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An inverted bell lying at the bottom of a lake $47.6\,m$ deep has $50\, cm^3$ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be ...... $cm^3$ (atmospheric pressure $= 70\, cm$ of $Hg$ and density of $Hg = 13.6\, g/cm^3$)
$350$
$300$
$250$
$22$
Solution
According to Boyle's law, pressure and volume are
inversely proportional to each other i.e. $\mathrm{P} \propto \frac{1}{\mathrm{V}}$
$\Rightarrow \mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$
$\Rightarrow\left(\mathrm{P}_{0}+\mathrm{h} \rho_{\mathrm{\omega}} \mathrm{g}\right) \mathrm{V}_{1}=\mathrm{P}_{0} \mathrm{V}_{2}$
$\Rightarrow \mathrm{v}_{2}=\left(1+\frac{\mathrm{h} \rho_{\mathrm{\omega}} \mathrm{g}}{\mathrm{P}_{0}}\right) \mathrm{V}_{1}$
$\Rightarrow \mathrm{V}_{2}=\left(1+\frac{47.6 \times 10^{2} \times 1 \times 1000}{70 \times 13.6 \times 1000}\right) \mathrm{V}_{1}$
$\Rightarrow \mathrm{V}_{2}=(1+5) 50 \mathrm{cm}^{3}=300 \mathrm{cm}^{3}$
$\left.\quad \text { [As } \mathrm{P}_{2}=\mathrm{P}_{0}=70 \mathrm{cm} \text { of } \mathrm{Hg}=70 \times 13.6 \times 1000\right]$
