n×34πr3=34πR3
∴n1/3r=R
∴ Total change in surface energy
=(n(4πr2)−4πR2)T
⇒4πT(nr2−R2)
∴ Heat energy
=J×34πR34πT(nr2−R2)=J3T(R3nr2−R1)
Put nr3=R3
∴J3T(r1−R1)
JEE Main 2021 — Physics Mechanics
A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:
Held on 26 Feb 2021 · Verified 6 Jul 2026.
rJ2T
rJ3T
J2T(r1−R1)
J3T(r1−R1)
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