Let radius of the small drop be r and the radius of the big drop be R. Then, total volume will remain constant, therefore
1000×[ρ(34πr3)]=34πR3ρ
⇒R=10r
Initial total surface energy, E1=1000×4πr2×S and
E2=4π(10r)2S
⇒E2E1=110Hence,x=10.
JEE Main 2024 — Physics Mechanics
A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, the E1 : E2 is x:1, where x=________.
Held on 30 Jan 2024 · Verified 6 Jul 2026.
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