27 drops of mercury coalesce to form a bigger | vedclass

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Question

(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.Since the volume remains conserved, $V_{big} = 27 	imes V_{small}$

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.Since the volume remains conserved, $V_{big} = 27 	imes V_{small}$.$\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3 \Rightarrow R^3 = 27r^3 \Rightarrow R = 3r$.The initial surface energy is $U_i = 27 	imes (S 	imes 4 \pi r^2) = 108 \pi r^2 S$.The final surface energy is $U_f = S 	imes 4 \pi R^2 = S 	imes 4 \pi (3r)^2 = 36 \pi r^2 S$.The relative increase in surface energy is given by $\frac{\Delta U}{U_i} = \frac{U_f - U_i}{U_i} = \frac{36 \pi r^2 S - 108 \pi r^2 S}{108 \pi r^2 S}$.$\frac{\Delta U}{U_i} = \frac{-72 \pi r^2 S}{108 \pi r^2 S} = -\frac{72}{108} = -\frac{2}{3}$.

$27$ drops of mercury coalesce to form a bigger drop. What is the relative increase in surface energy?

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