A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets is

Answer

8

Explanation

The surface area of the liquid drop is
`A = 4 pi R ^(2)`
Its surface energy is E When the drop splits in 512 droplets, the surface area of each droplet is `4pi r ^(2)`
`therefore` Total surface area `A _(2) = 512 xx 4pi r ^(2)`
The volume of bigger drop is `(4)/(3) pi R ^(3) and ` volume of small droplets is `512 xx (4)/(3) pi r ^(3)`
` therefore (4)/(3) pi R ^(3) = 512 xx (4)/(3) pi r ^(3) implies r = (R)/(8)`
`therefore A _(2) = 512 xx 4 pi r ^(2) = 512 xx 4pi ((R)/(8)) ^(2)`
`= 8 A _(1)`
Surface energy E = A.T (T is surface tension and A is area)
`therefore ( E _(n))/( E) = ( A _(2). T)/( A _(1). T) = (8. A _(1))/( A _(2)) = 8`
` therefore E _(n) = 8 E`