Two small drops of mercury,each of radius R,c | vedclass

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(C) Let $T$ be the surface tension of mercury.Total surface energy before coalescence $(E_1)$: Since there are two drops of radius $R$,$E_1 = 2 	imes

Vedclass · Accepted Answer

$2^{1/3}: 1$

Vedclass · Answer

(C) Let $T$ be the surface tension of mercury.Total surface energy before coalescence $(E_1)$: Since there are two drops of radius $R$,$E_1 = 2 	imes (4 \pi R^2 T) = 8 \pi R^2 T$.Let the radius of the new large drop be $R'$. By conservation of volume: $\frac{4}{3} \pi R'^3 = 2 	imes \frac{4}{3} \pi R^3$,which gives $R'^3 = 2R^3$,or $R' = 2^{1/3} R$.Total surface energy after coalescence $(E_2)$: $E_2 = 4 \pi R'^2 T = 4 \pi (2^{1/3} R)^2 T = 4 \pi (2^{2/3}) R^2 T$.The ratio of surface energies is $\frac{E_1}{E_2} = \frac{8 \pi R^2 T}{4 \pi (2^{2/3}) R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.Thus,the ratio is $2^{1/3} : 1$.

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

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