When a large air bubble rises from the bottom of a lake to the surface, its radius doubles. If the atmospheric is equal to that of a column of water of  height H, then depth of the lake is

Volume of air bubble $\mathrm{V}=\frac{4}{3}\pi r^3$. We get , $V\propto r^3$ If r is 2 times, V becomes 8 times at the surface of lake.Pressure at the surface of lake is givenP₁=1 atmosphere, ${\mathrm{V}}_1=8\mathrm{V}\Rightarrow {\mathrm{P}}_1=\mathrm{Hdg}$ where d=density of water.  Pressure at the bottom of lake, 

$P_2=$ Pressure of atmosphere+ Pressure of water ${\mathrm{P}}_2=\mathrm{Hdg}+\mathrm{hdg}=(\mathrm{H}+\mathrm{h})\mathrm{dg}$

Where, h=depth of lake. Let final volume, ${\mathrm{V}}_2=\mathrm{V}.$ Because temperature is constant, hence from Boyle’s law $P_1{V}_1=P_2{V}_2\Rightarrow Hdg\times 8\mathrm{V}=(H+h)dg\times V\Rightarrow h=7H$