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 )

 

Answers (1)
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.

-

 

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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