n×34πr3=34πR3
∴n=r3R3.....(i)
KE=Ui−Uf
=n×4πr2T−4πR2T
=4πT(nr2−R2)
=4πTR2(rR−1) from (i)
21(34πR3⋅ρ)υ2=4πTR2(rR−1)
6Rρυ2=T(rR−1)
6ρυ2=T(r1−R1)
υ=ρ6T(r1−R1)
JEE Main 2014 — Physics Mechanics
A large number of liquid drops each of radius r coalesce to form a single drop of the radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of the liquid T, density ρ)
Held on 19 Apr 2014 · Verified 6 Jul 2026.
ρ2T(r1−R1)
ρ6T(r1−R1)
ρ4T(r1−R1)
ρT(r1−R1)
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