The volume of an air bubble becomes three tim | vedclass

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(C) Let the depth of the lake be $h$ and the atmospheric pressure be $P_0 = 75\, cm$ of $Hg$.According to Boyle's Law,$P_1 V_1 = P_2 V_2$,where $P_1$

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$15$

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(C) Let the depth of the lake be $h$ and the atmospheric pressure be $P_0 = 75\, cm$ of $Hg$.According to Boyle's Law,$P_1 V_1 = P_2 V_2$,where $P_1$ and $V_1$ are pressure and volume at the bottom,and $P_2$ and $V_2$ are pressure and volume at the surface.At the surface,$P_2 = P_0 = 75\, cm$ of $Hg$ and $V_2 = 3V$.At the bottom,$P_1 = P_0 + hho_w g$ and $V_1 = V$.Given $ho_w = \frac{1}{10} ho_{Hg}$,we can express $P_1$ in terms of $cm$ of $Hg$ as $P_1 = 75 + \frac{h}{10}$.Substituting these into the equation: $(75 + \frac{h}{10})V = 75 	imes 3V$.$75 + \frac{h}{10} = 225$.$\frac{h}{10} = 150$.$h = 1500\, cm = 15\, m$.

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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