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Energy needed in breaking a drop of radi...

Energy needed in breaking a drop of radius `R` into `n` drops of radii `r` is given by

A

`4pi T (r^2 - nR^2)`

B

`4pi T(r^2 - nR^2)`

C

`4pi T (nr^2 - R^2)`

D

`4pi T (nr^2 + R^2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Energy required = T x change in area
Energy `= T (n.4 pi r^2 - 4 pi R^2) = 4pi T (nr^2 - R^2)`
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