A cylindrical container, fitted with a light piston, encloses a hydrogen gas. Initially, the piston is resting in the equilibrium and its height from the bottom of the container is $x= 0.8 \text{ cm}$ .

Now, a ball is dropped on the piston from a height $h$ above it. The ball makes multiple bounces before coming to rest. When all the motion ceases and equilibrium is again established, it is observed that the piston is again at the same height $x$ .

If all the mechanical energy lost by the ball and piston is absorbed by the gas, then, find the height $h$ (in $\text{cm}$ ) from which the ball was dropped.

The answer is 2.

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We can conserve the energy of the ball and the gas as a system.

At the final equilibrium the ball stays at rest on the piston.

If P0 be the atmospheric pressure then the final pressure = P0 + mg/A

For the gas dU = nCv ( T2 - T1)= f/2 ( P2V2 - P1V1)

Since V1 = V2 = Ax and P1 = P0 and f = 5

dU = 5/2 (mg/A)(Ax)

The decrease in potential energy of the ball contributes to increase in internal energy of the gas

mgh = dU

h = 5/2 x = 2 cm