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
(7)
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(a) – 5 J (b) – 10 J (c) – 15 J (d) – 20 J Answer-a (126)