Question

An air bubble of volume $V_0$ is released by a fish at a depth $h$ in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure $P$ above the lake. The volume of the bubble just before touching the surface will be (density of water is $\rho )$

• A. $V_0$
• B. $V_0\left(\frac{\rho gh}{P}\right)$
• C. $\frac{V_0}{(1+\frac{\rho gh}{P})}$
• D. $V_0(1+\frac{\rho gh}{P})$

Answer 1

The correct option is D $V_0(1+\frac{\rho gh}{P})$

According to Boyle's law, [at constant temperature]
$PV=$ constant
From the diagram, we can say that,
$P_1{V}_1=P_2{V}_2$


Given, $P$ is the atmospheric pressure at the top of the lake. Let the volume of the bubble just before touching the surface be $V$ and volume of the bubble at depth $h$ of the lake be $V_0$.
Then from $P_1{V}_1=P_2{V}_2$
$(P+h\rho g)V_0=PV$
$\Rightarrow V=\left(\frac{P+h\rho g}{P}\right)V_0$
$\therefore V=V_0[1+\frac{\rho gh}{P}]$
Thus, option (d) is the correct answer.