Explanation:
Suppose, in the mixture, the volumes of oxygen and hydrogen are \({V_{1}}\) and \({V_{2}}\), respectively. If the densities are \({d_{1}}\) and \({d_{2}}\), respectively, then their masses will be \({V_{1} d_{1}}\) and \({V_{2} d_{2}}\), respectively. If the density of the mixture is \({d}\), then
\(\begin{aligned}& d=\dfrac{\text { Total mass }}{\text { Total volume }}=\dfrac{V_{1} d_{1}+V_{2} d_{2}}{V_{1}+V_{2}} \\& =\dfrac{V_{2} d_{2}\left(\dfrac{V_{1} d_{1}}{V_{2} d_{2}}+1\right)}{V_{2}\left(\dfrac{V_{1}}{V_{2}}+1\right)}=\dfrac{d_{2}\left(\dfrac{V_{1} d_{1}}{V_{2} d_{2}}+1\right)}{\left(\dfrac{V_{1}}{V_{2}}+1\right)} \\& \text { Given: }=\dfrac{V_{1}}{V_{2}}=\dfrac{1}{4}\end{aligned}\)
\(\begin{aligned}\dfrac{d_{1}}{d_{2}} & =\dfrac{\text { Molecular weight of oxygen }}{\text { Molecular weight of hydrogen }} \\& =\dfrac{32}{2}=\dfrac{16}{1}\end{aligned}\)
\(\therefore d=\dfrac{d_{2}\left(\dfrac{1}{4} \times \dfrac{16}{1} \times 1\right)}{\dfrac{1}{4}+1}=4 d_{2} \text { or } \dfrac{d}{d_{2}}=4\)
Let \({v}\) be the speed of sound in the mixture and \({v_{2}}\) in hydrogen. Then by Laplace's formula, we have
\(\begin{gathered}v=\sqrt{\left(\dfrac{\gamma P}{d}\right)} \text { and } v_{2}=\sqrt{\left(\dfrac{\gamma P}{d_{2}}\right)} \\\therefore \dfrac{v}{v_{2}}=\sqrt{\left(\dfrac{d_{2}}{d}\right)}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2} \\\Rightarrow v=\dfrac{v_{2}}{2}=\dfrac{1270}{2} \\=635 {~ms}^{-1}\end{gathered}\)