JEE Physics Waves MCQs Set 10

Answer: (a) a combination of transverse and longitudinal motion of the water particles

Explanation:
1. Water particles on the surface move in small circular or elliptical paths.
2. This motion has both a component along and a component perpendicular to the direction of travel.

Teacher's Note:
1. Waves on the surface of a liquid are not purely one type.
2. Inside a liquid, sound waves are purely longitudinal.

Answer: (b) longitudinal

Explanation:
1. Polarisation restricts the direction of the vibrations perpendicular to the direction of travel.
2. In sound waves the particles vibrate along the direction of travel, so there is no perpendicular direction to restrict.

Teacher's Note:
1. Only transverse waves can be polarised.
2. Light can be polarised, sound cannot.

Answer: (a) a solid but not in a gas

Explanation:
1. A transverse mechanical wave needs a medium that can resist shear (rigidity).
2. Solids have rigidity; gases (and liquids in bulk) do not.

Teacher's Note:
1. Rigidity is required for shear waves.
2. Gases carry only longitudinal waves.

Answer: (a) wavelength 18 cm

Explanation:
1. Compare with y = A sin(ωt − kx + φ): A = 4 cm, ω = π/5 rad/s, k = π/9 rad/cm.
2. λ = 2π/k = 18 cm; f = ω/2π = 0.1 Hz; v = ω/k = 1.8 cm/s.

Teacher's Note:
1. Read ω and k from the coefficients of t and x.
2. Then λ = 2π/k and v = ω/k.

Answer: (a) wave speed 2 units

Explanation:
1. The constant 4 only shifts the mean position; the amplitude is 2.
2. v = ω/k = 6/3 = 2 units.

Teacher's Note:
1. A constant added to y does not change amplitude.
2. v = coefficient of t divided by coefficient of x.

Answer: (a) Energy and linear momentum

Explanation:
1. A wave carries energy and momentum from point to point.
2. The particles only oscillate about their mean positions, so no mass is transported.

Teacher's Note:
1. Waves transport energy and momentum, not matter.
2. Particles do not travel with the wave.

Answer: (d) 1000

Explanation:
1. v = 600/2 = 300 m/s, so λ = v/f = 300/500 = 0.6 m.
2. Number of wavelengths = 600/0.6 = 1000.

Teacher's Note:
1. Find the speed first, then the wavelength.
2. Number of waves = distance / λ.

Answer: (c) \( \frac32\pi \)

Explanation:
1. ω = 25π rad/s, k = π/2 rad/m, so v = ω/k = 50 m/s.
2. Maximum particle velocity = Aω = 3 × 25π = 75π m/s.
3. Ratio = 75π/50 = 3π/2.

Teacher's Note:
1. Particle velocity amplitude is Aω; wave speed is ω/k.
2. The ratio equals Ak.

Answer: (b) y = (0.02 m) sin(7.85x − 1005t)

Explanation:
1. λ = 4/5 = 0.8 m, so k = 2π/0.8 = 7.85 rad/m.
2. ω = kv = 7.85 × 128 = 1005 rad/s.
3. A wave in the +x direction has the form sin(kx − ωt).

Teacher's Note:
1. sin(kx − ωt) travels towards +x.
2. Check ω = 2πf with f = v/λ = 160 Hz.

Answer: (a) 1/2

Explanation:
1. v = √(γP/ρ). For the same γ and P, v ∝ 1/√ρ.
2. v1/v2 = √(ρ2/ρ1) = √(1/4) = 1/2.

Teacher's Note:
1. Same pressure and same γ: only density differs.
2. Denser gas, slower sound.

Answer: (a) f remains the same but λ decreases

Explanation:
1. The frequency is fixed by the source and does not change at a boundary.
2. λ = v/f and v falls from 1450 to 330 m/s, so λ decreases.

Teacher's Note:
1. Frequency is unchanged at a boundary.
2. Wavelength changes in proportion to speed.

Answer: (d) vm > vd

Explanation:
1. Water vapour is lighter than the nitrogen and oxygen it replaces, so moist air has lower density.
2. v ∝ 1/√ρ, so sound is faster in moist air.

Teacher's Note:
1. Lower density means faster sound at the same pressure and temperature.
2. Moist air is less dense than dry air.

Answer: (c) 6.25 N

Explanation:
1. f = 1/0.04 = 25 Hz and λ = 0.50 m, so v = fλ = 12.5 m/s.
2. T = μv² = 0.04 × 12.5² = 6.25 N.

Teacher's Note:
1. Use v = √(T/μ) in the form T = μv².
2. Read T and λ from the bracket.

Answer: (a) 5 m/s and 2.24 m/s

Explanation:
1. At a distance x from the lower end the tension is the weight of the rope below, T = μgx, so v = √(gx).
2. Upper end: √(10 × 2.5) = 5 m/s.
3. 0.5 m from the bottom: √(10 × 0.5) = 2.24 m/s.

Teacher's Note:
1. Tension increases upwards, so the speed increases upwards.
2. μ cancels in v = √(gx).

Answer: (b) becomes one fourth

Explanation:
1. Intensity ∝ (amplitude)² × (frequency)².
2. New/old = 2² × (1/4)² = 4/16 = 1/4.

Teacher's Note:
1. I ∝ a²f².
2. Square each factor separately.

Answer: (a) a frequency of 500 Hz with the intensity changing twice per second

Explanation:
1. The heard note has the mean frequency (499 + 501)/2 = 500 Hz.
2. The beat frequency is 501 − 499 = 2 Hz, so the intensity rises and falls twice per second.

Teacher's Note:
1. Tone = average frequency, beats = difference.
2. The beat frequency is the number of intensity maxima per second.

Answer: (d) 120 Hz

Explanation:
1. Frequencies form an AP with difference 8 Hz: n16 = n1 + 15 × 8.
2. n16 = 2n1 gives n1 + 120 = 2n1, so n1 = 120 Hz.

Teacher's Note:
1. Consecutive differences are equal.
2. There are 15 gaps between 16 forks.

Answer: (c) 252 Hz

Explanation:
1. B is 252 Hz or 260 Hz.
2. Wax lowers B. If B were 260 Hz, lowering it would reduce the beats, but the beats increase to 6.
3. So B = 252 Hz (lowered to 250 Hz, giving 6 beats).

Teacher's Note:
1. Adding wax lowers the frequency.
2. Beats increasing means the forks moved further apart.

Answer: (b) 252 Hz

Explanation:
1. B is 252 Hz or 260 Hz. Wax lowers B, and the beats rise to 6, so B was below A: B = 252 Hz.
2. Removing wax restores B towards 252 Hz, where the beats are again 4.

Teacher's Note:
1. Waxing lowers the frequency; scraping raises it back.
2. Beats return to 4 only when B returns to its original value.

Answer: (a) 284 Hz

Explanation:
1. A is 284 Hz or 292 Hz.
2. Loading lowers A; the beats fall, so A moves closer to 288 Hz, which means A was below 288 Hz.
3. A = 284 Hz.

Teacher's Note:
1. Wax lowers the frequency.
2. Fewer beats mean the forks are closer in frequency.

Answer: (d) 4 beats per second with intensity ratio 81 : 1 between waxing and waning

Explanation:
1. f1 = 250 Hz and f2 = 254 Hz, so beats = 4 per second.
2. Imax/Imin = ((5 + 4)/(5 − 4))² = 81/1.

Teacher's Note:
1. ω = 2πf.
2. Intensity ratio is the square of (A1 + A2)/(A2 − A1).

Answer: (c) 404 m/s

Explanation:
1. f1 − f2 = v(1/2 − 1/2.02) = v × 0.02/4.04 = 2.
2. v = 2 × 4.04/0.02 = 404 m/s.

Teacher's Note:
1. f = v/λ for each wave.
2. Beats = difference of frequencies.

Answer: (a) 1024 Hz

Explanation:
1. Closed pipe: f = v/4L = 512 Hz.
2. Open pipe: f = v/2L = 2 × 512 = 1024 Hz.

Teacher's Note:
1. For the same length an open pipe has twice the fundamental of a closed pipe.
2. Only the factor 4L → 2L changes.

Answer: (b) 31.25

Explanation:
1. For the fundamental, L = v/4f = 330/(4 × 264) m = 0.3125 m.
2. L = 31.25 cm.

Teacher's Note:
1. Closed-pipe lengths are odd multiples of λ/4.
2. The shortest length is λ/4.

Answer: (a) \( \frac{V+v}{4(L+\ell)} \)

Explanation:
1. Heating expands the tube, so its length becomes L + ℓ, and increases the speed of sound to V + v.
2. Fundamental of a closed tube: f = speed/(4 × length) = (V + v)/[4(L + ℓ)].

Teacher's Note:
1. Heating increases both length and speed.
2. Use f = v/4L with the new values.