Two small spherical metal balls, having equal masses, are made from...

The terminal velocity achieved by ball in a viscous fluid is $v_{\mathrm{t}}=\frac{2(\rho-\sigma) \mathrm{r}^2 \mathrm{ g}}{9 \eta}$ where, $\rho=$ density of metal of ball, $\sigma=$ density of viscous medium, $\mathrm{r}=$ radius of ball and $\eta=$ coefficient of viscosity of medium Terminal velocity of first ball, $$\begin{aligned}\mathrm{v}_{\mathrm{t}_1}& =\frac{2\left(\rho_1-\sigma\right) \mathrm{r}_1^2 \mathrm{ g}}{9 \eta}\\ & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{ g}}{9 \eta} \end{aligned}$$...($i) \left[\because \rho_1=8 \rho_2\right]$ Similarly, for second ball $\mathrm{v}_{\mathrm{t}_2}=\frac{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{ g}}{9 \eta}$ From Eq. (i) and (ii), we get $$\begin{aligned}\frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}& =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{ g}}{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{ g}} \times \frac{9 \eta}{9 \eta}\\ & = \left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^ 2 \end{aligned}$$..($iii) \left[\because \sigma=0.1 \rho_2\right]$ Here, $\mathrm{r}_{1}=1 \mathrm{ mm}and \mathrm{r}_{2}=2 \mathrm{ mm}Substituting these values$ in Eq. (iii), we get \Rightarrow \quad \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}=\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\r \right)^2=$\frac{79}{36}