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

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(A) Let the radius of the large drop be $R'$. Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the two

Vedclass · Accepted Answer

$2^{1/3} : 1$

Vedclass · Answer

(A) Let the radius of the large drop be $R'$. Since the volume 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 \frac{4}{3} \pi R^3$$R'^3 = 2 R^3 \implies R' = 2^{1/3} R$Total surface energy before $(E_i)$ = $2 	imes (4 \pi R^2 T)$,where $T$ is the surface tension.Total surface energy after $(E_f)$ = $4 \pi R'^2 T$.The ratio of surface energies is $\frac{E_i}{E_f} = \frac{2 	imes 4 \pi R^2 T}{4 \pi R'^2 T} = \frac{2 R^2}{R'^2}$.Substituting $R' = 2^{1/3} R$:Ratio = $\frac{2 R^2}{(2^{1/3} R)^2} = \frac{2 R^2}{2^{2/3} R^2} = 2^{1 - 2/3} = 2^{1/3}$.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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