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(B) Let the depth of the lake be $d$ and the volume of the bubble at the bottom be $V$. The volume at the surface is $5V$.According to Boyle's law, si

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$4$

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(B) Let the depth of the lake be $d$ and the volume of the bubble at the bottom be $V$. The volume at the surface is $5V$.According to Boyle's law, since the temperature is constant, $P_1 V_1 = P_2 V_2$.At the bottom, the pressure $P_1$ is the sum of atmospheric pressure and the pressure due to the water column of depth $d$: $P_1 = P_{atm} + ho g d = ho g H + ho g d = ho g(H + d)$.At the surface, the pressure $P_2$ is equal to the atmospheric pressure: $P_2 = P_{atm} = ho g H$.Substituting these into the equation: $ho g(H + d) 	imes V = ho g H 	imes 5V$.Dividing both sides by $ho g V$, we get: $H + d = 5H$.Therefore, $d = 5H - H = 4H$.

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