The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be $75 cm$ of $Hg $ and the density of water to be $1/10$ of the density of mercury, the depth of the lake is ....... $m$
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(c) ${P_1}{V_1} = {P_2}{V_2}$==> $({P_0} + h\rho g)V$=${P_0} \times 3V$
==> $h\rho g = 2{P_0}$==> $h = \frac{{2 \times 75 \times 13.6 \times g}}{{\frac{{13.6}}{{10}} \times g}}$ $= 15 m$
==> $h\rho g = 2{P_0}$==> $h = \frac{{2 \times 75 \times 13.6 \times g}}{{\frac{{13.6}}{{10}} \times g}}$ $= 15 m$
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