A bubble from bottom of lake rises to its surface. Its volume doubles in the process. Assuming isothermal conditions, atmospheric pressure $= 75\, cm$ of $Hg$ and ratio of densities of mercury and water $40/3$. The depth of lake will be ..... $m$
Medium
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Isothermal condition $\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$
$\left(\rho_{1} {g} 5\right)(2 \mathrm{V})=\left[\left(\rho_{1} g 75\right)+\rho_{2}{g} h\right] V$
$\Rightarrow \rho_{1} \mathrm{g} 75=\rho_{2} {g} \mathrm{h}$
$\Rightarrow \mathrm{h}=75 \times \frac{\rho_{1}}{\rho_{2}}=\frac{75 \times 40}{3}=1000 \mathrm{cm}=10 \mathrm{m}$
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