Eight mercury drops, each of radius r, coales | vedclass

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(B) Let the radius of the bigger drop be $R$.
Since the volume is conserved during the coalescence, $8 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Let the radius of the bigger drop be $R$.
Since the volume is conserved during the coalescence, $8 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.
Solving for $R$, we get $R^3 = 8r^3$, which implies $R = 2r$.
The initial surface energy of the eight small drops is $U_i = 8 	imes (4 \pi r^2 S) = 32 \pi r^2 S$.
The final surface energy of the bigger drop is $U_f = 4 \pi R^2 S = 4 \pi (2r)^2 S = 16 \pi r^2 S$.
The surface energy released in this process is $\Delta U = U_i - U_f = 32 \pi r^2 S - 16 \pi r^2 S = 16 \pi r^2 S$.

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

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