Fluid Mechanics

JEE Main level, single correct, medium

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Question 127 of 192

A water drop of diameter 2cm2cm is broken into 6464 equal droplets. The surface tension of water is 0.075Nm−10.075N{m}^{-1}. In this process the gain in surface energy will be

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Answer: (A) 2.8×10−4J2.8\times {10}^{-4}J

By conserving the total volume 43πR3=64×(43πr3)\dfrac{4}{3}\pi {R}^{3}=64\times \left(\dfrac{4}{3}\pi {r}^{3}\right)⇒R=4r\Rightarrow R=4r
Change in surface energy is
ΔU=T×ΔA\Delta U=T\times \Delta A, =T[64×4πr2−4πR2]=T\left[64\times 4\pi {r}^{2}-4\pi {R}^{2}\right]
=4πTR2[64×116−1]=12×3.14×0.075×10−4=4\pi T{R}^{2}\left[64\times \dfrac{1}{16}-1\right] =12\times 3.14\times 0.075\times {10}^{-4}
≈2.8×10−4J\approx 2.8\times {10}^{-4}J

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