MHTCET 2016 Physics Solved Paper

(c) Surface area of the liquid drop A = 4πR2
Let E be the surface energy of liquid drop. When the drop splits into 512 droplets, the surface area becomes
A2 = 512 x 4πr2 [r = radius of smaller drop]
Comparing the volumes of biggerand all smaller droplets, we get
i.e.

4

3

πR3=512×

4

3

πr3 ⇒ r=

R

8

Total area of smaller droplets is
A1=512 ×4×π×(

R

8

)2 = 8 A
Change in surface area A2−A1
= 4 π(

512×R2

64

−R2)
= 4 π (8R2−R2)=7R2
Surface energy, E=AT[A= area, T= tension ]
So,

En

E0

=

A1×T

A×T

=

8A

A

=8
∴ En=8E