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 $Hg = 13.6$ $cm^3$)
Diffcult
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(b)According to Boyle's law, pressure and volume are inversely proportional to each other i.e. $P \propto \frac{1}{V}$
$⇒$ ${P_1}{V_1} = {P_2}{V_2}$
$⇒$ $({P_0} + h{\rho _w}g){V_1} = {P_0}{V_2}$
$ \Rightarrow {V_2} = \left( {1 + \frac{{h{\rho _w}g}}{{{P_0}}}} \right){V_1}$
$⇒$ ${V_2} = \left( {1 + \frac{{47.6 \times {{10}^2} \times 1 \times 1000}}{{70 \times 13.6 \times 1000}}} \right)\;\,{V_1}$
$ \Rightarrow {V_2} = (1 + 5)50\,c{m^3} = 300\,c{m^3}.$
$⇒$ ${P_1}{V_1} = {P_2}{V_2}$
$⇒$ $({P_0} + h{\rho _w}g){V_1} = {P_0}{V_2}$
$ \Rightarrow {V_2} = \left( {1 + \frac{{h{\rho _w}g}}{{{P_0}}}} \right){V_1}$
$⇒$ ${V_2} = \left( {1 + \frac{{47.6 \times {{10}^2} \times 1 \times 1000}}{{70 \times 13.6 \times 1000}}} \right)\;\,{V_1}$
$ \Rightarrow {V_2} = (1 + 5)50\,c{m^3} = 300\,c{m^3}.$
[As ${P_2} = {P_0} = 70 \,cm$ of $Hg $ $ = 70 \times 13.6 \times 1000$]
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