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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 $\rho$
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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 $(\mathrm{P}+\rho \mathrm{gh}) \mathrm{V}_{0}=\mathrm{PV}$
$$
\text { or } \mathrm{V}=\mathrm{V}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{P}}\right)
$$
$$
\text { or } \mathrm{V}=\mathrm{V}_{0}\left(1+\frac{\rho \mathrm{gh}}{\mathrm{P}}\right)
$$
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