A spherical liquid drop of radius R is divided into 8 equal droplets. If the surface tension is T, then work done in the process will be

Options

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

Correct Answer:

4πR²T

Explanation:

Given: Surface tension = T
Radius of liquid drop = R
Let r be the radius of each small droplet
.·. Volume of big drop = volume of small droplets
(4/3) πR³ = 8 x (4/3) πr²
or R = 2r or r = R/2 ——(i)
Surface area of big drop = 4πR²
Surface are of 8 small droplets = 8 x 4πr²
8 x 4π x (R/2)² = 2(4πR²)
Increase in surface area = 2(4πR²) – 4πR² = 4πR²
Work done = T x increase in surface area
= T x 4πR² = 4πR²T.