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Energy needed in breaking a liquid drop of radius ' $R$ ', into ' $n$ 'smaller drops each of radius ' $r$ ' is $(T -$ surface tension of the liquid)

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$(\frac{4}{3} π r^{3} n - \frac{4}{3} π R^{3}) T$

$(4 π R^{2} - 4 π r^{3}) / T$

$(4 π r^{2} n - 4 π R^{2}) T$

$(4 π R^{2} - 4 π r^{3}) n T$

answer is A.

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Detailed Solution

$| Δ U | = (4 π R^{2} T) - (n T \times 4 π r^{2}) (W = T A) | Δ U | = (4 π r^{2} n - 4 π R^{2}) T$

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