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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 241-270 of 524 questions

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241 Medium
241 of 524
Analyzing +4 | -1
Two tuning forks AA and BB vibrating simultaneously produce 55 beats. Frequency of BB is 512512. It is seen that if one arm of AA is filled, then the number of beats increases. Frequency of AA will be
View answer and explanation
Correct option: C
517517
Option C (517) is correct because filing the tuning fork increases its frequency, and since the beat frequency increases, the original frequency of A must have been higher than B.
242 Medium
242 of 524
Understanding +4 | -1
Speed of sound at constant temperature depends on
View answer and explanation
Correct option: D
None of the above
Option D is correct because the speed of sound at a constant temperature is independent of both pressure and density individually; it depends on the ratio of pressure to density, which remains constant.
243 Medium
243 of 524
Application +4 | -1
On the same path, the source and observer are moving in such a way that the distance between these two increases with the time. The speeds of source and observer are same and equal to 10 ms−110 \mathrm{~ms}^{-1} with respect to the ground while no wind is blowing. The apparent frequency received by observer is 1950 Hz1950 \mathrm{~Hz}, then the original frequency must be (Take, the speed of sound in present medium is 340 ms−1340 \mathrm{~ms}^{-1} )
View answer and explanation
Correct option: A
2068 Hz2068 \mathrm{~Hz}
Option A (2068 Hz2068 \mathrm{~Hz}) is correct because the apparent frequency is calculated using the Doppler effect formula, considering the source and observer moving apart.
244 Medium
244 of 524
Understanding +4 | -1
Two tones of frequencies n1\mathrm{n}_1 and n2\mathrm{n}_2 are sounded together. The beats can be heard distinctly when
View answer and explanation
Correct option: D
0<(n1−n2)<100 < \left(\mathrm{n}_1-\mathrm{n}_2\right) < 10
Option D (0<(n1−n2)<100 < (n_1 - n_2) < 10) is correct because the beat frequency must be less than 10 Hz to be distinctly heard.
245 Medium
245 of 524
Remembering +4 | -1
The harmonics which are present in a pipe, open at one end are
View answer and explanation
Correct option: A
Odd harmonics
Option A (Odd harmonics) is correct because a pipe open at one end only supports odd harmonics due to the boundary conditions.
246 Medium
246 of 524
Understanding +4 | -1
Statement-I: The Doppler effect can be observed in case of light and sound, but Doppler effect formula for light differs from that for sound.
Statement-II: Light does not require a material medium for its transmission whereas sound requires medium for its transmission.
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 Doppler effect applies to both light and sound, but the formulas differ due to light's independence from a medium.
247 Medium
247 of 524
Understanding +4 | -1
The equation y=a  sin⁡2π(tT−xλ)y = a\,\,\sin 2\pi \left( {\frac{t}{T} - \frac{x}{\lambda }} \right), where the symbols carry the usual meaning and aa, TT and &lambda;&lambda; are positive, represents a wave of
View answer and explanation
Correct option: D
Speed (λ/T\lambda /T)
Option D is correct because the speed of a wave is given by the wavelength divided by the period, which is λ/T\lambda/T.
248 Medium
248 of 524
Analyzing +4 | -1
A tuning fork of frequency 100 Hz100 \mathrm{~Hz} when sound together with another tuning fork of unknown frequency produces 2 beats per second. On loading the tuning fork whose frequency is not known and sounded together with the same tuning fork produces one beat, then the frequency of the unknown tuning fork (in Hz\mathrm{Hz}) is
View answer and explanation
Correct option: A
102
Option A (102 Hz) is correct because the unknown frequency is either 98 Hz or 102 Hz. Loading the fork decreases its frequency; only 102 Hz decreasing to 101 Hz would produce a 1 Hz beat.
249 Medium
249 of 524
Analyzing +4 | -1
Two persons AA and BB, each carrying a source of frequency 300 Hz300 \mathrm{~Hz}, are standing a few metre apart. AA starts moving towards BB with velocity 30 m/s30 \mathrm{~m} / \mathrm{s}. If the speed of sound is 300 m/s300 \mathrm{~m} / \mathrm{s}, which of the following is true?
View answer and explanation
Correct option: D
Both (a) and (b) are wrong
Option D (Both (a) and (b) are wrong) is correct because the number of beats heard by A and B are different and neither is 30 Hz.
250 Medium
250 of 524
Analyzing +4 | -1
Assertion (A): When a travelling wave is sent along a particular string by oscillating one end, the speed of the wave will increase if we increase the frequency of oscillations.
Reason (R): If a travelling wave sent along a particular string by oscillating its one end, it is the wave length of the wave that decreases.
View answer and explanation
Correct option: D
Assertion is false but Reason is true
Option D is correct because the assertion is false (frequency doesn't change wave speed on a string), but the reason is true (wavelength decreases with increasing frequency for a fixed wave speed).
251 Medium
251 of 524
Understanding +4 | -1
Assertion: If a wave pulse propagating on a string is reflected from free end then it is not inverted.
Reason: Transverse force is absent and medium is discontinuous at free end.
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 a wave pulse reflected from a free end is not inverted due to the absence of a transverse force and discontinuity of the medium.
252 Medium
252 of 524
Understanding +4 | -1
In stationary waves
View answer and explanation
Correct option: B
Energy is minimum at nodes and maximum at antinodes
Option B is correct because in stationary waves, energy is stored and oscillates between nodes (minimum energy) and antinodes (maximum energy).
253 Medium
253 of 524
Analyzing +4 | -1
Fundamental frequency of pipe is 100  Hz100\,\,Hz and other two frequencies are 300  Hz300\,\,Hz and 500  Hz500\,\,Hz, then
View answer and explanation
Correct option: C
One end is open and another end is closed
Option C (One end is open and another end is closed) is correct because the frequencies are odd multiples of the fundamental frequency, which is characteristic of a closed pipe.
254 Medium
254 of 524
Understanding +4 | -1
If sound wave travel from air to water, which of the following remain unchanged?
View answer and explanation
Correct option: C
Frequency
Option C (Frequency) is correct because frequency is a property of the source and does not change when a wave travels from one medium to another.
255 Medium
255 of 524
Application +4 | -1
In the fundamental mode, time taken by the wave to reach the closed end of the air filled pipe is 0.01  s0.01\,\,s. the fundamental frequency is
View answer and explanation
Correct option: A
25
Option A (25) is correct because the fundamental frequency is calculated using the time taken by the wave to reach the closed end of the pipe.
256 Medium
256 of 524
Remembering +4 | -1
The number of waves contained in unit length of the medium is called
View answer and explanation
Correct option: B
Wave number
Option B (Wave number) is correct because wave number is defined as the number of waves per unit length.
257 Medium
257 of 524
Analyzing +4 | -1
The transverse displacement y(x,t)y\left( {x,t} \right) of a wave on a string is given by y(x,t)=e−(ax2+bt2+2abxt)y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} xt} \right)}} This represent a
View answer and explanation
Correct option: A
Wave moving in xx- direction with speed ba\sqrt {\frac{b}{a}}
Option A is correct because the given equation can be rewritten in the form of a wave equation, and by comparing it with the standard form, the speed can be determined.
258 Medium
258 of 524
Application +4 | -1
Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 44 beats s−1{\mathrm{s}^{ - 1}}. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies are
View answer and explanation
Correct option: D
72  and  36\begin{array}{llllllllllllllllllll} {72\,\,and\,\,36} \end{array}
Option D (72 and 36) is correct because the highest frequency is twice the lowest, and the difference between adjacent forks is 4 Hz.
259 Medium
259 of 524
Understanding +4 | -1
Among the following equations, which one represents the spherical progressive wave?
View answer and explanation
Correct option: B
y=arsin⁡(ωt−kx)y=\frac{a}{r} \sin (\omega t-k x)
Option B is correct because in a spherical progressive wave, the amplitude decreases inversely with the distance r, hence the term a/r.
260 Medium
260 of 524
Understanding +4 | -1
In the musical octave ‘Sa’, ‘Re’, ‘Ga’
View answer and explanation
Correct option: B
The frequency of the note ‘Sa’ is smaller than that of ‘Re’, ‘Ga’
Option B (The frequency of the note ‘Sa’ is smaller than that of ‘Re’, ‘Ga’) is correct because in the musical octave, the frequency increases as you go from ‘Sa’ to ‘Re’ to ‘Ga’.
261 Medium
261 of 524
Understanding +4 | -1
The equation of a cylindrical progressive wave is
View answer and explanation
Correct option: C
y=arsin⁡(ωt−kr)y = \frac{a}{{\sqrt r }}\sin \left( {\omega t - kr} \right)
Option C (y=arsin⁡(ωt−kr)y = \frac{a}{{\sqrt r }} \sin \left( {\omega t - kr} \right)) is correct because the amplitude of a cylindrical wave decreases with the square root of the distance from the source due to energy spreading over an increasing circumference.
262 Medium
262 of 524
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.
263 Medium
263 of 524
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.
264 Medium
264 of 524
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.
265 Medium
265 of 524
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.
266 Medium
266 of 524
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.
267 Medium
267 of 524
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).
268 Medium
268 of 524
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.
269 Medium
269 of 524
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.
270 Medium
270 of 524
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.
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