\[\begin {array} {1 1} (a)\;8.95 \times 10^{-5}\;J \\ (b)\;2.45 \times 10^{-5}\;J \\ (c)\;3.25 \times 10^{-4}\;J \\ (d)\;2.10 \times 10^{-4}\;J \end {array}\]

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$R$- radius of big drop

$r$- radius of smaller drops

$\large\frac{4}{3} $$\pi R^3=n \large\frac{4}{3} $$\pi r^3$

$R^3=nr^3$

$n= \bigg( \large\frac{R^3}{r^3}\bigg)$

$=>n=\bigg(\large\frac{R}{r}\bigg)^3$

Initial area of drop $=4 \pi R^2$

Final area of drops $=n 4 \pi r^2$

Changes in area $=\Delta A= n \pi r^2-4 \pi R^2$

$\qquad= 4 \pi(nr^2-R^2)$

Work done $= 4 \pi T(nr^2-R^2)$

$W= 4 \pi R^3T \bigg(\large\frac{1}{r} -\frac{1}{R}\bigg)$

or $\qquad= 4 \pi R^2 T [n^{1/3}-1]$

$\qquad= 4 \pi \times (10^{-3})^2 \times 72 \times 10^{-3} [10^{6/3}-1]$

$\qquad= 8.95 \times 10^{-5}Joules$

Hence a is the right answer.

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