The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to $$h$$ meter of mercury column and relative density of mercury is $$\rho $$ then the depth of lake in metre is

Solution :
From Boyle's law ($$T$$ = constant)
$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2} \cr & \therefore \left( {H{d_{{\text{water}}}} + h{d_{{\text{mercury}}}}} \right)g\left( {\frac{4}{3}\pi {r^3}} \right) = h{d_{{\text{mercury}}}}g\left( {\frac{4}{3}\pi {{\left( {2r} \right)}^3}} \right) \cr & \Rightarrow H{d_{{\text{water}}}} = 8h{d_{{\text{mercury}}}} - h{d_{{\text{mercury}}}} \cr & \Rightarrow H = 7h\frac{{{d_{{\text{mercury}}}}}}{{{d_{{\text{water}}}}}} \cr & \therefore H = 7\;h\rho \cr} $$