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

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(B) Let the radius of each small drop be $R$. The volume of two small drops is $V_{initial} = 2 	imes (\frac{4}{3}\pi R^3) = \frac{8}{3}\pi R^3$.Let

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$2^{1/3}:1$

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(B) Let the radius of each small drop be $R$. The volume of two small drops is $V_{initial} = 2 	imes (\frac{4}{3}\pi R^3) = \frac{8}{3}\pi R^3$.Let the radius of the large drop be $R'$. Since the volume remains constant,$\frac{4}{3}\pi (R')^3 = \frac{8}{3}\pi R^3$,which gives $R' = 2^{1/3}R$.The surface energy of a drop is given by $E = T 	imes A$,where $T$ is surface tension and $A$ is surface area.Initial surface energy $E_i = 2 	imes (T 	imes 4\pi R^2) = 8\pi R^2 T$.Final surface energy $E_f = T 	imes 4\pi (R')^2 = 4\pi (2^{1/3}R)^2 T = 4\pi 2^{2/3} R^2 T$.The ratio of initial to final surface energy is $\frac{E_i}{E_f} = \frac{8\pi R^2 T}{4\pi 2^{2/3} R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.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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