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(B) Let $T$ be the surface tension of the liquid.Let $R$ be the radius of the original large drop and $r$ be the radius of each small drop.Since the v

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$1$

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(B) Let $T$ be the surface tension of the liquid.Let $R$ be the radius of the original large drop and $r$ be the radius of each small drop.Since the volume remains constant during the splitting process:$\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 of the original drop is $u_i = T 	imes 4 \pi R^2$.The total surface energy of the $1000$ small drops is $u_f = 1000 	imes (T 	imes 4 \pi r^2)$.Taking the ratio:$\frac{u_f}{u_i} = \frac{1000 	imes 4 \pi r^2}{4 \pi R^2} = 1000 	imes \left(\frac{r}{R}ight)^2$.Substituting $R = 10r$:$\frac{u_f}{u_i} = 1000 	imes \left(\frac{r}{10r}ight)^2 = 1000 	imes \frac{1}{100} = 10$.Given $\frac{u_f}{u_i} = \frac{10}{x}$,we have $10 = \frac{10}{x}$,which implies $x = 1$.

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $u_i$ is the surface energy of the original drop and $u_f$ is the total surface energy of the resulting drops (ignoring evaporation),and $\frac{u_f}{u_i} = \left(\frac{10}{x}\right)$,then the value of $x$ is $......$

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