By conserving the total volume 34πR3=64×(34πr3)⇒R=4r
Change in surface energy is
ΔU=T×ΔA, =T[64×4πr2−4πR2]
=4πTR2[64×161−1]=12×3.14×0.075×10−4
≈2.8×10−4J
A water drop of diameter 2cm is broken into 64 equal droplets. The surface tension of water is 0.075Nm−1. In this process the gain in surface energy will be
Held on 28 Jun 2022 · Verified 6 Jul 2026.
2.8×10−4J
1.5×10−3J
1.9×10−4J
9.4×10−5J
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