First slide

Surface tension

Question

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

Moderate

A

2 ${πR}^{2} T$
B

$3 {πR}^{2} T$
C

$4 {πR}^{2} T$
D

$2 {πRT}^{2}$

Solution

Radius of the larger drop = R
Suppose, radius of the droplets = r
Since, volume will be remain constant,
$\frac{4}{3} {πR}^{3} = 8 \times \frac{4}{3} {πr}^{2}$
[No. of droplets = 8]
Work done = (Increase in surface area) $\times$ Surface tension.
= $[8 [4 π {(\frac{R}{2})}^{2}} - 4 {πR}^{2}] \times T$
= (8 ${πR}^{2} - 4 {πR}^{2}) \times T = 4 {πR}^{2} T$

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