Fluid Mechanics

JEE Main level, single correct, medium

10 of 192

Question 10 of 192

An air bubble of volume 1cm31{cm}^{3} rises from the bottom of a lake 40m40m deep to the surface at a temperature of 12∘C.12^{\circ}C. The atmospheric pressure is 1×105Pa,1\times {10}^{5}Pa, the density of water is 1000kgm−31000kg{m}^{-3} and g=10ms−2.g=10m{s}^{-2}.There is no difference of the temperature of water at the depth of 40m40m and on the surface. The volume of air bubble when it reaches the surface will be

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Answer: (D) 5cm35{cm}^{3}

The pressure at the bottom of the lake can be calculated as follows:
Pbottom=P0+ρgh=(105+1000×10×40)Pa=5×105Pa\begin{matrix}{P}_{bottom} & = & {P}_{0}+\rho gh \\ & = & \left({10}^{5}+1000\times 10\times 40\right)Pa \\ & = & 5\times {10}^{5}Pa\end{matrix}
As the temperature of water remains constant, it can be written that
PbottomVbottom=PtopVtop...(1){P}_{bottom}{V}_{bottom}={P}_{top}{V}_{top}...(1)
Substitute the values of the known parameters into equation (1) to calculate the required volume of the bubble when it reaches the surface.
5×105×1=1×105×Vtop⇒Vtop=5×105×11×105cm3=5cm3\begin{matrix}5\times {10}^{5}\times 1 & = & 1\times {10}^{5}\times {V}_{top} \\ & \Rightarrow & {V}_{top}=\dfrac{5\times {10}^{5}\times 1}{1\times {10}^{5}}{cm}^{3} \\ & = & 5{cm}^{3}\end{matrix}

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