Q49 / 54 · JEE Advanced 2020 (Paper 2) · Physics
A spherical bubble inside water has radius . Take the pressure inside the bubble and the water pressure to be The bubble now gets compressed radially in an adiabatic manner so that its radius becomes . For the magnitude of the work done in the process is given by where is a constant and . The value of is______
Correct answer: 2.05
Solution
\(\begin{aligned}
&\mathrm{W}=(\Delta \mathrm{P})_{\text {avg }} \times 4 \pi \mathrm{R}^2 \mathrm{a} \\
&=\left|\frac{\mathrm{dP}}{2} \cdot 4 \pi \mathrm{R}^2 \mathrm{a}\right|
\end{aligned}\)
(for small change \((\Delta \mathrm{P})_{\text {avg }}<\mathrm{P}>\) arithmetic mean)
\(=\mathrm{PV} \text { gamma }=\mathrm{c} \Rightarrow \mathrm{dP}=-\gamma \frac{\mathrm{P}}{\mathrm{V}} \mathrm{dV}=-\frac{\gamma \mathrm{P}_0}{\mathrm{~V}} 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\frac{\gamma \mathrm{P}_0}{2 \mathrm{~V}} \times 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\frac{\gamma \mathrm{P}_0}{2 \times 4 \pi \mathrm{R}^3} 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\left(4 \mathrm{pRP} \times \mathrm{a}^2\right) \frac{3 \gamma}{2}\)
\(\therefore \mathrm{x}=2.05\)
Hence, the correct answer is 2.05.