The volume of an air bubble is doubled as it rises from the bottom of lake to its surface. The atmospheric pressure is $75 \,cm$ of mercury. The ratio of density of mercury to that of lake water is $\frac{40}{3}$. The depth of the lake in metre is .........
Medium
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(a)
$2 P_0=P_0+\rho g h$
$\Rightarrow P_0=\rho g h$
$\Rightarrow P_0=75 \,cm \text { mercury } \quad \text { [Atmospheric pressure] }$
$\Rightarrow \rho_{\text {mercury }} \times g \times \frac{75}{100}=\rho_{\text {water }} \times g \times h$
$\Rightarrow \frac{\rho_m}{\rho_w} \times \frac{75}{100}=h$
$\Rightarrow \frac{40}{3} \times \frac{75}{100}=h \quad\left[\because \frac{\rho_m}{\rho_w}=\frac{40}{3} \text { (given) }\right]$
$\Rightarrow h=10 \,m$
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