JEE Physics Waves MCQs Set 19

Answer: (a) vu, vi and va are equal

Explanation:
1. The speed of a mechanical wave depends only on the elasticity and density of the medium (and its temperature).
2. It does not depend on the frequency, so all three travel at the same speed in a given medium.

Teacher's Note:
1. Dispersion of sound in air is negligible.
2. Frequency fixes the wavelength, not the speed.

Answer: (b) 480 Hz

Explanation:
1. The fork forces the wire to vibrate, so the wire performs forced vibrations.
2. A body in forced vibration always oscillates at the frequency of the driving source, here 480 Hz.

Teacher's Note:
1. Forced vibrations take the driver's frequency; resonance only gives a large amplitude when the two match.
2. The natural frequency of the wire affects only the amplitude.

Answer: (d) 6 s slow

Explanation:
1. Time taken by sound to reach him: \( t=\frac{2000}{330}\approx6\ \text{s} \).
2. He starts his watch 6 s after the whistle actually blew, so his watch shows an earlier time than the true time: it is slow by 6 s.

Teacher's Note:
1. The delay always makes the watch lag behind the true time.
2. Light reaches him almost instantly, sound does not.

Answer: (a) 10 cm/s

Explanation:
1. Wavelength = distance between consecutive crests = 5 cm; frequency = 2 Hz.
2. \( v=f\lambda=2\times5=10\ \text{cm/s} \)

Teacher's Note:
1. v = fλ is the basic wave relation.
2. Crest-to-crest distance is one wavelength.

Answer: (b) Frequency

Explanation:
1. The frequency is set by the source and the number of crests arriving per second cannot change at the boundary.
2. The speed changes (higher in water) and so does the wavelength, since \( \lambda=\frac vf \).

Teacher's Note:
1. Frequency is the same on both sides of a boundary.
2. Speed and wavelength change together.

Answer: (a) v

Explanation:
1. The speed depends on the medium only, not on the frequency.
2. Increasing f to 4f makes the wavelength one fourth, but the speed remains v.

Teacher's Note:
1. Check with v = fλ: f increases and λ decreases in the same proportion.
2. The medium sets the speed.

Answer: (c) 9 m/s

Explanation:
1. Frequency: \( f=\frac{54}{60}=0.9\ \text{Hz} \).
2. \( v=f\lambda=0.9\times10=9\ \text{m/s} \)

Teacher's Note:
1. Convert waves per minute to hertz first.
2. Then use v = fλ.

Answer: (c) 2 : 1

Explanation:
1. For a gas, \( v=\sqrt{\frac{\gamma P}{\rho}} \). Both are monatomic, so γ is the same (5/3), and P is the same.
2. \( \frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt4=2 \)

Teacher's Note:
1. Speed varies inversely as the square root of the density at fixed pressure and γ.
2. The lighter gas carries sound faster.

Answer: (c) 1092 K

Explanation:
1. Speed is proportional to the square root of absolute temperature: \( v\propto\sqrt T \).
2. Doubling v needs four times the absolute temperature: \( T=4\times273=1092\ \text{K} \)

Teacher's Note:
1. Always use kelvin for v ∝ √T.
2. Double speed means four times T.

Answer: (a) transverse waves

Explanation:
1. In a transverse wave the particle displacement is perpendicular to the direction in which the wave travels (for example, waves on a string).
2. In a longitudinal wave the vibrations are along the direction of travel.

Teacher's Note:
1. Transverse waves show crests and troughs.
2. Sound in air is longitudinal.

Answer: (a) directly on the square of the wave amplitude and the square of the wave frequency

Explanation:
1. Energy per unit volume of a wave is \( \tfrac12\rho\omega^2A^2 \).
2. Multiplying by the wave speed gives the power per unit area: \( \tfrac12\rho v\omega^2A^2 \), proportional to \( A^2f^2 \).

Teacher's Note:
1. Power is proportional to the square of the amplitude and the square of the frequency.
2. Doubling the amplitude quadruples the power.

Answer: (d) sound wave in a gas

Explanation:
1. X-rays, gamma rays and visible light are electromagnetic waves, which are transverse.
2. A gas cannot support shear, so a sound wave in a gas is longitudinal.

Teacher's Note:
1. Electromagnetic waves are always transverse.
2. Fluids carry only longitudinal sound waves.

Answer: (a) \( \sqrt{\frac{32}{17}} \)

Explanation:
1. Take molar masses 2 (H₂) and 32 (O₂). Equal volumes means equal moles, so the mean molar mass is \( \frac{2+32}{2}=17 \).
2. Both gases are diatomic, so γ is the same and \( v\propto\frac{1}{\sqrt M} \).
3. \( \frac{v_{mix}}{v_{O_2}}=\sqrt{\frac{32}{17}} \)

Teacher's Note:
1. Equal volumes under the same conditions contain equal numbers of moles.
2. The mean molar mass of the mixture is the simple average.

Answer: (a) \( A\omega \)

Explanation:
1. Particle velocity: \( \frac{\partial y}{\partial t}=-A\omega\cos(kx-\omega t) \).
2. Its maximum magnitude is \( A\omega \).

Teacher's Note:
1. ω/k is the wave speed, not the particle speed.
2. Particle velocity depends on the amplitude and the angular frequency.

Answer: (b) 300 m/s

Explanation:
1. Compare with \( y=A\sin(\omega t-kx+\phi) \): ω = 600 rad/s and k = 2 rad/m.
2. \( v=\frac\omega k=\frac{600}{2}=300\ \text{m/s} \)

Teacher's Note:
1. The wave speed is the ratio of the coefficients of t and x.
2. The phase constant does not affect the speed.

Answer: (b) 12.5 cm

Explanation:
1. Wavelength: \( \lambda=\frac vf=\frac{300}{400}=0.75\ \text{m} \).
2. A phase difference of 60° is one sixth of a cycle: \( \Delta x=\frac{\lambda}{6}=0.125\ \text{m}=12.5\ \text{cm} \)

Teacher's Note:
1. Path difference = (phase difference / 2π) × λ.
2. 60° = π/3 rad.

Answer: (d) 500 Hz

Explanation:
1. Source moving toward a stationary observer: \( f'=f\frac{v}{v-v_s} \).
2. \( f'=450\times\frac{330}{330-33}=450\times\frac{330}{297}=500\ \text{Hz} \)

Teacher's Note:
1. Approaching source: denominator is v − v_s.
2. Receding source would give v + v_s.

Answer: (a) 436 Hz to 586 Hz

Explanation:
1. Speed of the whistle: \( u=\omega r=\frac{400\times2\pi}{60}\times1.2\approx50.3\ \text{m/s} \).
2. Lowest frequency (receding): \( f_{min}=\frac{500\times340}{340+50.3}\approx436\ \text{Hz} \).
3. Highest frequency (approaching): \( f_{max}=\frac{500\times340}{340-50.3}\approx587\ \text{Hz} \), about 586 Hz in the rounded form of the option.

Teacher's Note:
1. The extremes occur when the whistle moves directly toward or away from the listener.
2. Use the tangential speed of the whistle.

Answer: (d) \( \frac{19}{18} \)

Explanation:
1. \( f_1=\frac{fv}{v-34},\ f_2=\frac{fv}{v-17} \).
2. \( \frac{f_1}{f_2}=\frac{340-17}{340-34}=\frac{323}{306}=\frac{19}{18} \)

Teacher's Note:
1. A faster source raises the frequency more.
2. The ratio is of the denominators, inverted.

Answer: (b) 1000

Explanation:
1. Sound level \( L=10\log_{10}\frac{I}{I_0} \).
2. 30 = 10 log(I/I0) gives log(I/I0) = 3, so \( \frac{I}{I_0}=10^3=1000 \)

Teacher's Note:
1. Every 10 dB is a factor of 10 in intensity.
2. 30 dB is three factors of 10.

Answer: (b) 1.90 Å

Explanation:
1. For v much less than c, \( \frac{\Delta\lambda}{\lambda}=\frac vc=\frac{10^5}{3\times10^8}=\frac13\times10^{-3} \).
2. \( \Delta\lambda=5700\times\frac13\times10^{-3}=1.9\ \text{Å} \) (red shift, since the star recedes).

Teacher's Note:
1. Receding sources shift lines to longer wavelengths.
2. Use Δλ/λ = v/c.

Answer: (b) varies with the distance separating them

Explanation:
1. In a progressive wave each particle repeats the motion of the particle before it after a delay, so the phase depends on position: \( \Delta\phi=\frac{2\pi}{\lambda}\Delta x \).
2. At a given pair of positions the difference stays fixed in time.

Teacher's Note:
1. Phase difference is determined by the separation, not by the instant considered.
2. Particles one wavelength apart are in phase.

Answer: (b) 0.12 N

Explanation:
1. Wave speed: \( v=\frac\omega k=\frac{30}{1}=30\ \text{m/s} \).
2. \( T=\mu v^2=1.3\times10^{-4}\times900=0.117\ \text{N}\approx0.12\ \text{N} \)

Teacher's Note:
1. For a string, v = √(T/μ).
2. k = 1 rad/m here.

Answer: (b) 1.07 rad

Explanation:
1. Write the cosine as a sine: \( y_2=10^{-6}\sin\left[100t+\frac x{50}+\frac\pi2\right] \).
2. Phase difference: \( \frac\pi2-0.5=1.5708-0.5\approx1.07\ \text{rad} \)

Teacher's Note:
1. Convert both waves to the same trigonometric function before subtracting phases.
2. cos θ = sin(θ + π/2).

Answer: (b) 98 m

Explanation:
1. Observer moving away from the source: \( f'=f\frac{v-u}{v}=0.94f\Rightarrow u=0.06\times330=19.8\ \text{m/s} \).
2. Distance covered from rest: \( s=\frac{u^2}{2a}=\frac{19.8^2}{2\times2}=98\ \text{m} \)

Teacher's Note:
1. First find the speed at which the frequency drops to 94%.
2. Then use v² = 2as.