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The energy needed for breaking a drop of...

The energy needed for breaking a drop of radius R into n drops each of radius r is

A

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

B

`(4piT)/3(nr^(2)-R^(2))`

C

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

D

`4/3piT(R^(3)-nr^(3))`

Text Solution

Verified by Experts

The correct Answer is:
A
`R^(3) = nr^(3) " " :. r = (R/n)^(1//3)`
Energy needed `= T [ 4pir^(2)n- 4piR^(2)]`
`= 4piT [nr^(2)-R^(2)]`
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