The speed of sound in a gas, in which two waves of wavelength $1.0\,\, m$ and $1.02\,\, m$ produce $6$ beats per second, is approximately .... $m/s$
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Let, $n_{1}$ and $n_{2}$ be the frequencies of two waves.

$\lambda_{1}$ and $\lambda_{2}$ be the wavelengths of two waves.

$v$ be the speed of sound.

The number of beats per second are

$n_{1}-n_{2}=6$

$\frac{v}{\lambda_{1}}-\frac{v}{\lambda_{2}}=6$

$\frac{v}{1}-\frac{v}{1.02}=6$

$\frac{0.02 v}{1.02}=6$

$0.02 v=(6)(1.02)$

$v=\frac{6.12}{0.02}=306 \mathrm{ms}^{-1} \approx 300 \mathrm{m} / \mathrm{s}$

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