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**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.

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Topics: Properties of Bulk Matter
(130)

Subject: Physics
(2479)

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