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Question

A large number of liquid drops each of radius 'a' are merged to form a single spherical drop of radius 'b'. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of the big drop is
[ρ= density of liquid, T= surface tension of liquid]

A
[6Tρ(1a1b)]12
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B
[6Tρ(1b1a)]12
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C
[ρ6T(1a1b)]12
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D
[ρ6T(1b1a)]12
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Solution

The correct option is A [6Tρ(1a1b)]12
Let there are n small drops that merged to form a bigger single drop. Mass of the liquid is conserved before and after the merging.
ρ(n×4π3a3)=ρ(4π3b3)
We get n=b3a3
Total surface area of the small drops A1=n(4πa2)
Surface area of the bigger drop A2=4πb2
Change in surface area ΔA=A1A2
Or ΔA=4π(na2b2)
Or ΔA=4π(b3a3a2b2)=4πb3[1a1b]
Energy released E=TΔA= 4πTb3[1a1b]
Kinetic energy of the bigger drop K.E=12(ρ4π3b3)v2
12(ρ4π3b3)v2= 4πTb3[1a1b]
v=[6Tρ(1a1b)]1/2

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