Properties Of Solids And Liquids Question 328

Question: Energy needed in breaking a drop of radius R into n drops of radii r is given by [CPMT 1982, 97]

Options:

A) 4πT(nr2R2)

B)43π(r3nR2)

C) 4πT(R2nr2)

D)4πT(nr2+R2)

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Answer:

Correct Answer: A

Solution:

Energy needed = Increment in surface energy = (surface energy of n small drops) - (surface energy of one big drop) =n4πr2T4πR2T=4πT(nr2R2)



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