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) Read More
