Fluid Mechanics

JEE Main level, single correct, medium

106 of 192

Question 106 of 192

Pressure inside a soap bubble is greater than the pressure outside by an amount :
(given : R=\mathrm{R}= Radius of bubble S=\mathrm{S}= Surface tension of bubble)

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Answer: (D) 4 SR\frac{4 \mathrm{~S}}{\mathrm{R}}

There are two liquid-air surfaces in bubble so
ΔP=2(2 SR)=4 SR\Delta \mathrm{P}=2\left(\frac{2 \mathrm{~S}}{\mathrm{R}}\right)=\frac{4 \mathrm{~S}}{\mathrm{R}}

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