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NEET • PHYSICS • WAVES

Waves practice · Mixed difficulty

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Showing 181-210 of 215 questions

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181 Medium
181 of 215
Understanding +4 | -1
A progressive wave y=a sin[(kx−ωt)]\mathrm{y = a}\,\mathrm{sin}\left[ {\left( {\mathrm{kx} - \mathrm{\omega t}} \right)} \right] is reflected by a rigid wall at x=0\mathrm{x} = 0. Then the reflected wave can be represented by
View answer and explanation
Correct option: D
y=−asin(kx+ωt)\begin{array}{llllllllllllllllllll} {\mathrm{y = } - \mathrm{asin}\left( {\mathrm{kx + \omega t}} \right)} \end{array}
Option D (y=−asin(kx+ωt)\begin{array}{llllllllllllllllllll}{\mathrm{y = } - \mathrm{asin}\left( {\mathrm{kx + \omega t}} \right)}\end{array}) is correct because reflection from a rigid wall introduces a phase change of π\pi, and the wave travels in the opposite direction.
182 Medium
182 of 215
Application +4 | -1
A tuning fork of frequency 512 Hz512 \mathrm{~Hz} is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is
View answer and explanation
Correct option: A
506 Hz506 \mathrm{~Hz}
Option A (506 Hz506 \mathrm{~Hz}) is correct because when the tension increases, the frequency increases, meaning the original frequency was lower than the tuning fork's frequency.
183 Medium
183 of 215
Remembering +4 | -1
The speed of sound in a gas of density ρ\rho at a pressure PP is proportional to
View answer and explanation
Correct option: D
Pρ\sqrt {\frac{P}{\rho }}
Option D is correct because the speed of sound in a gas is directly proportional to the square root of the pressure divided by the density, as given by the formula v=Pρv = \sqrt{{\frac{{P}}{{\rho}}}}
184 Medium
184 of 215
Understanding +4 | -1
A sonometer wire of density ρ\rho and radius r{r} is held between two sharp edges at a distance ' ll '. The wire has a tension T\mathrm{T}. The frequency of ppth overtone will be
View answer and explanation
Correct option: D
n=12lTπr2ρn=\frac{1}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}}
Option C (n=p2lTπr2ρn=\frac{p}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}}) is correct because the frequency of the pth overtone is p times the fundamental frequency, and the linear density is related to the radius and density.
185 Medium
185 of 215
Analyzing +4 | -1
A tuning fork AA produces 4  beats/sec⁡4\,\,beats/\sec with another tuning fork BB of frequency 320  Hz320\,\,Hz. On filing the fork AA, 4  beats/sec⁡4\,\,beats/\sec are again heard. The frequency of fork AA, after filing is
View answer and explanation
Correct option: A
324  Hz324\,\,Hz
Option A (324 Hz324{\mathrm{ }}\,Hz) is correct because filing the fork increases its frequency, and since the beat frequency remains the same, the original frequency must have been lower than B.
186 Medium
186 of 215
Application +4 | -1
If the distance between successive compressions and rarefraction in a sound wave is 2m and velocity of sound is 360 m/s, then the frequency is
View answer and explanation
Correct option: D
90 Hz
Option D (90 Hz) is correct because the frequency is calculated using the wavelength (twice the distance between compression and rarefaction) and the velocity of sound.
187 Easy
187 of 215
Understanding +4 | -1
The frequency of a tuning fork is 256. It will not resonate with a fork of frequency
View answer and explanation
Correct option: A
738
Option A (738) is correct because resonance occurs when the frequencies are integer multiples of each other. 738 is not an integer multiple of 256.
188 Medium
188 of 215
Analyzing +4 | -1
The frequency of fundamental note in an organ pipe is 240  Hz240\,\,Hz. On blowing air, frequencies 720  Hz720\,\,Hz and 1200  Hz1200\,\,Hz are heard. This indicates that organ pipe is
View answer and explanation
Correct option: A
A pipe closed at one end
Option A (A pipe closed at one end) is correct because the given frequencies are odd multiples of the fundamental frequency.
189 Medium
189 of 215
Application +4 | -1
A transverse wave is represented by y=Asin⁡(ωt−kx)y = A\sin \left( {\omega t - kx} \right). For what value of wavelength, the wave velocity is half of maximum particle velocity
View answer and explanation
Correct option: A
πA\pi A
Option A (πA\pi A) is correct because the wave velocity is ω/k\omega/k and the maximum particle velocity is AωA\omega. Setting the wave velocity to half the maximum particle velocity and solving for the wavelength yields λ=πA\lambda = \pi A.
190 Medium
190 of 215
Application +4 | -1
5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :
View answer and explanation
Correct option: A
195 Hz
Option A (195 Hz) is correct because the frequency of the fork is calculated using the beat frequency and the relationship between frequency and length of the sonometer wire.
191 Medium
191 of 215
Application +4 | -1
The transverse displacement of string fixed at both ends is given by y=0.06 sin⁡(2πx3)cos⁡(120 πt)y = 0.06\,\sin \left( {\frac{{2\pi x}}{3}} \right)\cos \left( {120\,\pi t} \right), where x and y are in metre and t is in second. The length of the string is 1.5m and its mass is 3×10−2  kg3 \times {10^{ - 2}}\,\,kg . The tension in the string is equal to
View answer and explanation
Correct option: A
648 N
Option A (648 N) is correct because the tension is calculated using the wave speed derived from the given wave equation and the mass per unit length.
192 Medium
192 of 215
Understanding +4 | -1
On which principle does sonometer works?
View answer and explanation
Correct option: C
Resonance
Option C (Resonance) is correct because a sonometer works on the principle of resonance, where the frequency of the vibrating wire matches the frequency of the tuning fork, leading to maximum amplitude.
193 Medium
193 of 215
Analyzing +4 | -1
What is the base frequency if a pipe gives notes of frequencies 425, 255 and 595 and decide whether it is closed at one end or open at both ends
View answer and explanation
Correct option: B
85, closed
Option B (85, closed) is correct because the given frequencies are odd multiples of 85, which is characteristic of a closed pipe.
194 Medium
194 of 215
Remembering +4 | -1
Ultrasonic waves are those waves
View answer and explanation
Correct option: B
Man can’t hear
Option B (Man can’t hear) is correct because ultrasonic waves have frequencies above the human hearing range.
195 Medium
195 of 215
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 it represents the possible wavelengths based on the given relationship between crests and troughs.
196 Medium
196 of 215
Understanding +4 | -1
Quality of a musical note depends on
View answer and explanation
Correct option: A
Harmonics present
Option A (Harmonics present) is correct because the quality (timbre) of a musical note is determined by the number and relative intensities of the harmonics present.
197 Medium
197 of 215
Understanding +4 | -1
Two Cu wires of radii R1  and  R2{\mathrm{R}_1}\,\,\mathrm{and}\,\,{\mathrm{R}_2} such that (R1>R2)\left( {{R_1} > {R_2}} \right). Then which of the following is true?
View answer and explanation
Correct option: B
Transverse wave travels faster in thinner wire
Option B (Transverse wave travels faster in thinner wire) is correct because the speed of a transverse wave on a string is inversely proportional to the square root of the linear mass density, which is higher for thicker wires.
198 Easy
198 of 215
Application +4 | -1
An observer standing by the side of a road hears the siren of an ambulance, which is moving away from him. If the actual frequency of the siren is 2,000 Hz2,000 \mathrm{~Hz}, then the frequency heard by the observer will be
View answer and explanation
Correct option: B
1990 Hz1990 \mathrm{~Hz}
Option B (1990 Hz1990 \mathrm{~Hz}) is correct because the frequency heard by the observer decreases as the ambulance moves away, according to the Doppler effect.
199 Easy
199 of 215
Application +4 | -1
The phase difference between current and voltage in an AC\mathrm{AC} circuit is π/4\pi / 4 radian. If the frequency of AC\mathrm{AC} is 50 Hz50 \mathrm{~Hz}, then the phase difference is equivalent to the time difference :
View answer and explanation
Correct option: D
2.5 ms2.5 \mathrm{~ms}
Option D (2.5 ms2.5 \text{{ ms}}) is correct because the time difference is calculated by dividing the phase difference by the angular frequency, which is 2π2\pi times the frequency.
200 Hard
200 of 215
Application +4 | -1
Speed of sound in mercury at a certain temperature is 1450 m/s. Given the density of mercury as 13.6×103kgm313.6 \times {10^3}\frac{{kg}}{{{m^3}}}, the bulk modulus for mercury is [ in Nm3\frac{N}{{{m^3}}}]
View answer and explanation
Correct option: A
2.86×10102.86 \times {10^{10}}
Option A (2.86×10102.86 \times {10^{10}}) is correct because the bulk modulus is calculated using the formula relating speed of sound, density, and bulk modulus.
201 Medium
201 of 215
Application +4 | -1
The equation of a wave travelling on a string is y=10sin⁡[π(5t−x4)]y = 10\sin \left[ {\pi \left( {5t - \frac{x}{4}} \right)} \right], where x, y are in cm and t is in second. The velocity of wave is
View answer and explanation
Correct option: C
20 cm/s, in x-direction
Option C (20 cm/s, in x-direction) is correct because by comparing the given equation with the standard wave equation, the wave velocity can be calculated as ω/k\omega/k, and the sign indicates the direction of propagation.
202 Medium
202 of 215
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 both statements accurately describe the relationship between open and closed organ pipes and the formation of nodes and antinodes.
203 Medium
203 of 215
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 encompasses all terms inside the trigonometric function.
204 Easy
204 of 215
Application +4 | -1
A resonance air column of length 20 cm20 \mathrm{~cm} resonates with a tuning fork of frequency 450 Hz450 \mathrm{~Hz}. Ignoring the correction, the velocity of sound in air will be
View answer and explanation
Correct option: D
360 m/s360 \mathrm{~m} / \mathrm{s}
Option D (360 m/s360 \, m/s) is correct because the first resonance in an air column closed at one end occurs when the length of the column is one-quarter of the wavelength, allowing us to calculate the speed of sound.
205 Medium
205 of 215
Application +4 | -1
A string under a tension of 129.6 N129.6 \mathrm{~N} produces 10 beats/s when it is vibrated along with a tuning fork. When the tension in the string is increased to 160 N160 \mathrm{~N}, it sounds in unision with the same tuning fork. Calculate the fundamental frequency of the tuning fork.
View answer and explanation
Correct option: A
100
Option A (100) is correct because the fundamental frequency of the tuning fork is calculated by considering the change in frequency of the string due to the change in tension and the resulting beat frequency.
206 Hard
206 of 215
Application +4 | -1
A stone hangs in air from one end of a wire which is stretched over a sonometer. The wire is in unison with a certain tuning fork when the bridges of the sonometer are 45 cm45 \mathrm{~cm} apart. Now the stone hangs immersed in water at 4∘C4^{\circ} \mathrm{C} and the distance between the bridges has to be altered by 9 cm9 \mathrm{~cm} to re-establish unison of the wire with the same fork. Calculate the density of the stone.
View answer and explanation
Correct option: B
2.778 g cm−32.778 \mathrm{~g} \mathrm{~cm}^{-3}
Option B (2.778 g cm−32.778 \mathrm{~g} \mathrm{~cm}^{-3}) is correct because it is calculated using the change in length required to maintain the same frequency when the stone is immersed in water, considering the buoyant force.
207 Medium
207 of 215
Remembering +4 | -1
The harmonic 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 multiples of the fundamental frequency due to the boundary conditions.
208 Medium
208 of 215
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 is correct because matter waves are associated with the motion of microscopic particles.
209 Medium
209 of 215
Understanding +4 | -1
Out of the given waves (1), (2), (3) and (4)
y=asin⁡(kx+ωt)y = a\sin \left( {kx + \omega t} \right) ...(11)
y=asin⁡(ωt−kx)y = a\sin \left( {\omega t - kx} \right) …(22)
y=acos⁡(kx+ωt)y = a\cos \left( {kx + \omega t} \right) …(33)
y=acos⁡(ωt−kx)y = a\cos \left( {\omega t - kx} \right) …(44)
Emitted by four different sources S1,S2,S3  and  S4{S_1},{S_2},{S_3}\,\,and\,\,{S_4} respectively, interference phenomena would be observed in space under appropriate conditions when
View answer and explanation
Correct option: C
Source S2{S_2} emits wave (22) and S4{S_4}  emits wave (44)
Option C (Source S2{S_2} emits wave (22) and S4{S_4}  emits wave (44)) is correct because for interference to occur, the waves must be coherent, meaning they have a constant phase difference. Waves (2) and (4) are both traveling in the +x direction and have the same form, thus they can interfere.
210 Medium
210 of 215
Understanding +4 | -1
The musical interval between two tones of frequencies 320  Hz320\,\,Hz and 240  Hz240\,\,Hz is
View answer and explanation
Correct option: B
(43)\left( {\frac{4}{3}} \right)
Option B ((4/3)) is correct because the musical interval is the ratio of the two frequencies.
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