JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Fluid Mechanics

A spherical bubble inside water has radius

R

. Take the pressure inside the bubble and the water pressure to be

p_{0}

. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes

(R - a)

. For

a ≪ R

the magnitude of the work done in the process is given by

(4 π p_{0} R a^{2}) X

, where

X

is a constant and

γ = C_{p} / C_{V} = 41/30

. The value of

X

is ____.

Answer: 2.05

Solution

Work done:

W = (Δ P)_{avg} \times 4 π R^{2} a ≃ \frac{d P}{2} \cdot 4 π R^{2} a

For adiabatic process:

P V^{γ} = c

d P = - γ \frac{P}{V} d V = - \frac{γ P _{0}}{V} \times 4 π R^{2} a

W = \frac{γ P _{0}}{2 V} \times 4 π R^{2} a \times 4 π R^{2} a

= \frac{γ P _{0}}{2 \times \frac{4 π R ^{3}}{3}} \times (4 π R^{2} a)^{2}

= (4 π P_{0} R a^{2}) \times \frac{3 γ}{2}

Therefore:

X = \frac{3 γ}{2} = \frac{3 \times 41/30}{2} = \frac{41}{20} = 2.05

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer