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Question

Physics Question on mechanical properties of fluid

The radius of a spherical drop of water is 1 mm1 \,mm. If surface tension of water be 70×10−3Nm−1,70\times10^{-3} Nm^{-1}, the pressure difference inside and outside the drop will be

A

70 Nm−270\, Nm^{-2}

B

140 Nm−2140\, Nm^{-2}

C

280 Nm−2280\, Nm^{-2}

D

zero

Answer

140 Nm−2140\, Nm^{-2}

Explanation

Solution

The excess pressure pp is given by p=2TR p = \frac {2T}{R} where TT is surface tension and RR is radius of bubble Given T=70×10−3Nm−1T=70\times10^{-3}Nm^{-1}. R=1 mm=10−3mR=1\,mm=10^{-3}m =p=2×70×10−310−3=p=\frac{2\times70\times10^{-3}}{10^{-3}} =140 Nm−2=140\, Nm^{-2}