Fluid Mechanics Viscosity Question 201
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} $ Þ $ P_{1}V_{1}=P_{2}V_{2} $ Þ$ (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} $ Þ $ 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,cm^{3}=300,cm^{3}. $ [As $ P_{2}=P_{0}=70,cm $ of Hg $ =70\times 13.6\times 1000 $ ]
BETA
