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(B) Let $n$ be the number of small drops of radius $r$ that coalesce to form a big drop of radius $R$. Here,$n = 2$.By conservation of volume: $n 	im

Vedclass · Accepted Answer

$2^{-\frac{1}{3}}: 1$

Vedclass · Answer

(B) Let $n$ be the number of small drops of radius $r$ that coalesce to form a big drop of radius $R$. Here,$n = 2$.By conservation of volume: $n 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.Substituting $n = 2$: $2r^3 = R^3$,which gives $R = 2^{1/3} r$.The surface energy $E$ is given by $E = T 	imes A$,where $T$ is surface tension and $A$ is surface area.Initial surface energy $E_1 = n 	imes (4 \pi r^2 T) = 2 	imes 4 \pi r^2 T = 8 \pi r^2 T$.Final surface energy $E_2 = 4 \pi R^2 T = 4 \pi (2^{1/3} r)^2 T = 4 \pi (2^{2/3} r^2) T = 4 	imes 2^{2/3} \pi r^2 T$.The ratio of total surface energies after and before the change is $\frac{E_2}{E_1} = \frac{4 	imes 2^{2/3} \pi r^2 T}{8 \pi r^2 T} = \frac{2^{2/3}}{2} = 2^{2/3 - 1} = 2^{-1/3}$.Thus,the ratio is $2^{-1/3} : 1$.

Small drops of liquid of the same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is:

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