A bubble is at the bottom of a lake of depth | vedclass

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(A) According to Boyle's law,$P_1 V_1 = P_2 V_2$ at constant temperature.At the bottom of the lake,the total pressure is $P_1 = P_{atm} + P_{water} =

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$26 \, \ell$

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(A) According to Boyle's law,$P_1 V_1 = P_2 V_2$ at constant temperature.At the bottom of the lake,the total pressure is $P_1 = P_{atm} + P_{water} = ho g \ell + ho g h = ho g(h + \ell)$.The volume of the bubble is $V_1 = \frac{4}{3} \pi r^3$.At the surface,the pressure is $P_2 = P_{atm} = ho g \ell$.The radius becomes $3r$,so the volume is $V_2 = \frac{4}{3} \pi (3r)^3 = 27 	imes \frac{4}{3} \pi r^3$.Substituting these into the equation: $ho g(h + \ell) 	imes \frac{4}{3} \pi r^3 = ho g \ell 	imes 27 	imes \frac{4}{3} \pi r^3$.Simplifying,we get $h + \ell = 27 \ell$.Therefore,$h = 26 \ell$.

$A$ bubble is at the bottom of a lake of depth $h$. As the bubble rises to the surface,its radius increases by a factor of $3$. If the atmospheric pressure is equal to the pressure exerted by a water column of height $\ell$,then $h$ is equal to:

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