Waves JEE Main PYQs (2020–2026)

Solved Waves PYQs

3 recent questions with answers. The full set of 95 is free to practise.

1. JEE Main 2026EasyBeats
   Q1

   Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________ Hz.

   Numerical answer type

   Show answer & solution

   Correct answer: 384

   Before waxing the fork, A
   
   The beat frequency is the number of beats per second.
   
   $$f_b=\frac{\text{ Number of beats }}{\text{ Time }}=\frac{8\text{ beats }}{2s}=4Hz$$
   
   This means the absolute difference between the frequencies of fork $A\left(f_A\right)$ and fork $B\left(f_B\right)$ is 4 Hz :
   
   $$\left|f_A-f_B\right|=4Hz$$
   
   Since $f_B=380Hz$, the original frequency of fork A could be either :
   
   Case 1 : $f_A=380+4=384Hz$
   
   Case 2 : $f_A=380-4=376Hz$
   
   When a tuning fork is loaded with wax, its mass increases, which causes its frequency to decrease. Let the new frequency of fork $A$ be $f_A^{\prime }$, where $f_A^{\prime }<f_A$.
   
   The new beat frequency ( $f_b^{\prime }$ ) is :
   
   $$f_b^{\prime }=\frac{4\text{ beats }}{2s}=2Hz$$
   
   Initially, $f_A$ is higher than $f_B$ by $4Hz(384-380=4)$.
   
   If $f_A$ decreases (to say 382 Hz ), the gap between $f_A$ and $f_B$ decreases $(382-380=2)$.
   
   The original frequency of tuning fork A must have been higher than fork B to allow the beat frequency to decrease as $f_A$ dropped.
   
   Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is $384$.
2. JEE Main 2026MediumSpeed of Mechanical Waves
   Q2

   The velocity of sound in air is doubled when the temperature is raised from $0^{\circ }C$ to ${}^{\circ }C$ . The value of $\alpha$ is $\_\_\_\_$ .

   Numerical answer type

   Show answer & solution

   Correct answer: 819

   The velocity of sound in an ideal gas depends on the absolute temperature of the medium.
   
   The speed of sound (v) in air is given by the formula :
   
   $$v=\sqrt{\frac{\gamma RT}{M}}$$
   
   From this, we can see that :
   
   $$v\propto \sqrt{T}$$
   
   where T is the absolute temperature in Kelvin $(K)$ .
   
   Initial State :
   
   Temperature $=T_1=0^{\circ }C=(0+273)K=273K$
   
   Velocity $=v_1$
   
   Final State :
   
   Temperature $=T_2={\alpha }^{\circ }C=(\alpha +273)K$
   
   Velocity $=v_2=2v_1($ since the velocity is doubled $)$ .
   
   Using the proportionality $v\propto \sqrt{T}$ :
   
   $$\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$$
   
   Substituting the values :
   
   $$\frac{2v_1}{v_1}=\sqrt{\frac{\alpha +273}{273}}$$
   
   $$\Rightarrow 4=\frac{\alpha +273}{273}$$
   
   $$\Rightarrow 4\times 273=\alpha +273$$
   
   $$\Rightarrow \alpha =1092-273$$
   
   $$\Rightarrow \alpha =819$$
   
   Therefore, the value of $\alpha$ is 819 . Hence, the correct answer is $19$ .
3. JEE Main 2026EasyWave Motion and Equation of a Wave
   Q3

   The equation of a plane progressive wave is given by $y=5\cos \pi \left(200t-\frac{x}{150}\right)$ where $x$ and $y$ are in cm and $t$ is in second. The velocity of the wave is ________ m/s.

   • A120
   • B150
   • C200
   • D300

   Show answer & solution

   Correct answer: (D)

   A plane progressive wave moving in the positive $x$ -direction can be represented as:
   
   $$y=Acos(\omega t-kx)$$
   
   Where, $\omega$ is the angular frequency, $k$ is the propagation constant (wave number).
   
   The given equation is
   
   $$y=5\cos \pi \left(200t-\frac{x}{150}\right).$$
   
   $$y=5\cos \left(200\pi t-\frac{\pi x}{150}\right)$$
   
   By comparing this with $y=Acos(\omega t-kx),\omega =200\pi$ rad/s and $k=\frac{\pi }{150}$ rad/cm
   
   Wave velocity is the distance travelled by the wave per unit time. It is related to angular frequency and wave number by the relation:
   
   $$v=\frac{\omega }{k}$$
   
   Substituting the values of $\omega$ and $k$ :
   
   $$v=\frac{200\pi }{\pi /150}$$
   
   $$v=200\times 150=30,000\text{cm/s}=300\text{m/s}$$
   
   Therefore, the velocity of the wave is $300\text{m/s}$ . Thus, the correct option is (D).

Topic-wise question count

Where JEE Main has asked from Waves, most frequent first.

• Standing Waves and Resonance30
• Wave Motion and Equation of a Wave22
• Doppler Effect17
• Speed of Mechanical Waves11
• Superposition and Interference of Waves10
• Beats5

Year-wise Waves questions in JEE Main

Number of questions from this chapter in each year's paper(s).