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(C) The terminal velocity of a sphere of radius $r$ falling through a viscous fluid is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2 (\rho - \sig

Vedclass · Accepted Answer

$9$

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(C) The terminal velocity of a sphere of radius $r$ falling through a viscous fluid is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2 (ho - \sigma) g}{\eta}$,where $ho$ is the density of the sphere and $\sigma$ is the density of the fluid.From this formula,we see that $v_T \propto r^2$.Let the radius of the large sphere be $R$ and the radius of each small sphere be $r$. Since the volume remains constant when the sphere is broken into $27$ smaller identical spheres:$\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$$R^3 = 27 r^3 \Rightarrow R = 3r \Rightarrow r = R/3$.Now,the ratio of terminal velocities is:$\frac{
u_1}{
u_2} = \frac{R^2}{r^2} = \frac{R^2}{(R/3)^2} = \frac{R^2}{R^2/9} = 9$.Thus,the ratio $(
u_1/
u_2)$ is $9$.

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