Let radius of the newly formed drops be r.
As total volume will remain constant, therefore
34πR3=27×(34×πr3)
⇒R3=27r3
⇒R=3r
⇒r=3R
⇒r2=9R2
Now, work done =T×ΔA
=T[27(4πr2)]−T[4πR2]=T[27(4π9R2)]−T[4πR2]=TπR2(12−4)
=8πR2T
JEE Main 2024 — Physics Mechanics
A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :
Held on 29 Jan 2024 · Verified 6 Jul 2026.
8πR2T
3πR2T
81πR2T
4πR2T
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