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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

`4piT(nr^(2)-R^(2))`

B

`4/3pi(r^(3)n-R^(2))`

C

`4piT(R^(2)-nr^(2))`

D

`4piR(nr^(2)+R^(2))`

Text Solution

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The correct Answer is:
A
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