question_answer

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]

A)            350 cm3                                  

B)            300 cm3                                  

C)            250 cm3                                  

D)            22 cm3

Correct Answer: B

Solution :

           According to Boyle's law, pressure and volume  are 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 \[{{V}_{2}}=\left( 1+\frac{47.6\times {{10}^{2}}\times 1\times 1000}{70\times 13.6\times 1000} \right)\ \,{{V}_{1}}\]                       \[\Rightarrow {{V}_{2}}=(1+5)50\,c{{m}^{3}}=300\,c{{m}^{3}}.\] [As \[{{P}_{2}}={{P}_{0}}=70\,cm\] of Hg \[=70\times 13.6\times 1000\]]