An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 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 / $c m^{3}$ )

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Solution

The correct option is B

According to Boyle's law, pressure and volume are inversely proportional to each other i.e. $p \propto \frac{1}{v} \Rightarrow P_{1} V_{1} = P_{2} V_{2} \Rightarrow (P_{0} + h p_{w} g) V_{1} = P_{0} V_{2} \Rightarrow V_{2} = (1 + \frac{h p_{w} g}{P_{0}}) V_{1} \Rightarrow V_{2} = (1 + \frac{47.6 \times 1 \times 1000}{70 \times 13.6 \times 1000}) V_{1} [A s P_{2} = P_{0} = 70 c m o f H g = 70 \times 13.6 \times 1000] \Rightarrow V_{2} = (1 + 5) 50 c m^{3} = 300 c m^{3}$

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An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 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)

The correct option is B According to Boyle's law, pressure and volume are inversely proportional to each other i.e. p∝1v⇒P1V1=P2V2⇒(P0+hpwg)V1=P0V2⇒V2=(1+hpwgP0)V1⇒V2=(1+47.6×1×100070×13.6×1000)V1[As P2=P0=70 cm of Hg=70×13.6×1000]⇒V2=(1+5)50 cm3=300 cm3

An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 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 / $c m^{3}$ )