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

Answer

C

Explanation

Here, `R=(2.8)/(2)=1.4mm=0.14cm`
Let r be the radius of each small drop.
Then, `125xx(4)/(3)pir^(3)=(4)/(3)piR^(3)`
(or) `r=(R)/(5)=(0.14)/(5)=0.023cm`
`{:("change in"),("energy"):}}={{:("surface tension"),(xx"increase in area"):}`
`=75xx[125xx4pir^(2)-4piR^(2)]`
`=75xx4pixx[125xx(0.028)^(2)-(0.14)^(2)]`
`=74erg`