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An ideal monatomic helium gas at pressure P0 and volume V0 is enclosed in a vertical cylinder whose top is open to atmosphere. A frictionless horizontal piston of mass m and cross-section A sits on the gas. A spring of constant k = m*g*A/V0 connects the piston to the upper wall. Initially the system is in equilibrium with the spring unstretched. All processes are adiabatic with gamma = 5/3 for helium. When the piston is displaced slightly from equilibrium and released, its frequency of small oscillations is (1 / (2*pi)) * sqrt(eta * g * A / V0). If atmospheric pressure Pa = m*g/A (so that P0 = 2*m*g/A), find the value of eta.

1. 8/3
2. 11/3
3. 13/3
4. 19/3

Correct answer: 13/3

Solution

Displacing the piston upward by small distance x expands the gas adiabatically. The net restoring force is F = -(gamma*P0*A²/V0 + k)*x. So omega² = (gamma*P0*A²/V0 + k)/m. Substituting k = m*g*A/V0 and P0 = 2*m*g/A (using Pa = m*g/A): omega² = (5/3 * 2*m*g/A * A²/V0 + m*g*A/V0)/m = g*A/V0*(10/3 + 1) = 13*g*A/(3*V0). So eta = 13/3. However, if Pa = 0 (vacuum above piston) then P0 = m*g/A and eta = 5/3 + 1 = 8/3. The standard JEE treatment with Pa = m*g/A gives eta = 13/3.

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