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← IIT-JEE 2008 · Paper 1
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IIT-JEE 2008 · Paper 1, Physics question 40

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Paragraph for Question Nos. 38 to 40

A small spherical monoatomic ideal gas bubble (γ=53)\left(\gamma=\frac53\right) is trapped inside a liquid of density ρℓ\rho_\ell (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains nn moles of gas. The temperature of the gas when the bubble is at the bottom is T0T_0, the height of the liquid is HH and the atmospheric pressure is P0P_0 (Neglect surface tension).

A container of liquid of height H open to the atmosphere (pressure P₀ at the top); a small gas bubble is at height y above the bottom.

The buoyancy force acting on the gas bubble is (Assume RR is the universal gas constant)

Choose one option
Answer and worked solution

Official answer: B

Volume. V=nRTPV=\dfrac{nRT}P with TT from the previous part and P=P0+ρℓg(H−y)P=P_0+\rho_\ell g(H-y):
V=nRT0(P0+ρℓgH)2/5 [P0+ρℓg(H−y)]3/5.V=\frac{nRT_0}{(P_0+\rho_\ell gH)^{2/5}\,[P_0+\rho_\ell g(H-y)]^{3/5}}.

Buoyancy. F=ρℓgVF=\rho_\ell gV:
F=ρℓnRgT0(P0+ρℓgH)2/5 [P0+ρℓg(H−y)]3/5.F=\frac{\rho_\ell nRgT_0}{(P_0+\rho_\ell gH)^{2/5}\,[P_0+\rho_\ell g(H-y)]^{3/5}}.

(B)