Two organ pipes are of same size. Hydrogen and oxygen gases are filled in them. Taking the elasticity of gases to be same, the ratio of their fundamental frequencies will be

1 : 4

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4 : 1

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8 : 1

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16 : 1

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Solution

The correct option is C $4 : 1$
Velocity of sound in gas $v = \sqrt{\frac{γ R T}{M}}$ where

$M$ is the molecular mass of the gas.

For $T a n d γ$ to be constant, $\frac{v_{H_{2}}}{v_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{H_{2}}}} = \sqrt{\frac{32}{2}} = 4$

Now fundamental frequency $ν = \frac{v}{2 L}$

For the same length of the pipes $ν \propto v$

Thus $ν_{H_{2}} : ν_{O_{2}} = v_{H_{2}} : v_{O_{2}} = 4 : 1$

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Two organ pipes are of same size. Hydrogen and oxygen gases are filled in them. Taking the elasticity of gases to be same, the ratio of their fundamental frequencies will be

The correct option is C 4:1Velocity of sound in gas v=√γRTM where M is the molecular mass of the gas.For Tandγ to be constant, vH2vO2=√MO2MH2=√322=4 Now fundamental frequency ν=v2LFor the same length of the pipes ν∝vThus νH2:νO2=vH2:vO2=4:1