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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 301-330 of 524 questions

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301 Medium
301 of 524
Application +4 | -1
The equation of a simple harmonic wave is given by y=6  sin⁡  2π(2t−0.1x)y = 6\,\,\sin \,\,2\pi \left( {2t - 0.1x} \right) where x and y are in mm and t is in seconds. The phase difference between two particles 2 mm apart at any instant is
View answer and explanation
Correct option: B
72∘72^\circ
Option B (72∘72^\circ) is correct because the phase difference is calculated using the wave number and the distance between the particles.
302 Medium
302 of 524
Application +4 | -1
A plane wave is described by the equation y=3cos⁡(x4−10t−π2)y = 3\cos \left( {\frac{x}{4} - 10t - \frac{\pi }{2}} \right). The maximum velocity of the particles of the medium due to this wave is
View answer and explanation
Correct option: A
30
Option A (30) is correct because the maximum particle velocity is the product of the amplitude and the angular frequency.
303 Medium
303 of 524
Remembering +4 | -1
Echo is based on which amongst the given phenomenon that sound wave follows?
View answer and explanation
Correct option: B
Reflection
Option B (Reflection) is correct because an echo is formed when sound waves bounce off a surface and return to the listener, which is the definition of reflection.
304 Medium
304 of 524
Understanding +4 | -1
When a tuning fork produces sound waves in air, which one of the following is same in the material of tuning fork as well as in air?
View answer and explanation
Correct option: B
frequency
Option B (frequency) is correct because the frequency of a sound wave is determined by the source (tuning fork) and remains constant regardless of the medium it travels through.
305 Medium
305 of 524
Application +4 | -1
A sound wave of frequency vv propagating through air with a velocity cc, is reflected from a surface which is moving away from the source with a constant speed vv. the frequency of the reflected wave, measured by the observed at the position of the source, is
View answer and explanation
Correct option: C
v(c+2v)c+v\frac{{v\left( {c + 2v} \right)}}{{c + v}}
Option C (v(c+2v)c+v\frac{{v\left( {c + 2v} \right)}}{{c + v}}) is correct because it correctly applies the Doppler effect formula for a wave reflected from a moving surface, considering both the moving source and moving observer.
306 Medium
306 of 524
Application +4 | -1
The fundamental note produced by a closed organ pipe is of frequency ff. The fundamental note produced by an open organ pipe of same length will be of frequency
View answer and explanation
Correct option: C
2f2f
Option C (2f2f) is correct because the fundamental frequency of an open organ pipe is twice that of a closed organ pipe of the same length.
307 Medium
307 of 524
Understanding +4 | -1
Doppler shift in frequency does not depend upon
View answer and explanation
Correct option: D
Distance from the source to the listener
Option D (Distance from the source to the listener) is correct because the Doppler shift in frequency depends on the relative velocities of the source and observer, and the frequency of the wave, but not on the distance between them.
308 Medium
308 of 524
Remembering +4 | -1
The ratio of speed of source to the speed of sound is called
View answer and explanation
Correct option: B
Mach number
Option B (Mach number) is correct because the Mach number is defined as the ratio of the speed of a source to the speed of sound.
309 Medium
309 of 524
Understanding +4 | -1
Beats are produced by two travelling waves each of loudness II and nearly equal frequencies n1{n_1} and n2{n_2}. The beat frequency will be and maximum loudness hard will be
View answer and explanation
Correct option: B
(n1−n2),4I\left( {{n_1} - {n_2}} \right),4I
Option B ((n1−n2),4I\left( {{n_1} - {n_2}} \right),4I) is correct because the beat frequency is the difference in frequencies and the maximum loudness is four times the individual loudness.
310 Medium
310 of 524
Analyzing +4 | -1
Assertion: Where two vibrating tuning forks having frequencies 256 Hz256 \mathrm{~Hz} and 512 Hz512 \mathrm{~Hz} are held near each other, beats cannot be heard.
Reason: The principle of superposition is valid only if the frequencies of the oscillators are nearly equal.
View answer and explanation
Correct option: C
If assertion is true, but reason is false
Option C (If assertion is true, but reason is false) is correct because while beats are not heard, the principle of superposition applies regardless of frequency difference.
311 Medium
311 of 524
Application +4 | -1
Two air columns (of resonance tubes) 100 cm100 \mathrm{~cm} and 101 cm101 \mathrm{~cm} long give 17 beats in 20 seconds, when each is sounding its fundanental mode. Calculate the velocity of sound.
View answer and explanation
Correct option: B
343.4 ms−1343.4 \mathrm{~ms}^{-1}
Option B (343.4 ms−1343.4 \mathrm{~ms}^{-1}) is correct because the velocity of sound is calculated using the beat frequency and the lengths of the air columns in the resonance tubes.
312 Medium
312 of 524
Remembering +4 | -1
Frequency range of the audible sound is
View answer and explanation
Correct option: B
20 Hz−20 kHz20 \mathrm{~Hz}-20 \mathrm{~kHz}
Option B (20 Hz−20 kHz20 \mathrm{~Hz}-20 \mathrm{~kHz}) is correct because this is the generally accepted range of frequencies that humans can hear.
313 Medium
313 of 524
Application +4 | -1
A wave of frequency 500  Hz500\,\,Hz has velocity 360  m/sec360\,\,m/sec. The distance between two nearest points 600{60^0} out of phase, is
View answer and explanation
Correct option: B
12  cm12\,\,cm
Option B (12  cm12\,\,cm) is correct because the distance between two points 600{60^0} out of phase is calculated using the formula Δ=λ2π×ϕ\Delta = \frac{\lambda }{{2\pi }} \times \phi.
314 Medium
314 of 524
Remembering +4 | -1
Maximum number of beats frequency heard by a human being is
View answer and explanation
Correct option: A
1010
Option A (10) is correct because the persistence of hearing is approximately 0.1 seconds, limiting the maximum distinguishable beat frequency to around 10 Hz.
315 Medium
315 of 524
Application +4 | -1
The fundamental frequency of a closed end organ pipe is nn. Its length is doubled and radius is halved. Its frequency will be become nearly
View answer and explanation
Correct option: A
n/2n/2
Option A (n/2n/2) is correct because the fundamental frequency of a closed pipe is inversely proportional to its length, and doubling the length halves the frequency.
316 Medium
316 of 524
Application +4 | -1
A standing wave is maintained in a homogeneous string of cross-sectional area ss and density ρ\rho. It is formed by the superposition of two waves travelling in opposite directions given by the equation y1=asin⁡(ωt−kx)y_1=a \sin (\omega t-k x) and y2=2asin⁡(ωt+kx)y_2=2 a \sin (\omega t+k x). The total mechanical energy confined between the sections corresponding to the adjacent antinodes is
View answer and explanation
Correct option: C
5πspω2a22k\frac{5 \pi s p \omega^2 a^2}{2 k}
The correct answer is C. The total mechanical energy is calculated by summing the energy densities of the two waves and multiplying by the volume between adjacent antinodes.
317 Medium
317 of 524
Application +4 | -1
Calculate the fundumental frequency of a sonometer wire of length =20 cm=20 \mathrm{~cm} tension 25 N25 \mathrm{~N}, crosssectional aren =10−2 cm2=10^{-2} \mathrm{~cm}^2 and density of the material of wire =104 kg m−3=10^4 \mathrm{~kg} \mathrm{~m}^{-3}.
View answer and explanation
Correct option: B
125 Hz125 \mathrm{~Hz}
Option B (125 Hz125 \mathrm{~Hz}) is correct because it is the fundamental frequency calculated using the given parameters of the sonometer wire.
318 Medium
318 of 524
Understanding +4 | -1
If waves are originating from point source, then amplitude (A)(A) at a distance (r)(r) from the source are related as
View answer and explanation
Correct option: C
A∝1rA \propto \frac{1}{r}
Option C (A∝1rA \propto \frac{{1}}{{r}}) is correct because the intensity of a wave from a point source decreases with the square of the distance, and intensity is proportional to the square of the amplitude.
319 Medium
319 of 524
Analyzing +4 | -1
Assertion: For same length but different gas COP fundamental wavelength remain same.
Reason: Here length is same.
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 the fundamental wavelength depends only on the length of the tube for the same mode of vibration.
320 Medium
320 of 524
Understanding +4 | -1
I. In a standing wave, energy is transmitted from one region of space to other.

II. In a standing wave, all the particles cross their mean position together.

Which of the above statement(s) is/are correct?
View answer and explanation
Correct option: B
Only II
Option B (Only II) is correct because in a standing wave, energy is not transmitted, and all particles at a node cross their mean position together.
321 Medium
321 of 524
Application +4 | -1
A wave travelling in stretched string is described by the equation y=Asin(kx−ωt)y = \mathrm{Asin}\left( {kx - \omega t} \right).The maximum particle velocity is
View answer and explanation
Correct option: A
AωA\omega
Option A (AωA\omega) is correct because it represents the maximum particle velocity in a wave, which is the product of the amplitude and angular frequency.
322 Medium
322 of 524
Understanding +4 | -1
If the tension and diameter of a sonometer wire having fun
View answer and explanation
Correct option: D
f′=ff^{\prime}=f
Option D (f′=ff^{\prime}=f) is correct because the changes in tension, density, and radius compensate each other, resulting in the same frequency.
323 Medium
323 of 524
Application +4 | -1
A particle on a stretched string supporting a travelling wave, takes 5.0 ms5.0 \mathrm{~ms} to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm2.0 \mathrm{~cm}. Find the frequency, the wavelength and the wave speed.
View answer and explanation
Correct option: A
50 Hz; 4 cm; 2 m/s
Option A (50 Hz; 4 cm; 2 m/s) is correct because it accurately calculates the frequency, wavelength, and wave speed based on the given time and distance.
324 Medium
324 of 524
Application +4 | -1
A plane sound wave passes from medium 1 into medium 2 . The speed of sound in medium 1 is 200 m/s200 \mathrm{~m} / \mathrm{s} and in medium 2 is 100 m/s100 \mathrm{~m} / \mathrm{s}. The ratio of amplitude of the transmitted wave to that of incident wave is
View answer and explanation
Correct option: D
23\frac{2}{3}
Option D (23\frac{2}{3}) is correct because it is calculated using the formula for the ratio of transmitted wave amplitude to incident wave amplitude based on the speeds of sound in the two media.
325 Medium
325 of 524
Application +4 | -1
A 20  cm20\,\,cm long string, having a mass of 1.0  g1.0\,\,g, is fixed at both the ends. The tension in the string is 0.5  N0.5\,\,N. the string is set into vibration using an external vibrator of frequency 100  Hz100\,\,Hz. Find the separation (in cmcm) between the successive nodes on the string
View answer and explanation
Correct option: A
5
Option A (5 cm) is correct because it is the calculated separation between successive nodes using the given parameters and the formula for wave speed on a string.
326 Medium
326 of 524
Understanding +4 | -1
Assertion: When two vibrating tuning forks having frequencies 240 Hz and 300 Hz are held near each other, beats cannot be heard by us
Reason: This is because beats cannot be distinctly heard due to the property of persistence of hearing.
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 the assertion and reason are both true, and the reason correctly explains why beats are not distinctly heard at those frequencies due to persistence of hearing.
327 Medium
327 of 524
Understanding +4 | -1
Assertion: Corresponding to a path difference xx, the phase difference is 2πλx\frac{2 \pi}{\lambda} x.
Reason: The phase difference between displacement and velocity of a particle executing SHM is π\pi.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option C is correct because the assertion is true, but the phase difference between displacement and velocity in SHM is π/2\pi/2, not π\pi.
328 Medium
328 of 524
Application +4 | -1
An open organ pipe has fundamental frequency 100  Hz100\,\,Hz. What frequency will be produced if its one end is closed?
View answer and explanation
Correct option: B
50,150,250……\begin{array}{llllllllllllllllllll} {50,150,250 \ldots \ldots } \end{array}
Option B (50,150,250……50,150,250 \ldots \ldots) is correct because closing one end of an open organ pipe halves the fundamental frequency and only odd harmonics are produced.
329 Medium
329 of 524
Application +4 | -1
When the temperature of an ideal gas is increased by 600  K600\,\,K, the velocity of sound in the gas becomes 3\sqrt 3 times the initial velocity in it. The initial temperature of the gas is
View answer and explanation
Correct option: B
270C{27^0}C
Option B (270C{27^0}C) is correct because it is calculated using the relationship between the velocity of sound and temperature in an ideal gas.
330 Medium
330 of 524
Remembering +4 | -1
The minimum audible wavelength at room temperature is about
View answer and explanation
Correct option: D
20  mm20\,\,mm
Option D (20  mm20\,\,mm) is correct because it is calculated using the maximum audible frequency and the speed of sound.
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