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JEE • PHYSICS • WAVES

Waves practice · Mixed difficulty

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Showing 151-180 of 543 questions

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151 Medium
151 of 543
Understanding +4 | -1
Assertion: The flash of lightening is seen before the sound of thunder is heard.
Reason: Speed of sound is greater than speed of light.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because light travels faster than sound.
152 Medium
152 of 543
Application +4 | -1
A string is producing transverse vibration whose equation is y=0.021sin⁡(x+30t)y = 0.021\sin \left( {x + 30t} \right), Where xx and yy are in meters and tt is in seconds. If the linear density of the string is 1.3  ×  10−4  kg/m1.3\,\, \times \,\,{10^{ - 4}}\,\,kg/m, then the tension in the string is NN will be
View answer and explanation
Correct option: D
0.117
Option D (0.117) is correct because it accurately calculates the tension in the string using the wave equation and linear density.
153 Medium
153 of 543
Application +4 | -1
A whistle with frequency 1020 Hz1020 \mathrm{~Hz} is blown at a station. A man travelling in train moving towards the station at 30 ms−130 \mathrm{~ms}^{-1} hears the sound of the whistle. If the speed of sound is 340 ms−1340 \mathrm{~ms}^{-1}, the apparent frequency heard by him is
View answer and explanation
Correct option: B
1110 Hz1110 \mathrm{~Hz}
Option B (1110 Hz1110 \mathrm{~Hz}) is correct because it is the apparent frequency calculated using the Doppler effect formula when the observer is moving towards the source.
154 Medium
154 of 543
Application +4 | -1
Two whistles A and B have frequencies 600 Hz600 \mathrm{~Hz} and 590 Hz590 \mathrm{~Hz} reactively. An observer is standing in the middle of the line joining the two sources. Source BB and observer are moving towards right with velocity 30 m/s30 \mathrm{~m} / \mathrm{s} and A\mathrm{A} is standing at left side. If the velocity of sound in air is 300 m/s300 \mathrm{~m} / \mathrm{s}, the number of beats listened by the observer are:
View answer and explanation
Correct option: B
4
Option B (4) is correct because the beat frequency is the difference between the frequencies heard by the observer from the two whistles.
155 Medium
155 of 543
Application +4 | -1
An observer is standing 500  m500\,\,m away from a vertically hill. Starting between the observer and the hill a police van having a siren of frequency 1000  Hz1000\,\,Hz moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is 970  Hz970\,\,Hz, the frequency of the sound heard after reflection from the hill (in Hz) is about, (velocity of sound =330  ms−1330\,\,m{s^{ - 1}}
View answer and explanation
Correct option: B
10321032
Option B (10321032) is correct because it accounts for the Doppler effect twice: once as the van approaches the hill and again as the reflection approaches the observer.
156 Medium
156 of 543
Application +4 | -1
For an interference pattern, the maximum and minimum intensity ratio is 64:164: 1, then what will be the ratio of amplitudes?
View answer and explanation
Correct option: B
9:79: 7
Option B (9:79: 7) is correct because the ratio of amplitudes is derived from the ratio of maximum and minimum intensities in an interference pattern.
157 Medium
157 of 543
Understanding +4 | -1
Phase difference between a particle at a compression and a particle at the next rare fraction is
View answer and explanation
Correct option: C
π\pi
The correct answer is C (π\pi) because the phase difference between a compression and the next rarefaction in a longitudinal wave is half a wavelength, which corresponds to a phase difference of π\pi radians.
158 Easy
158 of 543
Understanding +4 | -1
The engineer of a train blows the train whistle as he approaches a crossing. A few moments later he hears a echo from the whistle. The engineer hears the echo of the whistle because of:
View answer and explanation
Correct option: A
Reflection
Option A (Reflection) is correct because an echo is formed due to the reflection of sound waves.
159 Medium
159 of 543
Analyzing +4 | -1
Two waves travelling in a medium in the x-direction are represented by , y1=Asin⁡(αt−βx){y_1} = A\sin \left( {\alpha t - \beta x} \right) and y2=Acos⁡(βx+αt−π/4){y_2} = A\cos \left( {\beta x + \alpha t - \pi /4} \right) where y1    and  y2{y_1}\;\,\,and\,\,\mathrm{ }{y_2} are the displacements of the particles of the medium, t is time, and  α and β\alpha \,and\,\beta are constants. The two waves have different
View answer and explanation
Correct option: B
Directions of propagation
Option B (Directions of propagation) is correct. The waves have different signs in front of the x and t terms, indicating opposite directions of propagation.
160 Medium
160 of 543
Application +4 | -1
a segment of wire vibrates with a fundamental frequency of 450  Hz450\,\,Hz under a tension of 9kg−wt9\mathrm{ }kg - wt. then tension at which the fundamental frequency of the same wire becomes 900  Hz900\,\,Hz is
View answer and explanation
Correct option: A
36kg−wt36\mathrm{ }kg - wt
Option A (36kg−wt36\mathrm{ }kg - wt) is correct because the frequency is proportional to the square root of the tension, so doubling the frequency requires quadrupling the tension.
161 Medium
161 of 543
Application +4 | -1
The path difference between two interfering waves at a point on screen is 171.5 times the wavelength. If the path difference is 0.01372 cm. Find the wavelength?
View answer and explanation
Correct option: D
8000 Ao8000\,\mathop \mathrm{A}\limits^\mathrm{o}
Option D (8000 Ao8000\,\mathop \mathrm{A}\limits^\mathrm{o}) is correct because the wavelength is calculated by dividing the path difference by 171.5.
162 Easy
162 of 543
Analyzing +4 | -1
If fundamental frequency is 50 and next successive frequencies are 150 and 250 then it is
View answer and explanation
Correct option: B
A pipe closed at one end
Option B (A pipe closed at one end) is correct because the given frequencies (50, 150, 250) are odd multiples of the fundamental frequency, which is characteristic of a pipe closed at one end.
163 Medium
163 of 543
Understanding +4 | -1
Statement I: Sonometer is used to determine frequency of unknown T.F.

Statement II: In sonometer riders used as indicator.
View answer and explanation
Correct option: B
If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I.
Option B is correct because both statements are true, but Statement II doesn't explain Statement I. A sonometer determines frequency, and riders indicate resonance, but the riders' function isn't the reason the sonometer works.
164 Medium
164 of 543
Understanding +4 | -1
Assertion: Transverse waves are not produced in liquids and gases.
Reason: Light waves are transverse-like waves.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not correct explanation of the assertion
Option B (If both assertion and reason are true but reason is not correct explanation of the assertion) is correct because transverse waves require elasticity of shape, which liquids and gases lack, and light waves are indeed transverse, but this doesn't explain the assertion.
165 Medium
165 of 543
Analyzing +4 | -1
Assertion: In reflection of waves, frequency of waves remains same but their phase changes continuously.
Reason: When same sound note is heard after reflection from an obstacle, then it is known as echo.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not the correct explanation of assertion
Option B (If both assertion and reason are true but reason is not the correct explanation of assertion) is correct because both statements are true independently, but the phenomenon of echo doesn't explain the phase change during reflection.
166 Medium
166 of 543
Application +4 | -1
The energy per unit area associated with a progressive sound wave will be doubled if
View answer and explanation
Correct option: C
The amplitude of the wave is increased by 41.4%
Option C (The amplitude of the wave is increased by 41.4%) is correct because the energy per unit area is proportional to the square of the amplitude, so to double the energy, the amplitude must increase by a factor of 2\sqrt{2}.
167 Medium
167 of 543
Analyzing +4 | -1
A fork of unknown frequency when sounded with one of frequency 288 Hz288 \mathrm{~Hz} gives 4 beats per second and when loaded with a piece of wax again gives 4 beats per second. How do you account for this and what was the unknown frequency?
View answer and explanation
Correct option: C
292 Hz292 \mathrm{~Hz}
Option C is correct because the frequency increases after loading with wax, indicating the original frequency was higher than 288 Hz.
168 Medium
168 of 543
Application +4 | -1
Transverse waves are generated in two uniform steel wires AA and BB by attaching their free ends to a fork of frequency 500 Hz500 \mathrm{~Hz}. The diameter of wire AA is half that of BB and tension in wire AA is half the tension in wire BB. What is the ratio of velocities of waves in AA and BB ?
View answer and explanation
Correct option: B
2:1\sqrt{2}: 1
Option B (2:1\sqrt{2}: 1) is correct because the ratio of velocities is proportional to the square root of the ratio of tensions and inversely proportional to the square root of the ratio of mass per unit length.
169 Medium
169 of 543
Application +4 | -1
A man stands in a narrow, steep-sided valley. When he shouts, he hears two echoes, one after 1 s1 \mathrm{~s} and other after 2 s2 \mathrm{~s}. If the velocity of sound in air is 330 m/s330 \mathrm{~m} / \mathrm{s}, the width of the valley is
View answer and explanation
Correct option: B
495 m495 \mathrm{~m}
Option B (495 m) is correct because the width of the valley is half the product of the speed of sound and the sum of the echo times.
170 Medium
170 of 543
Understanding +4 | -1
At a certain instant a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is
View answer and explanation
Correct option: D
Straight line
Option D (Straight line) is correct because at the instant of maximum kinetic energy, all particles are passing through their equilibrium positions.
171 Medium
171 of 543
Application +4 | -1
A transverse wave described by
y=(0.02 m)sin⁡[(1.0 m−1)x+(30 s−1)t]y=(0.02 \mathrm{~m}) \sin \left[\left(1.0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] propagates on a stretched string having a linear mass density of 1.2×10−4 kg m−11.2 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}. Find the tension in the string.
View answer and explanation
Correct option: A
0.108 N
Option A (0.108 N) is correct because it is calculated using the wave equation and the linear mass density to find the tension in the string.
172 Medium
172 of 543
Remembering +4 | -1
The waves associated with the motion of microscopic particles such as electrons, protons, neutrons are called
View answer and explanation
Correct option: A
Matter waves
Option A (Matter waves) is correct because the waves associated with microscopic particles are known as matter waves or de Broglie waves.
173 Medium
173 of 543
Analyzing +4 | -1
The equation of the standing wave in a string clamped at both ends, vibrating in its third harmonic is given by

y=0.4sin⁡(0.314x)cos⁡(600πt){y}=0.4 \sin (0.314 {x}) \cos (600 \pi {t}) where x{x} and y{y} are in cm and t{t} is in sec-
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because the frequency is 300 Hz, the length is 30 cm, and the nodes are located at the specified positions, all derived from the given standing wave equation.
174 Medium
174 of 543
Application +4 | -1
The rms speed of hydrogen at 27∘C27^{\circ} \mathrm{C} is vv. The rms speed of oxygen at 1200 K1200 \mathrm{~K} is
View answer and explanation
Correct option: C
v2\frac{v}{2}
Option C (v2\frac{v}{2}) is correct because the rms speed is inversely proportional to the square root of the molar mass and directly proportional to the square root of the temperature.
175 Medium
175 of 543
Application +4 | -1
The equation of a transverse wave on a stretched string is given by y=0.05sin⁡2π(t0.002−x0.1)y=0.05 \sin 2 \pi\left(\frac{t}{0.002}-\frac{x}{0.1}\right), where xx and yy are expressed in metre and tt in second. The speed of the wave is
View answer and explanation
Correct option: B
50 ms−150 \mathrm{~ms}^{-1}
Option B (50 ms−150 \mathrm{~ms}^{-1}) is correct because the speed of the wave is the ratio of the coefficient of t to the coefficient of x in the wave equation.
176 Medium
176 of 543
Application +4 | -1
If speed of sound in air at 27∘C27^{\circ} \mathrm{C} is vv then at what temperature speed of sound becomes 2v2 v ?
View answer and explanation
Correct option: D
927∘C927^{\circ} \mathrm{C}
Option D (927∘C927^{\circ} \mathrm{C}) is correct because the speed of sound is proportional to the square root of the absolute temperature.
177 Medium
177 of 543
Application +4 | -1
A stone is dropped into a lake from a tower 500  metre500\,\,metre high. The sound of the splash will be heard by the man approximately after
View answer and explanation
Correct option: A
11.5  seconds\begin{array}{llllllllllllllllllll} {11.5\,\,seconds} \end{array}
Option A (11.5 seconds) is correct because it accounts for both the time the stone takes to fall and the time the sound takes to travel back up.
178 Medium
178 of 543
Understanding +4 | -1
A sound carried by air from a guitar to a listener is a wave of following type
View answer and explanation
Correct option: D
Longitudinal progressive
Option D (Longitudinal progressive) is correct because sound waves are longitudinal, and they propagate through the air from the guitar to the listener.
179 Medium
179 of 543
Application +4 | -1
It takes 2.0s2.0\mathrm{ }s for a sound wave to travel between two fixed points when the day temperature is 10∘C10^\circ C. if the temperature rises to 30∘C30^\circ C the sound wave travels between the same fixed parts in
View answer and explanation
Correct option: A
1.9 s1.9\,s
Option A (1.9 s1.9\,s) is correct because the speed of sound is proportional to the square root of the absolute temperature, so as temperature increases, time decreases.
180 Medium
180 of 543
Application +4 | -1
Amplitude of a mass -spring system, which is executing simple harmonic motion decreases with time. If mass =500 g= 500\,g decay constant =20 g/s,= 20\,g/s, then how much time is required for the amplitude of the system to drop to half of its initial value? (In 2=0.693)(\mathrm{In}\,2 = 0.693)
View answer and explanation
Correct option: A
34.65 s
Option A (34.65 s) is correct because it calculates the time required for the amplitude to drop to half its initial value using the damped oscillation formula.
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