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A drop of mercury of radius 2 mm is spli...

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`

A

`23.4muJ`

B

`18.5muJ`

C

`26.8muJ`

D

`16.8muJ`

Text Solution

Verified by Experts

The correct Answer is:
A
Increase in surface energy
`=4piR^(2)T(n^(1//3)-1)=4pi (2xx10^(-3))^(2) (0.465 )(8^(1//3)-1) = 23.4xx10^(-6) J = 23.4 muJ`
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