The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12s, will be :

Waves PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q59 with Solution

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Correct answer: D

If two sound waves produce $40$ beats in $12\,\text{s}$ , then the beat frequency is

f_b = \frac{40}{12} = \frac{10}{3}\,\text{Hz}

So,

|f_1-f_2|=\frac{10}{3}

For sound in the same gas, velocity $v$ is the same for both waves, so

f_1=\frac{v}{\lambda_1}, \qquad f_2=\frac{v}{\lambda_2}

Given:

\lambda_1=4.08\,\text{m}, \qquad \lambda_2=4.16\,\text{m}

Therefore,

\left|\frac{v}{4.08}-\frac{v}{4.16}\right|=\frac{10}{3}

v\left|\frac{1}{4.08}-\frac{1}{4.16}\right|=\frac{10}{3}

\frac{1}{4.08}-\frac{1}{4.16} =\frac{4.16-4.08}{4.08\times 4.16} =\frac{0.08}{16.9728}

So,

v\left(\frac{0.08}{16.9728}\right)=\frac{10}{3}

Hence,

v=\frac{10}{3}\cdot \frac{16.9728}{0.08}

v=\frac{10}{3}\cdot 212.16 = 707.2\,\text{m/s}

\boxed{v=707.2\,\text{m/s}}

So the correct option is:

D: $707.2\,\text{m s}^{-1}$

Stored correct answer: D

Derived answer: D

They agree.