Question

An air bubble of volume 15 $c{m}^3$ is formed at a depth of 50 m in a lake. If the temperature of the bubble while rising remains constant then the volume of bubble at the surface will be ( g = $10m{s}^{{}^{-}2}$ and atmospheric pressure = 1.0 X 10${}^5$ Pa)

• A. $100c{m}^3$
• B. $90c{m}^3$
• C. $80c{m}^2$
• D. $40c{m}^2$

Answer 1

The correct option is B $90c{m}^3$

We have pressure at the depth of 50 m as $\rho gh=1000\times 10\times 60=6\times {10}^{5}kg/m{s}^{-2}$.
As the temperature remains constant, using Boyle's law we have $P_1{V}_1=P_2{V}_2$ or
$6\times {10}^5\times 15=1\times {10}^5\times V_2$
or
$V_2=90c{m}^3$