JEE Advanced 2020 · Paper 2 · Physics Q13

← JEE Advanced 2020 · Paper 2

PhysicsNumerical+4Question 13 of 54

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 $a \ll R$ the magnitude of the work done in the process is given by $(4\pi p_0 R a^2)X$ , where $X$ is a constant and $\gamma = C_p/C_V = 41/30$ . The value of $X$ is ______.

Your answer

Answer and worked solution

Official answer: 2.05

Second-order work. The surrounding water stays at $p_0$ , so to first order it supplies the work $p_0|\Delta V|$ . The extra work needed comes from the rise of the gas pressure above $p_0$ . For a small adiabatic change $\Delta V$ :
$p=p_0\left(1-\gamma\frac{\Delta V}{V_0}\right),\qquad W=\int(p-p_0)\,|dV|=\frac{\gamma p_0}{2V_0}(\Delta V)^2.$

Volume change. $\Delta V\approx4\pi R^2a$ and $V_0=\frac43\pi R^3$ , so
$W=\frac{\gamma p_0\,16\pi^2R^4a^2}{2\cdot\frac43\pi R^3}=6\pi\gamma p_0Ra^2=(4\pi p_0Ra^2)\cdot\frac{3\gamma}2.$

Value.
$X=\frac{3\gamma}2=\frac32\cdot\frac{41}{30}=2.05.$
Answer: 2.05

Topics:Work done by a gas Thermodynamic processes

← Previous question Next question →

JEE Advanced 2020 · Paper 2, Physics question 13