1000 small water drops of equal size combine | vedclass

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(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains conserved,the volume of $1000$ small drops

Vedclass · Accepted Answer

$1:10$

Vedclass · Answer

(B) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains conserved,the volume of $1000$ small drops equals the volume of the big drop: $1000 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.This simplifies to $R^3 = 1000 r^3$,so $R = 10r$.The initial surface energy $E_1$ of $1000$ small drops is $E_1 = 1000 	imes (4 \pi r^2 T)$,where $T$ is the surface tension.The final surface energy $E_2$ of the big drop is $E_2 = 4 \pi R^2 T$.The ratio of final surface energy to initial surface energy is $\frac{E_2}{E_1} = \frac{4 \pi R^2 T}{1000 	imes 4 \pi r^2 T} = \frac{R^2}{1000 r^2}$.Substituting $R = 10r$,we get $\frac{E_2}{E_1} = \frac{(10r)^2}{1000 r^2} = \frac{100 r^2}{1000 r^2} = \frac{1}{10}$.Thus,the ratio is $1:10$.

$1000$ small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

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