Physics · Thermodynamics

JEE Advanced 2020 — Paper 2 — Question 11

A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be $\mathrm{p}_{0}$ . The bubble now gets compressed radially in an adiabatic manner so that its radius becomes ( $\mathrm{R}-\mathrm{a}$ ). For $\mathrm{a} \ll \mathrm{R}$ the magnitude of the work done in the process is given by $\left(4 \pi \mathrm{p}_{0} \mathrm{Ra}^{2}\right) \mathrm{X}$ , where X is a constant and $\gamma=C_{p} / C_{V}=41 / 30$ . The value of $X$ is $\_\_\_\_$

Answer: 2.05

Numerical answer — enter this value.

Step-by-step solution

In adiabatic process

$d p=-\frac{\gamma \mathrm{p}}{\mathrm{V}} \mathrm{dV}=-\frac{\gamma \mathrm{p}_{0}}{\mathrm{~V}}\left(-4 \pi \mathrm{R}^{2} \mathrm{a}\right)$

Work done in the process $=-(d p)_{\text {avg }} d V=-\frac{d p}{2} d V$

$=-\frac{\gamma p_{0}}{2 V}\left(4 \pi R^{2} a\right)\left(-4 \pi R^{2} a\right)=\left(4 \pi p_{0} R a^{2}\right) \frac{3}{2} \times \frac{41}{30}$

$=(2.05)\left(4 \pi \mathrm{p}_{0} \mathrm{Ra}^{2}\right)$

$\Rightarrow \mathrm{x}=2.05$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic…

Answer: 2.05

Step-by-step solution

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