Physics - Fluid Question with Solution | TestHub

The terminal velocity of a sphere falling through a viscous liquid is given by Stokes' law:

$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

Since both spheres fall with the same uniform speed, $v_1=v_2$.

$\frac{2}{9}\frac{r_1^2({\rho }_1-\sigma )g}{\eta }=\frac{2}{9}\frac{r_2^2({\rho }_2-\sigma )g}{\eta }$

$r_1^2({\rho }_1-\sigma )=r_2^2({\rho }_2-\sigma )$

$\frac{r_1^2}{r_2^2}=\frac{{\rho }_2-\sigma }{{\rho }_1-\sigma }$

Substituting the given values ${\rho }_1=8\times 10^3{kg/m}^3$, ${\rho }_2=11\times 10^3{kg/m}^3$, and $\sigma =2\times 10^3{kg/m}^3$:

$\frac{r_1^2}{r_2^2}=\frac{11\times 10^3-2\times 10^3}{8\times 10^3-2\times 10^3}$

$\frac{r_1^2}{r_2^2}=\frac{9\times 10^3}{6\times 10^3}=\frac{3}{2}$

$\frac{r_1}{r_2}=\sqrt{\frac{3}{2}}$

Answer: $\sqrt{\frac{3}{2}}$