In an isothermal process 2 water drops of radius 1 mm are combined to form a bigger drop. Find the enegy change in this process if T=0.1 N/m

Options

A. `1muF`
B. `0.5muJ`
C. `0.25muF`
D. `0.75muF`

Answer

B

Explanation

`DeltaU=TDeltaA`
By volume conservation
`2xx(4)/(3)pi(1mm)^(3)=(4)/(3)piR^(3)impliesR=2^(1//3)` mm
`DeltaU=0.1N//mxx4pi[2xx(1mm)^(2)-(2^(1//3)mm)^(2)]`
`=0.4pixx[2-2^(2//3)]xx10^(-6)J`
`=0.4pixx0.41xx10^(-6)J=0.52muJ`