JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Fluid Mechanics

A small spherical monoatomic ideal gas bubble

(γ = \frac{5}{3})

is trapped inside a liquid of density

ρ_{ℓ}

(see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains

n

moles of gas. The temperature of the gas when the bubble is at the bottom is

T_{0}

, the height of the liquid is

H

and the atmospheric pressure is

P_{0}

(Neglect surface tension).

The buoyancy force acting on the gas bubble is (Assume

R

is the universal gas constant)

(A) $ρ_{ℓ} n R g T_{0} \frac{( P _{0} + ρ _{ℓ} g H ) ^{2/5}}{( P _{0} + ρ _{ℓ} g y ) ^{7/5}}$
(B) $\frac{ρ _{t} n R g T _{0}}{( P _{0} + ρ _{t} g H ) ^{2/5} [ P _{0} + ρ _{t} g ( H - y ) ] ^{3/5}}$ ✓ Correct
(C) $ρ_{ℓ} n R g T_{0} \frac{( P _{0} + ρ _{ℓ} g H ) ^{3/5}}{( P _{0} + ρ _{ℓ} g y ) ^{8/5}}$
(D) $\frac{ρ _{t} n R g T _{0}}{( P _{0} + ρ _{t} g H ) ^{3/5} [ P _{0} + ρ _{t} g ( H - y ) ] ^{2/5}}$

Answer: (B)

Solution

Buoyancy force is

F_{b} = ρ_{ℓ} V g

, where

V

is the volume of the bubble. Using the ideal gas law,

V = \frac{n R T _{2}}{P _{2}}

, where

T_{2} = T_{0} (\frac{P _{0} + ρ _{ℓ} g ( H - y )}{P _{0} + ρ _{ℓ} g H})^{2/5}

and

P_{2} = P_{0} + ρ_{ℓ} g (H - y)

. Thus

F_{b} = ρ_{ℓ} g \frac{n R T _{2}}{P _{2}} = \frac{ρ _{ℓ} n R g T _{0}}{( P _{0} + ρ _{ℓ} g H ) ^{2/5} [ P _{0} + ρ _{ℓ} g ( H - y ) ] ^{3/5}}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option