Waves NEET PYQs (2005–2026) with Solutions

10 solved Waves PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumHarmonics in open and closed pipes

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

   1. AOption A: $\frac{5}{9}$
   2. BOption B: $\frac{9}{5}$
   3. COption C: $1$
   4. DOption D: $\frac{3}{5}$

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   Correct answer: (C) $1$

   Explanation

   Open pipe (both ends open): there are antinodes at both ends, and the $p^{\text{th}}$ harmonic has $p$ nodes in between. For the 5th harmonic, $n = 5$. Closed pipe (one end closed): only odd harmonics exist, $\nu = (2k - 1)\frac{v}{4L}$. The 9th harmonic has $2k - 1 = 9$, i.e. $k = 5$, and this mode has $k = 5$ nodes (including the one at the closed end). So $m = 5$. $\frac{n}{m} = \frac{5}{5} = 1$, option (C).

2. Question 2 (NEET 2026)

   NEET 2026MediumProgressive Waves

   For a travelling harmonic wave y(x,t)=2.0 cos 2π(10t−0.0080x+0.35), where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:

   1. AOption A: 0.08π rad
   2. BOption B: 0.8π rad
   3. COption C: 8π rad
   4. DOption D: 0.008π rad

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   Correct answer: (B) 0.8π rad

   Explanation

   Given y(x,t)=2.0 cos 2π(10t−0.0080x+0.35). Wave number k=2π×0.008. For Δx=0.5 m=50 cm, phase difference Δϕ=kΔx=2π×0.008×50=0.8π rad. Hence option (B) is correct.

3. Question 3 (NEET 2025)

   NEET 2025EasyStanding Waves & Harmonics

   A pipe open at both ends has a fundamental frequency $f$ 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:

   1. AOption A: $\dfrac{f}{2}$
   2. BOption B: $f$
   3. COption C: $\dfrac{3f}{2}$
   4. DOption D: $2f$

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   Correct answer: (B) $f$

   Explanation

   Initially, for an open pipe of length $L$: $$f = \frac{v}{2L}$$ When half the pipe is dipped in water, the effective air column length becomes $L/2$and the pipe behaves as a closed organ pipe. Fundamental frequency for a closed pipe:$$f' = \frac{v}{4(L/2)} = \frac{v}{2L} = f$$ Hence, the new fundamental frequency remains unchanged. Therefore, option (B) is correct.

4. Question 4 (NEET 2023)

   NEET 2023MediumWave Basics

   The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is :

   1. AOption A: 1 : 2
   2. BOption B: 2 : 1
   3. COption C: 1 : 3
   4. DOption D: 3 : 1

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   Correct answer: (B) 2 : 1

   Explanation

   Fundamental frequency of open pipe: fopen = v/2L Fundamental frequency of closed pipe: fclosed = v/4L Therefore, fopen : fclosed = (v/2L) : (v/4L) = 2 : 1 Hence, option (B) is correct.

5. Question 5 (NEET 2022)

   NEET 2022EasyWave Equation & Speed

   If the tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is

   1. AOption A: 1 : 2
   2. BOption B: 1 : 1
   3. COption C: √2 : 1
   4. DOption D: 1 : √2

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   Correct answer: (D) 1 : √2

   Explanation

   Speed of transverse wave on a stretched string: $$v = \sqrt{\frac{T}{\mu}}$$ If tension becomes double: $$v' = \sqrt{\frac{2T}{\mu}} = \sqrt{2}v$$ Therefore, $$v : v' = 1 : \sqrt{2}$$

6. Question 6 (NEET 2020)

   NEET 2020HardBeats

   In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be :

   1. AOption A: 523 Hz
   2. BOption B: 524 Hz
   3. COption C: 536 Hz
   4. DOption D: 537 Hz

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   Correct answer: (B) 524 Hz

   Explanation

   Beat frequency = |fA - fB| = 6 Hz When tension in B decreases, its frequency decreases further and beat frequency increases to 7 Hz. Therefore, originally fB must have been less than 530 Hz. So, fB = 530 - 6 = 524 Hz.

7. Question 7 (NEET 2019)

   NEET 2019MediumStanding Waves & Harmonics

   A tuning fork with frequency 800 Hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. The speed of sound in air is:

   1. AOption A: 172 m/s
   2. BOption B: 500 m/s
   3. COption C: 156 m/s
   4. DOption D: 344 m/s

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   Correct answer: (D) 344 m/s

   Explanation

   For a closed organ pipe, the difference between two successive resonant lengths is λ/2. 31.25 − 9.75 = 21.5 cm ⇒ λ/2 = 21.5 cm ⇒ λ = 43 cm = 0.43 m Using v = fλ: v = 800 × 0.43 = 344 m/s

8. Question 8 (NEET 2018)

   NEET 2018MediumStanding Waves & Harmonics

   A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are obtained at 20 cm and 73 cm column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is:

   1. AOption A: 330 m/s
   2. BOption B: 339 m/s
   3. COption C: 300 m/s
   4. DOption D: 350 m/s

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   Correct answer: (B) 339 m/s

   Explanation

   For a resonance tube closed at one end, the difference between two successive resonant lengths is equal to half the wavelength: $$L_2 - L_1 = \frac{\lambda}{2}$$ Given: $$L_1 = 20\text{ cm}, \quad L_2 = 73\text{ cm}$$ $$L_2 - L_1 = 53\text{ cm} = 0.53\text{ m}$$ Hence, $$\lambda = 2 \times 0.53 = 1.06\text{ m}$$ Velocity of sound: $$v = f\lambda$$ $$v = 320 \times 1.06 = 339.2\text{ m/s}$$ Therefore, the velocity of sound in air is approximately $$339\text{ m/s}$$.

9. Question 9 (NEET 2018)

   NEET 2018MediumStanding Waves & Harmonics

   The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is:

   1. AOption A: 13.2 cm
   2. BOption B: 8 cm
   3. COption C: 16 cm
   4. DOption D: 12.5 cm

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   Correct answer: (A) 13.2 cm

   Explanation

   For a closed organ pipe, the third harmonic frequency is: $$f_c = \frac{3v}{4l}$$ For an open organ pipe, the fundamental frequency is: $$f_o = \frac{v}{2L}$$ Given: $$f_o = f_c$$ Therefore: $$\frac{v}{2L} = \frac{3v}{4l}$$ Cancelling $$v$$: $$\frac{1}{2L} = \frac{3}{4l}$$ $$L = \frac{2l}{3}$$ Given closed pipe length: $$l = 20\text{ cm}$$ $$L = \frac{2 \times 20}{3} = 13.33\text{ cm}$$ Hence, option (A) is correct.

10. Question 10 (NEET 2017)

    NEET 2017MediumStanding Waves & Harmonics

    The two nearest harmonics of a tube closed at one end and open at the other end are $220\,\text{Hz}$ and $260\,\text{Hz}$. What is the fundamental frequency of the system?

    1. AOption A: $40\,\text{Hz}$
    2. BOption B: $10\,\text{Hz}$
    3. COption C: $20\,\text{Hz}$
    4. DOption D: $30\,\text{Hz}$

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    Correct answer: (C) $20\,\text{Hz}$

    Explanation

    For a pipe closed at one end, only odd harmonics are present. Thus, $$\frac{nv}{4l} = 220$$ and $$\frac{(n+2)v}{4l} = 260$$ Subtracting the equations: $$\frac{2v}{4l} = 40$$ $$\frac{v}{4l} = 20\,\text{Hz}$$ Hence, the fundamental frequency is: $$20\,\text{Hz}$$