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(A) The speed of sound in an ideal gas is given by the formula $C = \sqrt{\frac{\gamma R T}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the univ

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$4:1$

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(A) The speed of sound in an ideal gas is given by the formula $C = \sqrt{\frac{\gamma R T}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.Since the temperature $T$ and the adiabatic index $\gamma$ (for diatomic gases like $H_2$ and $O_2$) are the same,the speed of sound is inversely proportional to the square root of the molar mass: $C \propto \frac{1}{\sqrt{M}}$.The molar mass of hydrogen $(H_2)$ is $M_{H_2} = 2 	ext{ g/mol}$.The molar mass of oxygen $(O_2)$ is $M_{O_2} = 32 	ext{ g/mol}$.Therefore,the ratio of the speed of sound in hydrogen to oxygen is $\frac{C_{H_2}}{C_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Thus,the ratio is $4:1$.

Column - $A$	Column - $B$
$(a)$ Speed of longitudinal waves in gas and liquid.	$(i)$ $v = \sqrt{\frac{B}{\rho}}$
$(b)$ Speed of longitudinal waves in solid rod.	$(ii)$ $v = \sqrt{\frac{Y}{\rho}}$
$(c)$ Speed of sound in air according to Newton.	$(iii)$ $v = \sqrt{\frac{P}{\rho}}$
$(d)$ Speed of sound in air according to Laplace.	$(iv)$ $v = \sqrt{\frac{\gamma P}{\rho}}$

The ratio of the speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

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