A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury `=0.465Jm^-2`

Answer

B, C, D

Explanation

`r=2mm=2xx10^-3m,`
`T=0.465Jm^2`
`Volume of initial drop =8x(vol. of small drops)`
`=(4/3)piR^3=(4/3)pr^3xx8`
`=r=R/2=2`
Increase is surfac energy
`=TA\'-TA=8xx4pi^2T-4piR^2T`
`=4piT[98xx(R^2/4)-R^2]`
`=4piTR^2`
`=4xx(3.14)xx(0.465)xx(4xx10^-6)`
`=23.36 muJ`