Question

A liquid drop having surface energy e is spread into 512 droplets of same size. The final surface energy of the droplets is a) 2e b) 4e c) 8e d) 12e

Answer 1

Answer:

E^s=8E

Explanation:

Let, suppose

Surface Tension = T

Density of Liquid = ρ

Radius of Bigger Drop = R

Radius of small drops = r

(Mass of Big Drop) =ρ(4π/3 R^3 )

Mass of small Drops= 512 ×ρ(4π/3 r^3 )

Hence  

(Mass of Big Drop)=512 × Mass of small Drops

So   R=8r

Surface Energy of Big Drop =E=AT=T(4πR^2 )

Surface Energy of 512  small Drops =E^s=AT=512 ×T(4πr^2 )

Then

E/E^s =T(4πR^2 )/(512 ×T(4πr^2 ) )=(8×T(4πr^2 ))/(512 ×T(4πr^2 ) )=8

E^s=8E