
The string vibrates in 4 loops when a 50 g50 \mathrm{~g} weight is placed in the pan of weight 15 g15 \mathrm{~g}. To make the string to vibrate in 6 loops the weight that has to be removed from the pan is
View answer and explanation
Correct option: C
0.036 kg−wt0.036 \mathrm{~kg}-\mathrm{wt}
Option C (0.036 kg-wt) is correct because it accurately calculates the weight that needs to be removed to change the vibration mode of the string.
276 Medium
276 of 543
Application +4 | -1
A stationary source emits a whistle at a frequency of 200 Hz200 \mathrm{~Hz}. If the velocity of propagation of sound is 340 ms−1340 \mathrm{~ms}^{-1}, then the observed frequency, if the observer is moving away from the source at 25 ms−125 \mathrm{~ms}^{-1}, will be
View answer and explanation
Correct option: B
215 Hz215 \mathrm{~Hz}
Option A (185 Hz185 \mathrm{~Hz}) is incorrect. The correct answer is derived from the Doppler effect formula when the observer is moving away from the source. The correct formula to use is f′=f(vv+vs)f' = f \left( \frac{{v}}{{v + v_s}} \right).
277 Medium
277 of 543
Analyzing +4 | -1
Statement-I: An open organ pipe of certain length have the same fundamental frequency as closed organ pipe of half the length
Statement-II: In the case of open organ pipe, at both the ends antinodes are formed, while in the closed organ pipe at one end antinode and at the other end node is formed
View answer and explanation
Correct option: C
Both statement-I &\& II are correct
Option C (Both statement-I & II are correct) is correct because the fundamental frequency of an open pipe is v/2Lv/2L and for a closed pipe is v/4Lv/4L, so for the same frequency, the closed pipe needs to be half the length of the open pipe, and the end conditions are correctly stated.
278 Medium
278 of 543
Understanding +4 | -1
When temperature is increased, the frequency of organ pipe
View answer and explanation
Correct option: A
increases
Option A (increases) is correct because the speed of sound increases with temperature, leading to a higher frequency in an organ pipe.
279 Medium
279 of 543
Application +4 | -1
Sound waves of wavelength λ\lambda travelling in a medium with a speed of v m/sv \mathrm{~m} / \mathrm{s} enter into another medium where its speed is 2v m/s2 v \mathrm{~m} / \mathrm{s}. Wavelength of sound waves in the second medium is
View answer and explanation
Correct option: C
2λ2 \lambda
Option C (2λ2 \lambda) is correct because the wavelength of a sound wave is directly proportional to its speed, and since the speed doubles, the wavelength also doubles.
280 Medium
280 of 543
Application +4 | -1
A transverse wave propagating on a stretched string of liner density 3×10−4kg−m−13 \times {10^{ - 4}}\mathrm{kg} - {\mathrm{m}^{ - 1}} is represented by the equation y=0.2sin⁡(1.5x+60t)y = 0.2\sin \left( {1.5x + 60t} \right) where xx is in meter and tt is in second. The tension in the string (in newton) is
View answer and explanation
Correct option: B
0.48
Option B (0.48 N) is correct because the tension in the string is calculated using the wave equation and the linear density.
281 Medium
281 of 543
Analyzing +4 | -1
Assertion(A): Speed of sound in O2\mathrm{O}_2 gas at temperature TT is vv. If temperature is increased to 2T2 T, the molecular oxygen breaks into atomic oxygen. Then, speed of sound under these conditions will become 2v2 v.

Reason(R): With increase in humidity in air, speed of sound increases.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
Option B is correct because both Assertion and Reason are correct, but the Reason does not explain the Assertion.
282 Medium
282 of 543
Understanding +4 | -1
When sound wave is refracted from air to water, which of the following will remain unchanged?
View answer and explanation
Correct option: C
Frequency
Option C is correct because frequency is a property of the source and does not change during refraction.
283 Medium
283 of 543
Analyzing +4 | -1
Assertion: The velocity of sound in air decreases if the pressure of air decreases at constant temperature.
Reason: According to Laplace's formula as modified by Newton, velocity of sound in air is given by V=γPρ.V = \sqrt {\frac{{\gamma P}}{\rho }}.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the velocity of sound in air increases with pressure at constant temperature, and the given formula is correct.
284 Medium
284 of 543
Analyzing +4 | -1
Given that y=Asin⁡[(2πλ(ct−x))]y = A\sin \left[ {\left( {\frac{{2\pi }}{\lambda }\left( {ct - x} \right)} \right)} \right], where yy and xx are measured in metres. Which of the following statements is true
View answer and explanation
Correct option: A
The unit of λ−1{\lambda ^{ - 1}} is same as that of 2πλ\frac{{2\pi }}{\lambda }
Option A (The unit of λ−1{\lambda ^{ - 1}} is same as that of 2πλ\frac{{2\pi }}{\lambda }) is correct because both represent inverse lengths, thus having the same units.
285 Medium
285 of 543
Understanding +4 | -1
Sound waves are travelling in a medium whose adiabatic elasticity is EE and isothermal elasticity E′E^{\prime}. The velocity of sound waves is proportional to:
View answer and explanation
Correct option: B
E\sqrt{E}
Option B (E\sqrt{E}) is correct because the velocity of sound waves is proportional to the square root of the adiabatic elasticity (E).
286 Medium
286 of 543
Remembering +4 | -1
The speed of sound in gas of density ρρ at a pressure pp is proportional to
View answer and explanation
Correct option: D
pρ\sqrt {\frac{p}{\mathrm{\rho }}}
Option D (pρ\sqrt {\frac{p}{\mathrm{\rho }}}) is correct because the speed of sound in a gas is proportional to the square root of the pressure divided by the density.
287 Medium
287 of 543
Understanding +4 | -1
In open organ pipe, if fundamental frequency is nn then the other frequencies are
View answer and explanation
Correct option: A
n,2n,3n,4nn,2n,3n,4n
Option A is correct because an open organ pipe produces all harmonics, both even and odd multiples of the fundamental frequency.
288 Medium
288 of 543
Understanding +4 | -1
If we study the vibration of a pipe open at both ends, then which of the following statements is not true ?
View answer and explanation
Correct option: C
Pressure change will be maximum at both ends
Option C (Pressure change will be maximum at both ends) is correct because in an open pipe, the pressure change is minimum at the open ends (nodes of pressure variation).
289 Medium
289 of 543
Application +4 | -1
Intensity level of sound of intensity II is 30  dB30\,\,dB. The ratio II0\frac{I}{{{I_0}}} is (Where I0{I_0} is the threshold of hearing)
View answer and explanation
Correct option: B
10001000
Option B (1000) is correct because it is the ratio of the sound intensity to the threshold of hearing, calculated using the decibel scale formula.
290 Medium
290 of 543
Application +4 | -1
Two trains, one coming towards and another going away from an observer both at 4  m/s4\,\,m/s produce whistle simultaneously of frequency 300  Hz300\,\,Hz. Find the number of beats produced
View answer and explanation
Correct option: C
77
Option C (77) is correct because it is calculated using the Doppler effect formula for both approaching and receding sources, then finding the difference in frequencies.
291 Medium
291 of 543
Understanding +4 | -1
Assertion: Standing wave does not transfer energy in the medium.
Reason: Every particle vibrates with its own energy and it does not share its energy with any other Particle.
View answer and explanation
Correct option: A
If both Assertion & reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & reason are True & the Reason is a correct explanation of the Assertion) is correct because standing waves do not transfer energy, and this is because each particle vibrates with its own energy without sharing it.
292 Medium
292 of 543
Analyzing +4 | -1
The wave function of a pulse is given by y=3(2x+3t)2y=\frac{3}{(2 x+3 t)^2} where xx and yy are in metre and tt is in second.

i) Identify the direction of propagation.

ii) Determine the wave velocity of the pulse.
View answer and explanation
Correct option: A
−1.5 m/s-1.5 \mathrm{~m} / \mathrm{s}
Option A (-1.5 m/s) is correct because the wave function is of the form f(x + vt), indicating propagation along the negative x-axis with a velocity of -1.5 m/s.
293 Medium
293 of 543
Analyzing +4 | -1
For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in metre) of the waves are
View answer and explanation
Correct option: B
11,13,15....\frac{1}{1},\frac{1}{3},\frac{1}{5}....
Option B (11,13,15....\frac{{1}}{{1}},\frac{{1}}{{3}},\frac{{1}}{{5}}....) is correct because these wavelengths satisfy the given conditions for the distance between crests and troughs.
294 Medium
294 of 543
Application +4 | -1
A fork of frequency 340 Hz340 \mathrm{~Hz} is allowed to vibrate just above the open of resonance tube of length 120 cm120 \mathrm{~cm}. If velocity of sound =340 m/s=340 \mathrm{~m} / \mathrm{s}. Find (a) maximum number of resonance (b) maximum and minimum liquid level at resonance condition.
View answer and explanation
Correct option: A
75;95,4575;95,45
Option A (75;95,4575;95,45) is correct because it correctly calculates the maximum number of resonances and the maximum and minimum liquid levels.
295 Medium
295 of 543
Analyzing +4 | -1
A tuning fork of frequency 480  Hz480\,\,Hz produces 10    beats  per  sec\mathrm{10}{\mkern 1mu} {\mkern 1mu} \,\,\mathrm{beats}\,\,\mathrm{per}\,\,\mathrm{sec} when sounded with a vibrating sonometer string. What must have been the frequency of string if slight increase in tension produces few beats s−1{\mathrm{s}^{ - 1}} than before?
View answer and explanation
Correct option: B
470  Hz \begin{array}{llllllllllllllllllll} {470\,\,Hz} \end{array}\,
Option B (470 Hz) is correct because increasing the tension would increase the string's frequency, reducing the beat frequency only if the original frequency was lower than the tuning fork's.
296 Medium
296 of 543
Remembering +4 | -1
Doppler phenomena is related with
View answer and explanation
Correct option: A
Pitch (frequency)
Option A (Pitch (frequency)) is correct because the Doppler effect relates to the change in frequency (pitch) of a wave due to the relative motion between the source and the observer.
297 Medium
297 of 543
Understanding +4 | -1
Statement-I: When a wave is transmitted from one medium to another, the frequency does not change.
Statement-II: A wave transports energy and momentum along with matter.
View answer and explanation
Correct option: A
Statement-I is correct
Statement-I is correct because the frequency of a wave remains constant when it travels from one medium to another. Statement-II is incorrect because waves transport energy and momentum but not matter.
298 Medium
298 of 543
Understanding +4 | -1
A tuning fork produces waves in a medium. If the temperature of the medium changes, then which of the following will change
View answer and explanation
Correct option: C
Wavelength
Option C (Wavelength) is correct because the speed of a wave in a medium depends on the temperature, and since frequency remains constant, the wavelength changes.
299 Medium
299 of 543
Application +4 | -1
A tuning fork vibrating with a sonometer having 20  cm20\,\,cm wire produces 55 beats per second. The beat frequency does not change if the length of the wire is changed to 21  cm21\,\,cm. The frequency of the tuning fork (in Hertz) must be
View answer and explanation
Correct option: C
205205
Option C (205) is correct because the frequency of the tuning fork is calculated by considering the beat frequencies produced with different lengths of the sonometer wire.
300 Medium
300 of 543
Remembering +4 | -1
The type of waves that can be propagated through solid is
View answer and explanation
Correct option: C
Both (aa) and (bb)
Option C (Both (a) and (b)) is correct because solids can support both transverse and longitudinal waves due to their rigidity and elasticity.
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