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PhysicsSound WaveMiscellaneousEasy2 minPYQ_2020

PhysicsEasynumerical

Under standard conditions, the density of a gas is $1.3 {mg cm}^{- 3}$ and the velocity of propagation of sound, in it, is $330 m s^{- 1}$ . The number of degrees of freedom of gas is

Answer:

5.00

Solution:

$V_{s o u n d} = \sqrt{\frac{γ P}{ρ}}$ and $γ = 1 + \frac{2}{f}$

$∴$ $1 + \frac{2}{f} = \frac{v_{s o u n d}^{2} \times ρ}{P}$

$1 + \frac{2}{f} = \frac{(330)^{2} \times (1.3)}{1.01 \times 10^{5}}$

$1 + \frac{2}{f} = 1.40$

$f = 5$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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