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Question: Answered & Verified by Expert
An air bubble of volume $\mathrm{v}_{0}$ is released by a fish at a depth $\mathrm{h}$ in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake. The volume of the bubble just before touching the surface will be (density) of water is $\rho$
PhysicsMechanical Properties of FluidsBITSATBITSAT 2011
Options:
  • A $\mathrm{v}_{0}$
  • B $\quad v_{0}(\rho g h / p)$
  • C $\frac{\mathrm{v}_{0}}{\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)}$
  • D $\mathrm{v}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)$
Solution:
1482 Upvotes Verified Answer
The correct answer is: $\mathrm{v}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{p}}\right)$
As the bubble rises the pressure gets reduced for constant temperature, if $\mathrm{P}$ is the standard atmospheric pressure, then $(\mathrm{P}+\rho \mathrm{gh}) \mathrm{V}_{0}=\mathrm{PV}$
$$
\text { or } \mathrm{V}=\mathrm{V}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{P}}\right)
$$

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