Question

An inverted bell lying at the bottom of a lake 47.6 m deep has 50 $c{m}^3$ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 $g/cm3$)

• A. 300 $c{m}^3$
• B. 800 $c{m}^3$
• C. 900 $c{m}^3$
• D. 200 $c{m}^3$

Answer 1

The correct option is A 300 $c{m}^3$

According to Boyle's law $P\alpha \frac{1}{V}$

We know that,

$V_2=(1+\frac{h\rho \omega g}{P_0})V_1{V}_2=(1+\frac{47.6\times {10}^2\times 1\times 1000}{70\times 13.6\times 1000})V_1{V}_2=(1+5)50c{m}^3=300c{m}^3$

$[P_0=P_2=70cmofHg=70\times 13.6\times 1000]$