Question

An inverted bell, lying at the bottom of lake $47.6m$ deep, has $50c{m}^3$ 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 = $70cm$ of $Hg$ and density of $Hg=13.6gc{m}^{-3}$ ]

• A. $600c{m}^3$
• B. $300c{m}^3$
• C. $250c{m}^3$
• D. $200c{m}^3$

Answer 1

Answer: (B)

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

$\Rightarrow P_1{V}_1=P_2{V}_2$
$\Rightarrow (P_0+h{\left(\rho \right)}_{\text{w}}g)V_1=P_0{V}_2$
$\Rightarrow V_2=(1+\frac{{\left(h\rho \right)}_{\text{w}}g}{P_0})V_1$
$$\begin{gathered} V_2=(1+\frac{47.6\star 1\star 1000\star 10}{70\star {\left(10\right)}^{-2}\star 13.6\star 1000\star 10}) \\ [As{P}_2=P_0=70cmofHg=70\star {\left(10\right)}^{-2}\star 13.6\star 1000\star 10] \end{gathered}$$
$\Rightarrow {\left(\text{V}\right)}_2=\left(1+5\right)50{\left(\text{cm}\right)}^3=300{\left(\text{cm}\right)}^3$