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:

8e

Explanation:

Let the density be d and surface tension be "t".

Now mass of both the drops are same.

Thus,

Mass=volume*densities

Thus,

d*(4/3pie(r1)^3)=d*(4/3pie(r2)^3)*512

r1=8*r2

The surface energy of the drop is

E = area * surfacetension(t)

E1 = t * ( 4pie * r1^2)

And E2 = t * ( 512 * 4pie*r2^2)

E2 = t * ( 512 * 4pie*r1^2/64)

E2 = t * ( 8 * 4pie*r1^2)

Thus, E1/E2 = 8

Answer 2

Answer:

same same

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