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Question

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

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

A

4πT(nr2R2)4 \pi T \left( n r ^ { 2 } - R ^ { 2 } \right)

B

43π(r3nR2)\frac { 4 } { 3 } \pi \left( r ^ { 3 } n - R ^ { 2 } \right)

C

4πT(R2nr2)4 \pi T \left( R ^ { 2 } - n r ^ { 2 } \right)

D

4πT(nr2+R2)4 \pi T \left( n r ^ { 2 } + R ^ { 2 } \right)

Answer

4πT(nr2R2)4 \pi T \left( n r ^ { 2 } - R ^ { 2 } \right)

Explanation

Solution

Energy needed = Increment in surface energy

= (surface energy of n small drops) –

(surface energy of one big drop)