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

Options

  1. A4 π T ( n r 2 − R 2 )
  2. B4 3 π ( r 3 n − R 2 )
  3. C4 π T ( R 2 − n r 2 )
  4. D4 π T ( n r 2 + R 2 )

Correct answer

A. 4 π T ( n r 2 − R 2 )

Step-by-step solution

Energy needed = Increment in surface energy = (surface energy of n small drops) – (surface energy of one big drop) = n 4 π r 2 T − 4 π R 2 T = 4 π T ( n r 2 − R 2 )

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