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

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$π σ D^{2} (n^{1 / 3} - 1)$

$π σ D^{2} (n^{2 / 3} - 1)$

$π σ D^{2} (n - 1)$

$π σ D^{2} (n^{4 / 3} - 1)$

answer is A.

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Detailed Solution

$E = T/4 π R^{2} (n^{1/3} -1) = σ .4 π \frac{D^{2}}{4} (n^{1/3} -1) E = σ π D^{2} (n^{1/3} -1);D= σ π D^{2} (n^{1/3} -1)$

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