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Q.

Energy needed in breaking a liquid drop of radius 'R', into 'n'smaller drops each of radius  'r' is  (T− surface tension of the liquid)

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a

(4πr2n−4πR2)T

b

(43πr3n−43πR3)T

c

(4πR2−4πr3)nT

d

(4πR2−4πr3)/T

answer is A.

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Detailed Solution

|ΔU|=(4πR2T)−(nT×4πr2)(W=TA) |ΔU|=(4πr2n−4πR2)T

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Energy needed in breaking a liquid drop of radius 'R', into 'n'smaller drops each of radius  'r' is  (T− surface tension of the liquid)