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
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$P_1 V_1 = P_2 V_2$ ⇒ ⇒ $(H_{\mathrm{Hg}}\rho_{\mathrm{Hg}} + H_{\mathrm{W}}\rho_{\mathrm{W}})V = H_{\mathrm{Hg}}\rho_{\mathrm{Hg}} \times 3V$
$\Rightarrow H_{\mathrm{Hg}}\rho_{\mathrm{Hg}} + H_{\mathrm{W}} \frac{\rho_{\mathrm{Hg}}}{10} = 3 H_{\mathrm{Hg}} \rho_{\mathrm{Hg}}$ $\Rightarrow H_{\mathrm{W}} = 2 H_{\mathrm{Hg}} \times 10 = \frac{2 \times 75 \times 10}{100} = 15 m$
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