Properties Of Solids And Liquids Question 282
Question: 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)[CPMT 1989]
Options:
A)350 cm3
B)300 cm3
C)250 cm3
D)22 cm3
Show Answer
Answer:
Correct Answer: B
Solution:
According to Boyle’s law, pressure and volumeare inversely proportional to each other
i.e. $ P\propto \frac{1}{V} $ therefore $ P _{1}V _{1}=P _{2}V _{2} $
therefore $ (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} $
therefore $ 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)50cm^{3}=300cm^{3}. $ [As $ P _{2}=P _{0}=70cm $ of Hg $ =70\times 13.6\times 1000 $ ]
