A solid sphere, of radius R acquires a terminal velocity v₁ when falling (due t gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v₂, when falling through the same fluid, the ratio (v₁/v₂) equals

We know that a spherical falling through liquid has the following terminal velocity:

${\mathrm{V}}_{\mathrm{T}}=\frac{2}{9}\frac{{\mathrm{r}}^2}{\mathrm{\eta }}({\mathrm{\rho }}_0-{\mathrm{p}}_{\mathrm{l}})\mathrm{g}$

All the parameters on the right hand side are constant for a given liquid and a given sphere of fixed density, with the exception of the sphere's radius. The terminal velocity of a particular sphere of radius r is precisely proportional to the square of the radius, according to this statement.
Thus, for the initial solid sphere ,

${\mathrm{v}}_1\propto {\mathrm{R}}^2$

While for the 27 each identical spheres,

${\mathrm{v}}_2\propto {\mathrm{r}}^2$

thus,

$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}\mathrm{\_\_\_\_\_\_\_}\left(1\right)$

The radii of the original sphere and the 27 identical spheres compare. The 27 identical spheres' volumes added together will make up the initial sphere's volume. As a result, we can compare their volume

$\frac{4}{3}{\mathrm{\pi R}}^3=27\times \frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow {\mathrm{R}}^3=27\times {\mathrm{r}}^3\Rightarrow \frac{\mathrm{R}}{\mathrm{r}}=3$

using equation 1,

$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}=9$

Hence the correct answer is 9.