Waves — NEET UG practice

62 questions

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Sample questions with solutions

Q1 · 2026

For sound waves, if the number of nodes for the 5th5^{\text{th}} harmonic of an open-ended pipe is nn and that for the 9th9^{\text{th}} harmonic of the same pipe with one of its ends closed is mm, the ratio nm\frac{n}{m} is

  • A.

    35\frac{3}{5}

  • B.

    59\frac{5}{9}

  • C.

    95\frac{9}{5}

  • D.

    11

Answer: D
  1. Recall that in an open pipe (open at both ends), every harmonic — 1st, 2nd, 3rd, and so on — is allowed, and the number of nodes formed always equals the harmonic number itself.

Given the pipe supports the 5th harmonic,

n=5n = 5
  1. Recall that in a pipe closed at one end, only odd harmonics (1st, 3rd, 5th, 7th, 9th, ...) can exist, and each successive odd harmonic adds one more node than the previous one.

Counting nodes for successive odd harmonics: 1st harmonic → 1 node, 3rd harmonic → 2 nodes, 5th harmonic → 3 nodes, 7th harmonic → 4 nodes, 9th harmonic → 5 nodes.

Therefore, for the 9th harmonic,

m=5m = 5
  1. Since both nn and mm are now known, find their ratio.

Hence,

nm=55=1\frac{n}{m} = \frac{5}{5} = 1

Hence, the answer is D.

Q2 · 2026

For a travelling harmonic wave y(x,t)=2.0cos2π(10t0.0080x+0.35)y(x, t)=2.0 \cos 2\pi(10t - 0.0080x + 0.35), where xx and yy are in cm and tt in ss. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:

  • A.

    0.08π rad0.08\pi \text{ rad}

  • B.

    0.8π rad0.8\pi \text{ rad}

  • C.

    8π rad8\pi \text{ rad}

  • D.

    0.008π rad0.008\pi \text{ rad}

Answer: B
  1. Compare the given equation with the standard form of a travelling wave to identify the wave number.

The general travelling wave equation is written as

y=Acos(ωtkx+ϕ0)y = A\cos(\omega t - kx + \phi_0)

Comparing with the given equation, the wave number is

k=2π×0.0080 rad/cmk = 2\pi \times 0.0080 \text{ rad/cm}
  1. Recall that the phase difference between two points separated by a distance Δx\Delta x is given by Δϕ=kΔx\Delta\phi = k\Delta x. Since kk is in per-cm units, first convert the given separation to cm.

Given Δx=0.5 m=50 cm\Delta x = 0.5\text{ m} = 50\text{ cm},

Δϕ=kΔx\Delta\phi = k\Delta x
  1. Substitute the known values into the phase difference formula.
Δϕ=2π×0.0080×50\Delta\phi = 2\pi \times 0.0080 \times 50
  1. Simplify to get the final phase difference.
Δϕ=0.8π rad\Delta\phi = 0.8\pi \text{ rad}

Hence, the answer is B.

Q3 · 2025

A pipe open at both ends has a fundamental frequency ff in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:

  • A.

    3f2\frac{3f}{2}

  • B.

    2f2f

  • C.

    f2\frac{f}{2}

  • D.

    ff

Answer: D
  1. Recall the fundamental frequency formula for a pipe open at both ends of length LL, where vv is the speed of sound.
f=v2Lf = \frac{v}{2L}
  1. When the pipe is dipped vertically into water up to half its length, the submerged end gets blocked by water, and the air column above the water (of length L/2L/2) now behaves as a pipe closed at one end.

For a closed pipe of length L=L/2L' = L/2, the fundamental frequency is

f=v4Lf' = \frac{v}{4L'}
  1. Substitute L=L/2L' = L/2 into the closed pipe formula.
f=v4(L/2)=v2Lf' = \frac{v}{4(L/2)} = \frac{v}{2L}
  1. Compare this result with the original fundamental frequency f=v2Lf = \frac{v}{2L} from step 1.

Since both expressions are identical,

f=ff' = f

Hence, the answer is D.

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