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(C) Let $r$ be the radius of each small droplet.The volume of the big drop is equal to the sum of the volumes of the $64$ small droplets:$\frac{4}{3}

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$12\pi R^2T$

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(C) Let $r$ be the radius of each small droplet.The volume of the big drop is equal to the sum of the volumes of the $64$ small droplets:$\frac{4}{3} \pi R^3 = 64 	imes \frac{4}{3} \pi r^3$$R^3 = 64 r^3 \implies R = 4r \implies r = \frac{R}{4}$.The initial surface area of the big drop is $A_i = 4\pi R^2$.The final surface area of the $64$ small droplets is $A_f = 64 	imes 4\pi r^2$.Substituting $r = \frac{R}{4}$:$A_f = 64 	imes 4\pi \left(\frac{R}{4}ight)^2 = 64 	imes 4\pi 	imes \frac{R^2}{16} = 16\pi R^2$.The increase in surface area is $\Delta A = A_f - A_i = 16\pi R^2 - 4\pi R^2 = 12\pi R^2$.The energy needed is given by $W = T 	imes \Delta A = T 	imes 12\pi R^2 = 12\pi R^2T$.

If $T$ is the surface tension of a liquid,the energy needed to break a liquid drop of radius $R$ into $64$ drops is:

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