JEE Physics Waves MCQs Set 20

Answer: (c) λ/6

Explanation:
1. Path difference = (phase difference / 2π) × λ.
2. \( \Delta x=\frac{60^\circ}{360^\circ}\lambda=\frac\lambda6 \)

Teacher's Note:
1. A full cycle of 360° corresponds to one wavelength.
2. 60° is one sixth of 360°.

Answer: (d) Only I and IV are true

Explanation:
1. For air treated as an ideal gas \( v=\sqrt{\frac{\gamma RT}{M}} \), which grows with temperature (I true, II false).
2. At constant temperature a change in pressure changes the density in proportion, so the ratio P/ρ is fixed: v does not depend on pressure (IV true, III false).

Teacher's Note:
1. Temperature matters; pressure alone does not.
2. Use P/ρ = RT/M.

Answer: (d) \( y=f(x^2-v^2t^2) \)

Explanation:
1. A travelling wave of fixed shape has displacement that depends on x and t only through the combination \( x-vt \) or \( x+vt \).
2. The first three are functions of \( x\mp vt \). The last is a function of \( x^2-v^2t^2=(x-vt)(x+vt) \), which is not a function of either combination alone, so it is not a travelling wave.

Teacher's Note:
1. Test: can the expression be written as f(x − vt) or f(x + vt)?
2. Products of both are not wave functions.

Answer: (b) a wave travelling in the negative x-direction with a velocity of 1.5 m/s

Explanation:
1. Positive sign between the x and t terms means the wave moves toward negative x.
2. Speed: \( v=\frac\omega k=\frac{15\pi}{10\pi}=1.5\ \text{m/s} \). Wavelength: \( \lambda=\frac{2\pi}{k}=\frac{2\pi}{10\pi}=0.2\ \text{m} \).
3. So the wave moves in the negative x-direction at 1.5 m/s with wavelength 0.2 m, which matches only option (b)

Teacher's Note:
1. Sign of kx and ωt together fixes the direction of travel.
2. Same sign means the wave moves toward negative x.

Answer: (b) 2

Explanation:
1. Observer approaching a stationary source: \( f'=f\frac{v+u}{v} \).
2. Train A: \( 5.5=5\left(1+\frac{u_A}{v}\right)\Rightarrow u_A=0.1v \). Train B: \( 6.0=5\left(1+\frac{u_B}{v}\right)\Rightarrow u_B=0.2v \).
3. \( \frac{u_B}{u_A}=2 \)

Teacher's Note:
1. Solve for the speed of each train as a fraction of the speed of sound.
2. The ratio of the speeds is the ratio of the fractional frequency rises.

Answer: (a) \( \text{amplitude is less than }\frac{\lambda}{2\pi} \)

Explanation:
1. Maximum particle speed: \( A\omega \). Wave speed: \( v=\frac\omega k \).
2. Condition \( A\omega<\frac\omega k\Rightarrow A<\frac1k=\frac{\lambda}{2\pi} \)

Teacher's Note:
1. Compare Aω with ω/k.
2. 1/k equals λ/2π.

Answer: (d) 600 Hz

Explanation:
1. When a wave passes from one medium to another, its frequency stays the same.
2. Only the speed and wavelength change (wavelength in air is 0.5 m, in water 2.5 m).

Teacher's Note:
1. Frequency is a property of the source.
2. Do not apply the Doppler formula; nothing is moving.

Answer: (c) sum of kinetic energy and potential energy

Explanation:
1. In SHM the kinetic and potential energies of one particle exchange continuously: each changes with time.
2. Their sum, the total mechanical energy \( \tfrac12m\omega^2A^2 \), stays constant.

Teacher's Note:
1. Individual energy forms oscillate at twice the particle frequency.
2. The total is conserved for ideal SHM.

Answer: (c) 0.5 m/s

Explanation:
1. The pulse at t = 0 is centred at x = 0 and at t = 2 s it is centred at x = 1 m.
2. The pulse has moved 1 m in 2 s: \( v=\frac{1}{2}=0.5\ \text{m/s} \)

Teacher's Note:
1. Replacing x by x − vt shifts a fixed shape.
2. Compare (x − 1) with (x − vt) at t = 2.

Answer: (c) 10 m

Explanation:
1. The level must fall by 20 dB, so the intensity must fall by a factor of 100.
2. Intensity from a point source varies as \( \frac1{r^2} \), so the distance must increase by a factor of 10: \( r=10\ \text{m} \)

Teacher's Note:
1. 20 dB corresponds to a factor 100 in intensity.
2. Inverse-square law gives the distance.

Answer: (d) 0.12 m

Explanation:
1. Tension at the lower end: \( T_1=2g=20\ \text{N} \); at the top: \( T_2=(2+6)g=80\ \text{N} \).
2. Wave speed \( v=\sqrt{T/\mu} \) doubles, and the frequency stays the same, so \( \lambda\propto v \) doubles: 0.12 m.

Teacher's Note:
1. Frequency does not change as the pulse travels.
2. Tension increases upward because it carries the weight of the rope below.

Answer: (b) 2

Explanation:
1. Situations (1) and (3) are single progressive waves (functions of 3x − 4t and 5x + 4t).
2. Situation (2) is a standing wave, which is the sum of two equal waves travelling in opposite directions: \( 2\sin3x\cos4t=\sin(3x+4t)+\sin(3x-4t) \)

Teacher's Note:
1. A standing wave always comes from two opposite waves.
2. The product of a function of x and a function of t signals a stationary wave.

Answer: (d) 7, 6

Explanation:
1. The nth harmonic has n loops, so 6 antinodes.
2. The number of nodes is one more than the number of loops: 7 (including the two fixed ends).

Teacher's Note:
1. Fixed ends are nodes.
2. Nodes = n + 1, antinodes = n.

Answer: (d) 9

Explanation:
1. Equal numbers of nodes and antinodes means the pipe is closed at one end (node at the closed end, antinode at the open end).
2. In a closed pipe with 5 loops of a quarter wavelength, \( L=\frac{9\lambda}{4} \), which is the ninth harmonic (only odd harmonics occur).

Teacher's Note:
1. Closed pipe: n nodes ⇒ harmonic number 2n − 1.
2. Open pipe has one more antinode than node.

Answer: (a) open pipe

Explanation:
1. A pipe closed at one end supports only odd harmonics (1, 3, 5, ...).
2. An open pipe supports all harmonics, so a tenth (even) harmonic can only occur in an open pipe.

Teacher's Note:
1. Even harmonics are missing in a closed pipe.
2. Open pipes have all harmonics.

Answer: (c) 150 Hz, 450 Hz

Explanation:
1. The given frequencies are all multiples of 150 Hz, so the fundamental is 150 Hz and the tube is open (all harmonics present).
2. Harmonics below 1000 Hz: 150, 300, 450, 600, 750, 900. Missing from the list: 150 Hz and 450 Hz.

Teacher's Note:
1. Find the highest common factor of the given frequencies.
2. In a closed tube 300 and 600 would not both appear.

Answer: (b) 292 cps

Explanation:
1. The unknown fork has frequency 284 or 292 Hz (4 beats from 288 Hz).
2. Wax lowers the frequency of the unknown fork. If it were 284 Hz, it would go lower and the beats would increase. If it were 292 Hz, it would move closer to 288 Hz and the beats would decrease to 2 (about 290 Hz).

Teacher's Note:
1. Loading a fork with wax lowers its frequency.
2. A fall in the beat frequency shows the two forks have moved closer.

Answer: (b) \( y=-a\cos(kx+\omega t) \)

Explanation:
1. A stationary wave needs a wave of equal amplitude travelling in the opposite direction, so the second wave contains \( kx+\omega t \).
2. Node at x = 0 for all t requires the sum to vanish there: \( a\cos(-\omega t)+y_2(0,t)=0\Rightarrow y_2(0,t)=-a\cos\omega t \).
3. So \( y_2=-a\cos(kx+\omega t) \), giving \( y=2a\sin kx\sin\omega t \)

Teacher's Note:
1. Opposite direction means the sign between kx and ωt flips.
2. Check the sum at x = 0.

Answer: (d) when two identical waves with a phase difference of π are moving in the same direction

Explanation:
1. A stationary wave needs two waves of the same frequency travelling in opposite directions.
2. Two waves moving in the same direction just combine into another travelling wave (here, with phase difference π, they cancel completely).

Teacher's Note:
1. Stationary waves come from reflection or from counter-propagating waves.
2. Same-direction waves never form nodes and antinodes.

Answer: (a) 1.21 Å

Explanation:
1. The atoms are at the end nodes: 3 nodes and 2 antinodes make 2 loops, each of length λ/2.
2. So \( 2\times\frac\lambda2=1.21\ \text{Å}\Rightarrow\lambda=1.21\ \text{Å} \)

Teacher's Note:
1. Count loops, not nodes.
2. Each loop spans half a wavelength.

Answer: (d) 10 m

Explanation:
1. A standing wave is flat twice per period, so the minimum time between flat instants is T/2 = 0.5 s and T = 1 s.
2. Wavelength: \( \lambda=vT=10\times1=10\ \text{m} \)

Teacher's Note:
1. The string is flat twice in each period.
2. Use λ = vT.

Answer: (c) 5 Hz

Explanation:
1. 5 segments (loops) in 10 m: \( 5\cdot\frac\lambda2=10\Rightarrow\lambda=4\ \text{m} \).
2. \( f=\frac v\lambda=\frac{20}{4}=5\ \text{Hz} \)

Teacher's Note:
1. Each segment is half a wavelength.
2. Then f = v/λ.

Answer: (c) 40 cm

Explanation:
1. Fundamental frequency \( f=\frac{1}{2L}\sqrt{\frac T\mu} \), so \( f\propto\frac1L \) at fixed tension.
2. \( L_2=50\times\frac{800}{1000}=40\ \text{cm} \)

Teacher's Note:
1. Shorter string, higher frequency.
2. Tension is assumed unchanged.

Answer: (a) 6

Explanation:
1. Approaching train: \( f_1=240\times\frac{320}{320-4}=243.04\ \text{Hz} \).
2. Receding train: \( f_2=240\times\frac{320}{320+4}=237.04\ \text{Hz} \).
3. Beats per second: \( f_1-f_2=6 \)

Teacher's Note:
1. Use the source-moving form of the Doppler formula for each train.
2. Beat frequency is the difference of the two frequencies.

Answer: (d) Sound travels faster in vacuum than in air

Explanation:
1. Sound is a mechanical wave and needs a material medium, so it cannot travel through a vacuum at all.
2. The other three statements are correct: sound is energy carried as waves that move along straight lines in a uniform medium.

Teacher's Note:
1. No medium means no sound.
2. Light, not sound, travels through a vacuum.