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

(a) Rn(2/3) σ

(b) (n(2/3) – 1) Rσ

(c) (n(1/3) – 1) Rσ

(d) 4πR2(n(1/3) – 1)σ

(e) [1/( n(1/3) – 1)] σR

Ans. d

(4)

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