Waves Previous Year Questions — JEE Main Physics
22 past-year Waves questions from JEE Main (Physics) — page 4.
Q43 (2026)
Two strings (A, B) having linear densities $\mu_{A} = 2 \times 10^{-4}\text{ kg/m}$ and $\mu_{B} = 4 \times 10^{-4}\text{ kg/m}$ and lengths $L_{A} = 2.5\text{ m}$ and $L_{B} = 1.5\text{ m}$ respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time $t_{1}$ and $t_{2}$ , respectively, to reach the joint. The ratio $t_{1}/t_{2}$ is :
- 1.08
- 1.90
- 1.67
- 1.18
### Related Formula
$$v = \sqrt{\frac{T}{\mu}}$$
$$t = \frac{L}{v}$$
### Core Logic
Given $L_{A} = 2.5\text{ m}$, $L_{B} = 1.5\text{ m}$, $T = 500\text{ N}$.
Velocity in string A:
$$v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = \sqrt{2500000} = 5 \sqrt{10} \times 10^{2}\text{ m/s}$$
Velocity in string B:
$$v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = \sqrt{1250000} = 5 \sqrt{5} \times 10^{2}\text{ m/s}$$
### Step 1: Calculate Ratio of Times
Time taken to reach joint:
$$t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5 \sqrt{10} \times 10^{2}}$$
$$t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5 \sqrt{5} \times 10^{2}}$$
Ratio:
$$\frac{t_{1}}{t_{2}} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{2.5}{1.5} \times \frac{\sqrt{5}}{\sqrt{10}} = \frac{5}{3} \times \frac{1}{\sqrt{2}}$$
$$\frac{t_{1}}{t_{2}} = \frac{1.666}{1.414} \approx 1.18$$
### Pattern Recognition
Time ratio is proportional to $\frac{L}{\sqrt{T/\mu}} = L \sqrt{\mu/T}$. Thus $t_1/t_2 = (L_1/L_2) \sqrt{\mu_1/\mu_2}$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q46 (2026)
Two loudspeakers $L_{1}$ and $L_{2}$ are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point A, at equidistance to both loud speakers, is moved by 25 m along the line AB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is \_\_\_\_ Hz
(Speed of sound in air is $324~\mathrm{m/s}$ and $\sqrt{5}=2.23$)
{{IMG1}}
### Related Formula
$$\Delta x = n\lambda, \quad v = f\lambda$$
### Core Logic
{{SOL_IMG1}}
Path difference at point B:
$$L_1 B = \sqrt{20^2 + 40^2} = 20\sqrt{5} \text{ m} = 44.6 \text{ m}$$
$$L_2 B = \sqrt{40^2 + 30^2} = 50 \text{ m}$$
$$\Delta x = 50 - 44.6 = 5.4 \text{ m}$$
$$\Delta x = n\lambda \implies 5.4 = 10 \times \lambda \implies \lambda = 0.54 \text{ m}$$
$$f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \text{ Hz}$$
### Pattern Recognition
Sees: Loudspeaker interference cycles along displacement line.
Shortcut: Compute path difference at endpoint B, relate to total interference cycles to find wavelength, then frequency.
Check: Numerical answer is 600 Hz. ✓
### Chapter Mix
Class 11 Physics: Waves and Sound
Q33 (2026)
In an open organ pipe $v_{3}$ and $v_{6}$ are $3^{\mathrm{rd}}$ and $6^{\mathrm{th}}$ harmonic frequencies, respectively. If $v_{6} - v_{3} = 2200$ Hz then length of the pipe is ____ mm.
(Take velocity of sound in air is 330 m/s.)
- $275$
- $225$
- $200$
- $250$
### Related Formula
$$f_n = n \left(\frac{v}{2L}\right)$$
### Core Logic
For an open organ pipe, harmonic frequency $v_n = n \cdot f_0 = n \left(\frac{v}{2L}\right)$.
Given $v_6 - v_3 = 2200$ Hz:
$$\frac{6v}{2L} - \frac{3v}{2L} = 2200$$
$$\frac{3v}{2L} = 2200$$
Substituting speed of sound $v = 330 \mathrm{~m/s}$:
$$\frac{3 \times 330}{2L} = 2200$$
$$L = \frac{990}{4400} = 0.225 \mathrm{~m} = 225 \mathrm{~mm}$$
### Step 1: Final Conclusion
The length of the pipe is $225 \mathrm{~mm}$.
### Pattern Recognition
Open organ pipe: Frequency difference $\Delta f = (6-3)f_0 = 3 f_0 = 2200 \implies f_0 = 2200/3$. Since $f_0 = v/(2L)$, solve directly for $L$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Q50 (2026)
The velocity of sound in air is doubled when the temperature is raised from $0^{\circ}$ C to $\alpha^{\circ}$ C. The value of $\alpha$ is ____.
### Related Formula
$$V = \sqrt{\frac{\gamma RT}{M}}$$
$$\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}$$
### Core Logic
The speed of sound in an ideal gas is directly proportional to the square root of its absolute temperature in Kelvin.
Initial state: $T_1 = 0^{\circ}\mathrm{C} = 273 \, \mathrm{K}$, $V_1 = V_0$
Final state: $T_2 = \alpha^{\circ}\mathrm{C} = (\alpha + 273) \, \mathrm{K}$, $V_2 = 2V_0$
### Step 1: Apply Ratio Equation
$$\frac{V_0}{2V_0} = \sqrt{\frac{273}{T_2}}$$
$$\frac{1}{4} = \frac{273}{T_2}$$
$$T_2 = 4 \times 273 = 1092 \, \mathrm{K}$$
### Step 2: Convert to Celsius
$$T_2 = \alpha + 273 = 1092$$
$$\alpha = 1092 - 273 = 819^{\circ}\mathrm{C}$$
### Pattern Recognition
If velocity doubles, absolute temperature must quadruple ($2^2 = 4$). $4 \times 273 = 1092 \, \mathrm{K}$, which translates perfectly to $819^{\circ}\mathrm{C}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Class 11 Physics: Kinetic Theory of Gases
Q34 (2026)
The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5/x. The value of x is ____.
- $4$
- $2$
- $1$
- $3$
### Related Formula
$$f_{n, \text{closed}} = \frac{n v}{4L_{\text{closed}}}$$ (where $n$ is an odd integer)
$$f_{n, \text{open}} = \frac{n v}{2L_{\text{open}}}$$ (where $n$ is any integer)
### Core Logic
We are given that the fifth harmonic of the closed organ pipe equals the first harmonic of the open pipe:
$$f_{5, \text{closed}} = f_{1, \text{open}}$$
$$\frac{5v}{4L_{\text{closed}}} = \frac{v}{2L_{\text{open}}}$$
### Step 1: Simplify Ratio
Isolating the length ratio:
$$\frac{L_{\text{closed}}}{L_{\text{open}}} = \frac{5}{4} \times 2 = \frac{10}{4} = \frac{5}{2}$$
Equating this to $5/x$ gives $x = 2$.
### Pattern Recognition
For a closed pipe, the fundamental is $v/4L$. For an open pipe, it's $v/2L$. If an odd harmonic $n$ of a closed pipe matches the $m$-th harmonic of an open pipe, $L_c / L_o = n/2m$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
