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) 2πR2T
(b) 3πR2T
(c) 4πR2T
(d) 2πRT2
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 = 4πR2T