Two small drops of mercury each of radius r c | vedclass

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(A) Let the radius of each small drop be $r$ and the radius of the large drop be $R$. Since the volume of mercury remains constant,the volume of the l

Vedclass · Accepted Answer

$2^{1/3} : 1$

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(A) Let the radius of each small drop be $r$ and the radius of the large drop be $R$. Since the volume of mercury remains constant,the volume of the large drop equals the sum of the volumes of the two small drops:$\frac{4}{3} \pi R^3 = 2 	imes \left( \frac{4}{3} \pi r^3 ight)$$R^3 = 2r^3 \implies R = 2^{1/3} r$The initial total surface energy $E_i$ of the two small drops is:$E_i = 2 	imes (4 \pi r^2 T) = 8 \pi r^2 T$The final surface energy $E_f$ of the single large drop is:$E_f = 4 \pi R^2 T = 4 \pi (2^{1/3} r)^2 T = 4 \pi (2^{2/3} r^2) T = 4 	imes 2^{2/3} \pi r^2 T$The ratio of the total surface energies before and after the change is:$\frac{E_i}{E_f} = \frac{8 \pi r^2 T}{4 	imes 2^{2/3} \pi r^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3} = 2^{1/3} : 1$

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

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