Answer

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**Hint:**The volume of the trapped air can be determined by using the relation between the volume of the fluid and the pressure of the fluid. The relation between the volume of the fluid and the pressure of the fluid is given by Boyle’s law.

**Useful formula**

The relation between the volume of the fluid and the pressure of the fluid is given by,

${V_2} = \left( {1 + \dfrac{{h\rho }}{{{P_0}}}} \right){V_1}$

Where, ${V_2}$ is the volume of the trapped air, $h$ is the height, $\rho $ is the density, ${P_0}$ is the pressure and ${V_1}$ is the volume of the lake.

**Complete step by step solution**

Given that,

The height of the bell inside the lake, $h = 47.6\,m = 47.6 \times {10^2}\,cm$,

The volume of the lake water, ${V_1} = 50\,c{m^3}$,

The atmospheric pressure, $70\,cm\,of\,Hg$

Density of $Hg = 13.6\,gc{m^{ - 3}}$

According to Boyle’s law, $P \propto \dfrac{1}{V}$

The pressure in the any fluid is inversely proportional to the volume of the fluid,

Now,

The relation between the volume of the fluid and the pressure of the fluid is given by,

${V_2} = \left( {1 + \dfrac{{h\rho }}{{{P_0}}}} \right){V_1}\,..................\left( 1 \right)$

By substituting the height, density of the water, pressure of the atmosphere and the volume of the lake in the above equation (1), then the equation (1) is written as,

${V_2} = \left( {1 + \dfrac{{47.6 \times {{10}^2} \times 1000}}{{70 \times 13.6 \times 1000}}} \right)50$

By multiplying the terms, then the above equation is written as,

${V_2} = \left( {1 + \dfrac{{4760}}{{952}}} \right)50$

On dividing the above equation, then

${V_2} = \left( {1 + 5} \right)50$

By adding the terms in the above equation, then the above equation is written as,

${V_2} = \left( 6 \right)50$

By multiplying the terms, then the above equation is written as,

${V_2} = 300\,c{m^3}$

**Hence, the option (A) is the correct answer.**

**Note:**According to Boyle's law, the pressure of any fluid is inversely proportional to the volume of the same fluid. As the volume of the fluid increases the pressure of the fluid decreases. If the volume of the fluid decreases the pressure of the fluid increases.

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