ExamBro
ExamBro
JEE Advanced · Physics · 13. Thermodynamics

A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R-a). For aR the magnitude of the work done in the process is given by 4πP0Ra2X, where X is a constant and γ=Cp/CV=41/30. The value of X is______

  1. A 2.05
  2. B 0.26
  3. C 0.456
  4. D 2.15
Verified Solution

Answer & Solution

Correct Answer

(A) 2.05

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
&\mathrm{W}=(\Delta \mathrm{P})_{\text {avg }} \times 4 \pi \mathrm{R}^2 \mathrm{a} \\
&=\left|\frac{\mathrm{dP}}{2} \cdot 4 \pi \mathrm{R}^2 \mathrm{a}\right|
\end{aligned}\)
(for small change \((\Delta \mathrm{P})_{\text {avg }}<\mathrm{P}>\) arithmetic mean)
\(=\mathrm{PV} \text { gamma }=\mathrm{c} \Rightarrow \mathrm{dP}=\) \(-~\gamma \frac{\mathrm{P}}{\mathrm{V}} \mathrm{dV}=-\frac{\gamma \mathrm{P}_0}{\mathrm{~V}} 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\frac{\gamma \mathrm{P}_0}{2 \mathrm{~V}} \times 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\frac{\gamma \mathrm{P}_0}{2 \times 4 \pi \mathrm{R}^3} 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}\)
\(=\left(4 \mathrm{pRP} \times \mathrm{a}^2\right) \frac{3 \gamma}{2}\)
\(\therefore \mathrm{x}=2.05\)
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app