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Question

Surface tension of water is 0.072Nm1 .The excess pressure inside a water drop of a diameter 1.2mm is:


A
240Nm2
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B
24Nm2
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C
0.06Nm2
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D
60Nm2
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Solution

The correct option is: D: 24Nm2

Given : Surface tension, S=0.072Nm1

Diameter of water drop, D=1.2mm=0.012m D=1.2mm=0.012D = 1.2mm = 0.012D=1.2mm=0.012mm
Thus radius of drop r=0.012m2
Excess pressure inside water drop, Δp=2Sr ΔP=2Sr\Delta P = \dfrac{2S}{r

Δp=2×0.0720.0122=24Nm2

∴\therefor ΔP=2×0.0720.006=24\Delta P = \dfrac{2\times 0.072}{0.006} = 24

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