An inverted bell, lying at the bottom of lake $47.6 \mathrm{m}$ deep, has $50 {\mathrm{cm}}^3$ of air trapped in it. The bell is brought to the surface of the lake, then the volume of the trapped air will become [atmospheric pressure = $70 \mathrm{cm}$ of $\mathrm{Hg}$ and density of $\mathrm{Hg}=13.6 \mathrm{g} {\mathrm{cm}}^{-3}$]

• A

  $600 {\mathrm{cm}}^3$
• Correct answer
  B

  $300 {\mathrm{cm}}^3$
• C

  $250 {\mathrm{cm}}^3$
• D

  $200 {\mathrm{cm}}^3$

Solution

According to Boyle's law, pressure and volume are inversely proportional to each other i.e. $\text{p}\propto \frac{1}{\text{v}}$



$\Rightarrow P_1{V}_1=P_2{V}_2$

$\Rightarrow (P_0+h{\mathrm{\rho }}_{\text{w}}g)V_1=P_0{V}_2$

$\Rightarrow V_2=(1+\frac{{\mathrm{h\rho }}_{\text{w}}g}{P_0})V_1$

$\begin{array}{l}{V}_2=(1+\frac{47.6*1*1000*10}{70*{10}^{-2}*13.6*1000*10})\\ \\ [As{P}_2=P_0=70\mathrm{cm}\mathrm{of}\mathrm{Hg}=70*{10}^{-2}*13.6*1000*10]\\ \end{array}$

$\Rightarrow {\text{V}}_2=\left(1+5\right)50 {\text{cm}}^3=300 {\text{cm}}^3$

Hence, the correct answer is option: 'B'.