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(A) The volume of the liquid remains constant during the process.Let $R$ be the radius of the large drop and $r$ be the radius of each of the $27$ sma

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$8 \pi R^2 T$

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(A) The volume of the liquid remains constant during the process.Let $R$ be the radius of the large drop and $r$ be the radius of each of the $27$ smaller drops.Volume of large drop = $27 	imes$ Volume of small drop$\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$$R^3 = 27 r^3$Taking the cube root on both sides,we get $R = 3r$,which implies $r = \frac{R}{3}$.The work done in the process is equal to the change in surface energy: $W = T \Delta A$.Initial surface area $A_i = 4 \pi R^2$.Final surface area $A_f = 27 	imes (4 \pi r^2) = 27 	imes 4 \pi \left(\frac{R}{3}ight)^2 = 27 	imes 4 \pi 	imes \frac{R^2}{9} = 12 \pi R^2$.Change in area $\Delta A = A_f - A_i = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2$.Therefore,the work done $W = T 	imes (8 \pi R^2) = 8 \pi R^2 T$.

$A$ small liquid drop of radius $R$ is divided into $27$ identical liquid drops. If the surface tension is $T$,then the work done in the process will be

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