JEE Main 2025 Apr 3 Shift 2, Physics Q2: Beats

JEE Main2025Apr 3, Shift 2Physics

In the resonance experiment, two air columns (closed at one end) of $100$ cm and $120$ cm long, give $15$ beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :

A
$335$ m/s
B
$370$ m/s
C
$340$ m/s
D
$360$ m/s

Solution

For a pipe closed at one end, fundamental frequency $f = \frac{v}{4 L}$ .

With $L_{1} = 1.0$ m and $L_{2} = 1.2$ m, the shorter pipe has the higher frequency, so the beat frequency is

$f_{1} - f_{2} = \frac{v}{4} (\frac{1}{1.0} - \frac{1}{1.2}) = \frac{v}{4} \cdot \frac{0.2}{1.2} = \frac{v}{24}$ .

Setting this equal to $15$ Hz: $\frac{v}{24} = 15 \Rightarrow v = 360$ m/s.

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JEE Main 2025 Apr 3 Shift 2, Physics Q2: Beats

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