2×34πR3=34πr3⇒r=21/3RUi=2×4πR2 TUf=4πr2 T=4πR2 T22/3∴ Heat lost =ui−uf=4πR2 T[2−22/3]
Two water drops each of radius 'r' coalesce to from a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :
Held on 2 Apr 2025 · Verified 6 Jul 2026.
4πr2 T[2−232]
4πr2 T[2−231]
4πr2 T[1+2]
4πr2T[2−1]
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