Q. An inverted bell lying at the bottom of a lake $47.6\ m$ deep has $50\ cm^{3}$ of air trapped in it. The ball is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure $= 70\ cm$ of $Hg$ and density of $Hg = 13.6\ g/cm^{3}$)
A) $350\ cm^{3}$
B) $300\ cm^{3}$
C) $250\ cm^{3}$
D) $22\ cm^{3}$
Explanation
(b) According to Boyle’s law, pressure and volume are inversely proportional to each other $i.e.\ P \propto \frac{1}{V}$
$\Rightarrow P_{1}V_{1} = P_{2}V_{2}$
$\Rightarrow \left( P_{0} + h\rho_{w}g \right)V_{1} = P_{0}V_{2}$
$\Rightarrow V_{2} = \left( 1 + \frac{h\rho_{w}g}{P_{0}} \right)V_{1}$
$\Rightarrow V_{2} = \left( 1 + \frac{47.6 \times 10^{2} \times 1 \times 1000}{70 \times 13.6 \times 1000} \right)V_{1}$
$\Rightarrow V_{2} = (1 + 5)50cm^{3} = 300cm^{3}$
[As $P_{2} = P_{0} = 70\ cm$ of $Hg = 70 \times 13.6 \times 1000\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.