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. 4E
C. 8E
D. 12E

Answer

C

Explanation

According to question, the surface area of the liquid drop is given by `A=4piR^2` and surface energy is E.
When the drop splits into 512 droplets, the surface area becomes `A_2 512xx = 4pir^2` [r= radius of smaller drop]
r can be calculated by comparing the total volume of bigger and all smaller droplets.
i.e. `4piR^2=512xx4/3pir^3Rightarrow r=R/8`
Hence, total area of smaller droplets is given by
`A_1 512xx4xxpixx(R/8)^2=8A`
Change in surface area is given by `A_2 - A_1`
=`4pi((512xxR^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_1xxT)/(AxxT)=(8A)/A=8` lt brgt `therefore E_n=8E`