The speed of sound in hydrogen gas at N . T . P is 1328 ms - 1 . If the density of hydrogen is 1 16 t h of that of air, then the speed of sound in air at N . T . P is

Here, v H 2 = 1328 m s − 1 ; ρ H 2 = ρ a i r 16 ; v = γ P 8 Let v air be the velocity of sound in air at N .T .P N o w , v a i r v H 2 = ρ H 2 ρ a i r { Since γ H 2 = γ a i r } o r v a i r = v H 2 ρ H 2 ρ a i r o r v a i r = 1328 ρ a i r 16 ρ a i r = 1328 1 16 = 1328 4 = 332 m s − 1

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PhysicsSound WaveMiscellaneousEasy2 minPYQ_2020

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The speed of sound in hydrogen gas at $N . T . P$ is $1328 {ms}^{- 1}$ . If the density of hydrogen is ${\frac{1}{16}}^{t h}$ of that of air, then the speed of sound in air at $N . T . P$ is

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$\begin{array}{l} Here, v_{H_{2}} = 1328 m s^{- 1}; ρ_{H_{2}} = \frac{ρ_{a i r}}{16}; v = \sqrt{\frac{γ P}{8}} \\ {Let v}_{air} be the velocity of sound in air at N .T .P \\ N o w, \frac{v_{a i r}}{v_{H_{2}}} = \sqrt{\frac{ρ_{H_{2}}}{ρ_{a i r}}} {\begin{cases} Since \\ γ_{H_{2}} = γ_{a i r} \end{cases}} \\ o r v_{a i r} = v_{H_{2}} \sqrt{\frac{ρ_{H_{2}}}{ρ_{a i r}}} \\ o r v_{a i r} = 1328 \sqrt{\frac{\frac{ρ_{a i r}}{16}}{ρ_{a i r}}} = 1328 \sqrt{\frac{1}{16}} = \frac{1328}{4} \\ = 332 m s^{- 1} \end{array}$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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