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

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$1$

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(D) Let $R$ be the radius of the original large drop and $r$ be the radius of each small drop.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $1000$ small drops:$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}{3} \pi r^3$$R^3 = 1000 r^3 \implies R = 10r$.The surface energy $E$ of a spherical drop is given by $E = T 	imes A$,where $T$ is the surface tension and $A$ is the surface area $(4\pi r^2)$.$E_1 = T(4\pi R^2)$$E_2 = 1000 	imes T(4\pi r^2)$Taking the ratio:$\frac{E_1}{E_2} = \frac{T(4\pi R^2)}{1000 	imes T(4\pi r^2)} = \frac{R^2}{1000 r^2}$Substituting $R = 10r$:$\frac{E_1}{E_2} = \frac{(10r)^2}{1000 r^2} = \frac{100 r^2}{1000 r^2} = \frac{1}{10}$Given $\frac{E_1}{E_2} = \frac{x}{10}$,comparing both sides,we get $x = 1$.

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $E_1$ is the surface energy of the original drop and $E_2$ is the total surface energy of the resulting drops,then $\frac{E_1}{E_2} = \frac{x}{10}$. The value of $x$ is:

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