Question

A water drop of radius `10^-2` m is brokenn into 1000 equal droplets. Calculate the gain in surface energy. Surface tension of water ils `0.075Nm^-1`

Answer 1

The volume of the origiN/Al drop is
`V=4/3piR^3, where R=10^-2m`
if r is the radius of each broken droplet the volume is also `V=1000x4/3pir^3`
Thus, `1000r^3=R^3`
`or r=R/10`
The surface area of the origiN/Al drop is `A_1=4piR^2 and ` the surface area of the 1000 droplets is
`A_2=1000xx4pir^2=40piR^2`
The increase in area is
`/_A=A_2-A_1=40piR^2-4piR^2=36piR^2`
The gain in surface energy is
`=/_U=(/_A)S=36piR^2S`
`=36x3.14xx(10^-4m^2)x(0.075Nm^-1)`
`=8.5xx10^-4J`.