Question

A water drop of radius r is broken into 1000 equal droplets. The surface tension of water is s. The gain in surface energy is

Answer 1

Answer:

Let us say r is the radius of each eight of the bubble. Since volume of water remains the same, we have

3

4

πR

3

=8×

3

4

πr

3

⇒r=

2

R

So the work done in this process will be = change in the surface energy i.e.

(8×4πr

2

×T)−(4πR

2

×T)

=(8×4π(

2

R

)

2

×T)−(4πR

2

×T)∵r=

2

R

=8πR

2

T−4πR

2

T=4πR

2

T