NEET Waves PYQs with Solutions | 2013–2025

NEET Waves Chapter-wise PYQ Solutions

2025

Q: A pipe open at both ends has a fundamental frequency f. If half its length is submerged in water, what will be the new fundamental frequency of the air column?

Solution:

• Original pipe: Open at both ends → fundamental λ = 2L, f = v/2L
• Half submerged → effectively behaves like a pipe closed at one end → fundamental λ’ = 4(L/2) = 2L, so f’ = v/2L = f
  Answer: f

Concept: Organ pipe fundamental frequency, open vs closed ends.

2024

Q1: Assertion (A): A glass tube partly filled with water behaves like an open organ pipe.
Reason (R): The open end corresponds to an antinode and the end touching water corresponds to a node.

Solution:

• Correct. The tube’s air column forms a standing wave. Open end = antinode, water surface = node.
  Answer: A & R true, R explains A

Concept: Organ pipes and nodes/antinodes.

Q2: For $y = C \sin\left(\frac{2\pi}{\lambda}(at – x)\right)$y=Csin(λ2π​(at−x)), find the frequency.

Solution:

• Wave form: $y = C \sin(kx – \omega t)$y=Csin(kx−ωt)
• Compare: $k = 2\pi/\lambda$k=2π/λ, $\omega = 2\pi a/\lambda$ω=2πa/λ → f = ω/2π = a/λ
  Answer: a/λ

Concept: Traveling wave relation.

2023

Q: Two consecutive harmonics of a tube closed at one end are 220 Hz and 260 Hz. Find the fundamental frequency.

Solution:

• Closed-end pipe → harmonics are odd multiples: f₁, 3f₁, 5f₁…
• Let f₁ = fundamental. Consecutive harmonics: 3f₁ and 5f₁
• Difference = 5f₁ − 3f₁ = 2f₁ → 260 − 220 = 40 → f₁ = 20 Hz
  Answer: 20 Hz

Concept: Harmonics in closed-end pipes.

2021

Q: Listener moving at 1 m/s between two sources of 660 Hz hears beats. Find beats per second.

Solution:

• Relative frequency change due to motion: f’ = f(v ± u)/v
• Two sources → Δf = |f₁ − f₂| = 4 Hz
  Answer: 4 beats/sec

Concept: Beats due to relative motion.

2020

Q: Two sound waves of slightly different frequencies interfere. What phenomenon is observed?

Solution:

• Interference of close frequencies → periodic variation of intensity → beats
  Answer: Beats

Concept: Beats in sound waves.

2019

Q: Which medium allows the highest speed of sound?

Solution:

• Speed of sound: v = √(B/ρ) → higher density and elasticity → solids > liquids > gases
  Answer: Solid

Concept: Speed of sound in different media.

2018

Q: Rope of length L carries transverse wave λ₁ at lower end. If mass distribution changes, wavelength at top = λ₂. Find λ₂/λ₁.

Solution:

• Wave velocity: v = √(T/μ) → λ = v/f
• μ changes along rope → v changes → λ₂/λ₁ = √(μ₁/μ₂)
  Answer: √(μ₁/μ₂)

Concept: Wave propagation on non-uniform medium.

2017

Q: Listener hears echo of siren 800 Hz from cliff, source moves at 15 m/s. Frequency heard in echo?

Solution:

• First Doppler shift (source to cliff): f₁ = f(v/(v − vs))
• Second shift (cliff to listener): f₂ = f₁(v/(v − vs))
• Substitute values (v = 340 m/s, vs = 15 m/s): f₂ ≈ 830 Hz
  Answer: ~830 Hz

Concept: Doppler effect with reflection.

2015

Q: Relationship between velocity v, frequency f, wavelength λ?

Solution:

• Wave relation: v = f λ
  Answer: v = f λ

Concept: Basic wave formula.

2014

Q: Listener hears beats due to two sources 660 Hz and 654 Hz. Find beat frequency.

Solution:

• Beat frequency: f_beat = |f₁ − f₂| = |660 − 654| = 6 Hz
  Answer: 6 beats/sec

Concept: Beats.

2013

Q: Two 660 Hz sources produce beats when listener moves at 1 m/s, wave speed = 330 m/s. Beats/sec?

Solution:

• Doppler shift: Δf = f (u/v) = 660 × (1/330) ≈ 2 Hz per source → total beat frequency ≈ 4 Hz
  Answer: 4 beats/sec

Concept: Doppler effect with motion.