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 ρ)

• A. $V_{0}$
• B. $V_{0}(\rho gh/P)$
• C. $\frac{V_{0}}{\left( 1 + \frac{\rho gh}{P} \right)}$
• D. $V_{0}\left( 1 + \frac{\rho gh}{P} \right)$

Answer 1

Answer: (D)

$V_{0}\left( 1 + \frac{\rho gh}{P} \right)$

Answer 2

According to Boyle’s law multiplication of pressure and volume will remains constant at the bottom and top.

If P is the atmospheric pressure at the top of the lake and the volume of bubble is V then from$P_{1}V_{1} = P_{2}V_{2}$

$(P + h\rho g)V_{0} = PV$ ⇒ $V = \left( \frac{P + h\rho g}{P} \right)V_{0}$

∴ $V = V_{0}\left\lbrack 1 + \frac{\rho gh}{P} \right\rbrack$