Q: A spherical liquid drop of radius R is divided into eight equal droplets. If the surface tension is T, then the work done in this process will be

(a) 2πR^{2}T

(b) 3πR^{2}T

(c) 4πR^{2}T

(d) 2πRT^{2}

Ans: (c)

Sol:

$ \displaystyle \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$

R = 2 r

$ \displaystyle W = T [8 \times 4 \pi r^2 – 4 \pi R^2 ]$

$ \displaystyle W = T [8 \times 4 \pi (R/2)^2 – 4 \pi R^2 ]$

$ \displaystyle W = T [8 \pi R^2 – 4 \pi R^2 ]$

W = 4πR^{2}T