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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 241-270 of 543 questions

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241 Medium
241 of 543
Understanding +4 | -1
A standing wave is produced in a string fixed at both ends. In this case
View answer and explanation
Correct option: C
All alternate antinodes vibrate in phase
Option C is correct because in a standing wave, all particles between two consecutive nodes vibrate in phase, and adjacent segments vibrate in opposite phases, so alternate antinodes vibrate in phase.
242 Medium
242 of 543
Application +4 | -1
If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maximum and minima is
View answer and explanation
Correct option: C
0.667361111
Option C (16:01) is correct because the ratio of maximum to minimum intensities is calculated using the squares of the sum and difference of the amplitudes.
243 Medium
243 of 543
Understanding +4 | -1
Two waves having sinusoidal waveforms have different wavelengths and different amplitude. They will be having
View answer and explanation
Correct option: C
Different quality and different intensity
The correct answer is C (Different quality and different intensity) because waves with different amplitudes have different intensities, and waves with different wavelengths have different qualities (timbres).
244 Medium
244 of 543
Application +4 | -1
A progressive wave of frequency 600 Hz600 \mathrm{~Hz} is travelling with a velocity of 300 m/s300 \mathrm{~m} / \mathrm{s}. Distance between two points of medium having phase difference of 30∘30^{\circ} will be nearly
(1) 0.021 m0.021 \mathrm{~m}
View answer and explanation
Correct option: C
0.041 m0.041 \mathrm{~m}
Option A (0.041 m) is correct because it is the calculated distance between two points with a 30-degree phase difference for a wave with the given frequency and velocity.
245 Medium
245 of 543
Analyzing +4 | -1
Assertion: Under given conditions of pressure and temperature, sound travels faster in a monoatomic gas than in diatomic gas.
Reason: Opposition for wave to travel is more in diatomic gas than monoatomic gas.
View answer and explanation
Correct option: C
If assertion is true, but reason is false
Option C is correct. The assertion is true because the velocity of sound is greater in monoatomic gases, but the reason is false as diatomic gases have more degrees of freedom, leading to more energy absorption and thus, a slower speed of sound.
246 Medium
246 of 543
Remembering +4 | -1
Ultrasonic waves are produced by
View answer and explanation
Correct option: A
Piezoelectric effect
Option A (Piezoelectric effect) is correct because ultrasonic waves are commonly produced using the piezoelectric effect, where certain materials generate an electric potential in response to mechanical stress, or vice versa.
247 Medium
247 of 543
Application +4 | -1
Ultrasonic waves of frequency 4.5MHz4.5 \mathrm{MHz} are used to detect tumour in soft tissues. The speed of sound in tissue is 1.5 km s−11.5 \mathrm{~km} \mathrm{~s}^{-1} and that in air is 340 ms−1340 \mathrm{~ms}^{-1}. Find the wavelength of this ultrasonic wave in air and in tissue.
View answer and explanation
Correct option: A
7.36×10−5 m;  3.3×10−4 m7.36 \times 10^{-5} \mathrm{~m};\,\,3.3 \times 10^{-4} \mathrm{~m}
Option A (7.36×10−5 m;  3.3×10−4 m7.36 \times 10^{-5} \mathrm{~m};\,\,3.3 \times 10^{-4} \mathrm{~m}) is correct because it correctly calculates the wavelengths in air and tissue using the given speeds and frequency.
248 Medium
248 of 543
Application +4 | -1
The displacement yy of a wave travelling in the xx -direction is given by y=10−4sin⁡(600 t−2x+π3)y = {10^{ - 4}}\sin \left( {600\,t - 2x + \frac{\pi }{3}} \right) meters, where xx is expressed in meters and tt is second. The speed of the wave-motion, in ms−1\mathrm{m}{\mathrm{s}^{ - 1}}, is
View answer and explanation
Correct option: B
300
Option B (300) is correct because the speed of the wave is calculated by dividing the angular frequency by the wave number.
249 Medium
249 of 543
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.
250 Medium
250 of 543
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.
251 Hard
251 of 543
Understanding +4 | -1
In the first overtone position of the resonance tube, the pressure anti-node is present at (L is the length of the tube) which positions from the closed end.
View answer and explanation
Correct option: A
L3,L\frac{\mathrm{L}}{3}, \mathrm{L}
Option A (L3,L\frac{\mathrm{{L}}}{{3}}, \mathrm{{L}}) is correct because in the first overtone of a closed pipe, pressure antinodes occur at L/3 and L from the closed end.
252 Medium
252 of 543
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.
253 Medium
253 of 543
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.
254 Medium
254 of 543
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.
255 Medium
255 of 543
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.
256 Medium
256 of 543
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.
257 Medium
257 of 543
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.
258 Medium
258 of 543
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.
259 Medium
259 of 543
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).
260 Medium
260 of 543
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.
261 Medium
261 of 543
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).
262 Medium
262 of 543
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.
263 Medium
263 of 543
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.
264 Medium
264 of 543
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.
265 Medium
265 of 543
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.
266 Medium
266 of 543
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.
267 Medium
267 of 543
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.
268 Medium
268 of 543
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.
269 Medium
269 of 543
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’.
270 Medium
270 of 543
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.
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