Phenomena on Liquid Surfaces. Help Me Solve This Question!?

Given, radius of a large water drop(R) = 5mm = 5x10^ -3m

also given that the large water drop becomes 64 small drops of same radius.

the radius of small water drops is =(r)

a) volume of large water drop is = 4/3x pi x R^3

volume of tiny drop is = 4/3x pi x r^3

if 64 is the number of tiny water drops formed then,

4/3x pi x R^3 = n x 4/3 x pi x r^3

4/3x22/7x (5x10^ -3)^3 = 64x4/3x22/7 x r^3

r^3 = (5x10^ -3)^3/64

r = 1.25x 10^ -3 m

or r = 0.00125mm

b) surface area of large water drop is = 4x pi x R^2

= 4 x 22/7 x (5x10^ -3)^2

= 31.4 x 10^ -5 m

surface tension of 64 tiny drops of water is = 64x 4 x pi x r^2

= 64x4x22/7x (1.25x 10^ -3)^2

= 125.7 x 10^ -5 m

increase in the surface area = 125.7x10^ -5 - 31.4 x 10^ -5

= 94.3x10^-5m

work done = increase in S.A x S.T (surface tension)

= 94.3 x 10^ -5 x 72x 10^ -3

= 6.8x10^ -5 J

this is the final answer.

(ENJOY PHYSICS)

R.GAUTHAM SHENOY