Q.

If σ be the surface tension, the work done in breaking a big drop of radius R into n drops of equal radius is

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a

Rn2/3σ

b

(n2/3−1)Rσ

c

(n1/3−1)Rσ

d

4πR2(n1/3−1)σ

answer is D.

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Detailed Solution

Volume will be conserved43πR3=n43πr3R=n1/3rworkdone = final energy - initial energy= [n4πn-2/3R2−4πR2]σ= 4πR2(n1/3−1)σ
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