A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (s = 75 dyne/cm).

Options

A. zero
B. 19 erg
C. 46 erg
D. 74 erg

Answer

D

Explanation

Here, `R = (2.8)/(2) = 1.4mm = 0.14cm`
Let r be the radius of each small drop. Then
`125 xx (4)/(3) pi r^(3) = (4)/(3) pi R^(3) or r = (R )/(5) = (0.14)/(5) = 0.028cm`
Change in energy = surface tension `xx` increase in area
`=75 xx [125 xx 4pi r^(2) - 4pi R^(2)]`
`=75 xx 4pi xx [125 xx (0.028)^(2)- (0.14)^(2)] = 74` erg