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A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the liquid is $' \sigma'$ is , the change in energy in this process is :

$\begin {array} {1 1} (1)\;\pi \sigma D^2(n^{\large\frac{1}{3}}-1) & \quad (2)\;\pi \sigma D^2(n^{\large\frac{2}{3}}-1) \\ (3)\;\pi \sigma D^2(n-1) & \quad (4)\;\pi \sigma D^2(n^{\large\frac{4}{3}}-1) \end {array}$

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Ans : (1)
$\pi \sigma D^2(n^{\large\frac{1}{3}}-1)$
answered Feb 27, 2014 by thanvigandhi_1
 

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