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

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(A) Let the radius of the two small drops be $R$ and the radius of the large drop be $R'$.Since the volume remains constant during coalescence:$2 	im

Vedclass · Accepted Answer

$2^{1/3} : 1$

Vedclass · Answer

(A) Let the radius of the two small drops be $R$ and the radius of the large drop be $R'$.Since the volume remains constant during coalescence:$2 	imes (\frac{4}{3} \pi R^3) = \frac{4}{3} \pi (R')^3$$2R^3 = (R')^3$$R' = 2^{1/3} R$Surface energy $U$ is given by $U = T 	imes A$,where $T$ is surface tension and $A$ is the surface area.Initial surface area $A_i = 2 	imes (4 \pi R^2) = 8 \pi R^2$.Final surface area $A_f = 4 \pi (R')^2 = 4 \pi (2^{1/3} R)^2 = 4 \pi (2^{2/3} R^2) = 2^{2/3} (4 \pi R^2)$.The ratio of initial surface energy $U_i$ to final surface energy $U_f$ is:$\frac{U_i}{U_f} = \frac{T 	imes A_i}{T 	imes A_f} = \frac{8 \pi R^2}{4 \pi 	imes 2^{2/3} R^2} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.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 total surface energy before and after the change is -

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