An air bubble of volume Vo 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 \(\rho\)) :
Answer & explanation
Correct answer: option 4
According to Boyle's law, since temperature is constant hence product of pressure and volume will remain constant at the bottom and the top :
If P is the atmospheric pressure at the top of the lake and the volume of bubble is V then from P1V1 = P2V2
$ (P_a + h \rho g)V_0= P_a V$
\(\Rightarrow V = \frac{P_a + h \rho g}{P_a} V_o\)
V = \(V_o[1 + \frac{\rho g h}{P_a}]\)