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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