Question

Derive the expression for extra surface energy if we increase the spherical drop of radius “r”

by “Δr”.​

Answer 1

Answer:

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Explanation:

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

Answer 2

Answer:

Due to surface tension, free liquid drops and bubbles are spherical, if the effect of gravity and air ... Let the radius of the drop increases from r to r + Δr, where Δr is very very small, hence the inside