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

Options

A. 2E
B. 4R
C. 8E
D. 12

Answer

C

Explanation

According to question, the surface area of the liquid drop is splits into 512 droplets, the surface area becomes
`A_(2) = 512 xx 4pi r^(2)` [r = radius of smaller drop]
r can be calculated by comparing the total volume of bigger and all smaller droplets.
i.e., `(4)/(3)pi R^(3) = 512 xx (4)/(3) pi r^(3) rArr r = (R)/(8)`
Hence, total area of smaller droplets is given by
`A_(1) = 512 xx 4 xx pi xx ((R)/(8))^(2) = 8A`
Change in surface area is given by `A_(2) - A_(1)`
`= 4pi ((512 xx R^(2))/(64) - R^(2))`
`4pi (8R^(2) - R^(2)) = 7R^(2)`
Surface energy, E = AT [ A area, T - tension]
So, `(E_(n))/(E_(o)) = (A_(1) xx T)/(A xx T) = (8A)/(A) = 8`
`:. E_(n) = 8E`