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Question

An air chamber of volume V has a neck of cross – sectional area a into which a light ball of mass m can move without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air in B, the time period of the resulting oscillation of the ball is given by

A
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B
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C
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D
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Solution

The correct option is D
Let P be the pressure of air in the chamber. When the ball is pressed down a distance x, the volume of air decreases from V to Sat V − Δ V. Hence the pressure increases from P to P+ Δ P. The change in volume is ΔV=ax
The excess pressure Δ P is related to the bulk modulus B
As Δ P =BΔVV
Now restoring force on ball = excess pressure × cross-sectional area
or, F=BaVΔV
or,F=Ba2Vx (ΔV=ax)
or, F=kx
where k=Ba2V
i.e., Fx
Hence the motion of the ball is simple harmonic. If m is the mass of the ball, the time period of the SHM is

T=2πmk
Or, T=2πmVBa2


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