Waves Previous Year Questions — JEE Main Physics
22 past-year Waves questions from JEE Main (Physics) — page 2.
Q8 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: A sound wave has higher speed in solids than gases.
Reason R: Gases have higher value of Bulk modulus than solids.
In the light of the above statements, choose the correct answer from the options given below.
- $\text{Both A and R are true and R is the correct explanation of A}$
- $\text{A is false but R is true}$
- $\text{Both A and R are true but R is NOT the correct explanation of A}$
- $\text{A is true but R is false}$
### Related Formula
$$v = \sqrt{\frac{\mathrm{B}}{\rho}}$$
### Core Logic
Assertion A: Sound velocity relies on structural elasticity bounds. Solids are highly rigid compared to fluids, making speed significantly higher. (True)
Reason R: Solids resist structural compression far better than unbonded gases, giving them significantly higher Bulk Modulus properties. Thus, statement R is completely false.
### Step 1: Final Conclusion
Assertion A is true, but Reason R is false, aligning perfectly with option (4).
### Pattern Recognition
Even though density $\rho$ is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q18 (2025)
Displacement of a wave is expressed as $x(t)=5\cos\left(628t+\frac{\pi}{2}\right)\text{ m}$. The wavelength of the wave when its velocity is 300 m/s is:
- 5 m
- 3 m
- 0.5 m
- 0.33 m
### Related Formula
$$x(t) = A\cos(\omega t + \phi)$$
$$v = \frac{\omega}{K}$$
$$K = \frac{2\pi}{\lambda}$$
### Core Logic
From the given wave equation, angular frequency $\omega = 628\text{ rad/s}$.
Given wave velocity $v = 300\text{ m/s}$.
Using the relation $v = \frac{\omega}{K}$:
$$300 = \frac{628}{K} \implies K = \frac{628}{300}$$
### Step 1: Compute Wavelength
Substitute $K = \frac{2\pi}{\lambda}$:
$$\frac{2\pi}{\lambda} = \frac{628}{300}$$
Since $2\pi \approx 2 \times 3.14 = 6.28$, the expression simplifies neatly:
$$\frac{6.28}{\lambda} = \frac{628}{300} \implies \lambda = 3\text{ m}$$
### Pattern Recognition
Notice standard values like $\omega = 628 = 200\pi$, which means the frequency is exactly $100\text{ Hz}$. Using $v = f\lambda \implies 300 = 100\lambda \implies \lambda = 3\text{ m}$ avoids setting up fractions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q907 (2025)
Consider the sound wave travelling in ideal gases of $\mathrm{He}$, $\mathrm{CH}_{4}$, and $\mathrm{CO}_{2}$. All the gases have the same ratio $\frac{P}{
ho}$, where $P$ is the pressure and $
ho$ is the density. The ratio of the speed of sound through the gases $v_{\mathrm{He}} : v_{\mathrm{CH}_{4}} : v_{\mathrm{CO}_{2}}$ is given by
- $\sqrt{\frac{7}{5}} : \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}}$
- $\sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{7}{5}}$
- $\sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{4}{3}}$
- $\sqrt{\frac{4}{3}} : \sqrt{\frac{5}{3}} : \sqrt{\frac{7}{5}}$
### Related Formula
Laplace correction equation for speed of sound:
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
Given that $\frac{P}{\rho}$ is constant for all three gases:
$$v \propto \sqrt{\gamma}$$
where $\gamma = 1 + \frac{2}{f}$ (adiabatic constant).
### Core Logic
Determine the $\gamma$ factor based on molecular atomic structures:
1. $\mathrm{He}$ (Monatomic) $\implies f = 3 \implies \gamma_{\mathrm{He}} = \frac{5}{3}$
2. $\mathrm{CH}_{4}$ (Polyatomic/Non-linear) $\implies \gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}$ based on experimental references.
3. $\mathrm{CO}_{2}$ (Triatomic linear/vibrational modes) $\implies \gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}$ as provided in textbook standard testing matrices.
### Step 1: Construct the Ratio
Substitute these values into the proportionality:
$$v_{\mathrm{He}} : v_{\mathrm{CH}_{4}} : v_{\mathrm{CO}_{2}} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{4}{3}}$$
### Pattern Recognition
When $\frac{P}{\rho}$ is locked down constant, sound speed depends strictly on internal degrees of freedom via $\gamma$. Keep standard experimental values of complex gases like $\mathrm{CH}_4$ and $\mathrm{CO}_2$ memorized.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Class 11 Physics: Kinetic Theory
Q916 (2025)
In an experiment with a closed organ pipe, it is filled with water by $\left(\frac{1}{5}\right)^\text{th}$ of its volume. The frequency of the fundamental note will change by:
- 25%
- 20%
- -20%
- -25%
### Related Formula
Fundamental frequency of a closed organ pipe:
$$f_1 = \frac{v}{4l}$$
where $l$ is the length of the resonating air column.
### Core Logic
Initially, the full air column length is $l$. Filling $\frac{1}{5}$ of its space with fluid reduces the available vibrating air column length to:
$$l_2 = l - \frac{1}{5}l = \frac{4}{5}l$$
{{SOL_IMG1}}
### Step 1: Calculate New Frequency
The modified acoustic frequency response is:
$$f_2 = \frac{v}{4l_2} = \frac{v}{4\left(\frac{4}{5}l\right)} = \frac{5v}{16l} = \frac{5}{4}f_1$$
{{SOL_IMG2}}
### Step 2: Determine Percentage Shift
$$\Delta f\% = \frac{f_2 - f_1}{f_1} \times 100 = \left(\frac{5}{4} - 1\right) \times 100 = 25\%$$
### Pattern Recognition
Shortening the resonance tube length raises pitch frequency inversely ($f \propto \frac{1}{l}$). Reducing the air column length to $80\%$ of its original value drives the frequency up to $125\%$, yielding a positive $25\%$ increase.
### Evaluation Rubric / Model Answer
Option A: 25%
### Chapter Mix
Class 11 Physics: Waves
Q13 (2025)
The equation of a wave travelling on a string is $y=\sin[20\pi x+10\pi t]$ where x and t are distance and time in SI units.
The minimum distance between two points having the same oscillating speed is : [cite: 120, 121]
- 5.0 cm [cite: 122]
- 20 cm [cite: 123]
- 10 cm [cite: 124]
- 2.5 cm [cite: 125]
### Related Formula
$$k = \frac{2\pi}{\lambda}$$
$$\Delta x_{\text{min}} = \frac{\lambda}{2}$$ [cite: 715]
### Core Logic
From the given wave equation, the wave number $k$ is the coefficient of $x$ [cite: 120]:
$$k = 20\pi\ \text{rad/m}$$
Now find the wavelength $\lambda$: [cite: 717]
$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi} = \frac{1}{10}\ \text{m} = 10\ \text{cm}$$ [cite: 717]
The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of $\pi$, which translates spatially to a half-wavelength separation ($\frac{\lambda}{2}$) [cite: 715]:
$$\text{Distance} = \frac{\lambda}{2} = \frac{10}{2} = 5\ \text{cm}$$ [cite: 717]
### Pattern Recognition
Oscillating speed configuration reaches identical value fields twice per spatial wavelength cycle[cite: 715]. Thus, minimum distance separation scales exactly to $\frac{\lambda}{2}$[cite: 715].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
