# Energy needed in breaking a drop of radius R into n drops of radii r is given by Option 1) $4\pi T \left ( nr^{2} - R^{2} \right )$ Option 2) $\frac{4}{3}\pi \left ( nr^{3} - R^{2} \right )$ Option 3) $4\pi T \left (R^{2} - nr^{2} \right )$ Option 4) $4\pi T \left (nr^{2} + R^{2}\right )$

P Prateek Shrivastava

As we learnt in

Surface energy -

The energy required to increase the surface area of the liquid by one unit is defined as surface energy.

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Energy needed = Increment in surface energy   = (surface energy of n small drops) – (surface energy of one big drop)

$= n4\pi r^{2}T-4\pi R^{2}T= 4\pi T\left ( nr^{2}-R^{2} \right )$

Option 1)

$4\pi T \left ( nr^{2} - R^{2} \right )$

Correct

Option 2)

$\frac{4}{3}\pi \left ( nr^{3} - R^{2} \right )$

Incorrect

Option 3)

$4\pi T \left (R^{2} - nr^{2} \right )$

Incorrect

Option 4)

$4\pi T \left (nr^{2} + R^{2}\right )$

Incorrect

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