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(D) Radius of the big drop $= R$.Number of small drops, $n = 64$.Let the radius of each small drop be $r$.In the process of breaking the drop, the vol

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$12 \pi R^2 T$

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(D) Radius of the big drop $= R$.Number of small drops, $n = 64$.Let the radius of each small drop be $r$.In the process of breaking the drop, the volume of the liquid remains constant.$V_i = V_f$$\frac{4}{3} \pi R^3 = n \left( \frac{4}{3} \pi r^3 ight)$$R^3 = 64 r^3 \Rightarrow R = 4r \Rightarrow r = \frac{R}{4}$The work done in breaking a big drop into $64$ small drops is equal to the increase in surface energy.$W = 	ext{Final Surface Energy} - 	ext{Initial Surface Energy}$$W = T(n \cdot 4 \pi r^2) - T(4 \pi R^2)$$W = 4 \pi T [64 r^2 - R^2]$Substituting $r = \frac{R}{4}$:$W = 4 \pi T [64 (\frac{R}{4})^2 - R^2]$$W = 4 \pi T [64 (\frac{R^2}{16}) - R^2]$$W = 4 \pi T [4 R^2 - R^2] = 4 \pi T [3 R^2] = 12 \pi R^2 T$.

The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$) into $64$ equal drops is

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