Fluid Mechanics

JEE Main level, numerical answer, medium

114 of 192

Question 114 of 192

There is an air bubble of radius 1.0mm1.0mm in a liquid of surface tension 0.075Nm–10.075N{m}^{–1} and density 1000kgm–31000kg{m}^{–3} at a depth of 10cm10cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is _______ Pa(g=10ms−2)Pa\left(g=10m{s}^{-2}\right).

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Answer: 1150

The pressure inside the bubble is given by
ΔP=P−P0=ρgh+2Tr\Delta P=P-{P}_{0}=\rho gh+\dfrac{2T}{r}
The data given is
r=10−3mT=0.075Nm−1ρ=1000kgm−3h=10cmr={10}^{-3}m T=0.075N{m}^{-1} \rho =1000kg{m}^{-3} h=10cm
Therefore, the pressure in the bubble is
ΔP=(1000×10×10×10−2)+2×0.07510−3=1000+150=1150Pa\Delta P=\left(1000\times 10\times 10\times {10}^{-2}\right)+\dfrac{2\times 0.075}{{10}^{-3}}=1000+150=1150Pa

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