Fluid Mechanics

JEE Main level, single correct, medium

146 of 192

Question 146 of 192

An air bubble of radius 0.1 cm0.1 \mathrm{~cm} is in a liquid having surface tension 0.06 N/m0.06 \mathrm{~N} / \mathrm{m} and density 103 kg/m310^3 \mathrm{~kg} / \mathrm{m}^3. The pressure inside the bubble is 1100Nm−21100 \mathrm{Nm}^{-2} greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? (g=9.8 ms−2)\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)

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Answer: (A) 0.1 m0.1 \mathrm{~m}

Given: Radius of air bubble,
r=0.1 cm=10−3 mr=0.1 \mathrm{~cm}=10^{-3} \mathrm{~m}
Surface tension of liquid,
S=0.06 N/m=6×10−2 N/mS=0.06 \mathrm{~N} / \mathrm{m}=6 \times 10^{-2} \mathrm{~N} / \mathrm{m}
Density of liquid, ρ=103 kg/m3\rho=10^3 \mathrm{~kg} / \mathrm{m}^3
Excess pressure inside the bubble,
ρexe =1100Nm−2\rho_{\text {exe }}=1100 \mathrm{Nm}^{-2}
Depth of bubble below the liquid surface, h=h= ?
As we know,
ρExcess =hρg+2sr⇒1100=h×103×9.8+2×6×10−210−3⇒1100=9800h+120⇒9800h=1100−120⇒h=9809800=0.1 m\begin{aligned} &\rho_{\text {Excess }}=h \rho g+\frac{2 s}{r} \\ &\Rightarrow \quad 1100=h \times 10^3 \times 9.8+\frac{2 \times 6 \times 10^{-2}}{10^{-3}} \\ &\Rightarrow \quad 1100=9800 h+120 \\ &\Rightarrow \quad 9800 h=1100-120 \\ &\Rightarrow \quad h=\frac{980}{9800}=0.1 \mathrm{~m} \end{aligned}

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