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(A) Let $R$ be the radius of the original drop and $r$ be the radius of each small drop.Since the volume remains constant,the volume of the original d

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$9$

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(A) Let $R$ be the radius of the original drop and $r$ be the radius of each small drop.Since the volume remains constant,the volume of the original drop equals the sum of the volumes of the $729$ small drops:$\frac{4}{3} \pi R^3 = 729 	imes \frac{4}{3} \pi r^3$$R^3 = 729 r^3$$R = 9r$ or $r = \frac{R}{9}$.The surface energy $E$ of the original drop is $E = T 	imes 4 \pi R^2$,where $T$ is the surface tension.The total surface energy $U$ of the $729$ small drops is $U = 729 	imes (T 	imes 4 \pi r^2)$.Substituting $r = \frac{R}{9}$ into the expression for $U$:$U = 729 	imes T 	imes 4 \pi \left(\frac{R}{9}ight)^2$$U = 729 	imes T 	imes 4 \pi 	imes \frac{R^2}{81}$$U = 9 	imes (T 	imes 4 \pi R^2) = 9E$.Therefore,$\frac{E}{U} = \frac{E}{9E} = \frac{1}{9}$.Comparing this with $\frac{E}{U} = \frac{1}{x}$,we get $x = 9$.

$A$ spherical liquid drop splits into $729$ identical spherical drops. If $E$ is the surface energy of the original drop and $U$ is the total surface energy of the resulting drops,then $\frac{E}{U} = \frac{1}{x}$. The value of $x$ is

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