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If the radius of a spherical liquid (of ...

If the radius of a spherical liquid (of surface tension S) drop increases from r to `r +Deltar` , the corresponding increase in the surface energy is

A

`8 pi r Delta rS`

B

`4pi r DeltarS`

C

`16 pi r DeltarS`

D

`8 pi r DeltarS`

Text Solution

Verified by Experts

The correct Answer is:
A
`W=4piS[(r+Deltar)^2-r^2]` neglecting `(Deltar)^2`
`= 8pir DeltarS`
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