Physics - Fluid Question with Solution | TestHub

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, $$\begin{array}{rl}{\mathrm{v}}_{{\mathrm{t}}_1}& =\frac{2\left({\rho }_1-\sigma \right){\mathrm{r}}_1^2\mathrm{g}}{9\eta }\\ & =\frac{2\left(8{\rho }_2-\sigma \right){\mathrm{r}}_1^2\mathrm{g}}{9\eta }\end{array}$$ ...(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 $$\begin{array}{rl}\frac{{\mathrm{v}}_{{\mathrm{t}}_1}}{{\mathrm{v}}_{{\mathrm{t}}_2}}& =\frac{2\left(8{\rho }_2-\sigma \right){\mathrm{r}}_1^2\mathrm{g}}{2\left({\rho }_2-\sigma \right){\mathrm{r}}_2^2\mathrm{g}}\times \frac{9\eta }{9\eta }\\ & =\left(\frac{8{\rho }_2-0.1{\rho }_2}{{\rho }_2-0.1{\rho }_2}\right){\left(\frac{r_1}{r_2}\right)}^2\end{array}$$ ..(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 $$\Rightarrow \frac{{\mathrm{v}}_{{\mathrm{t}}_1}}{{\mathrm{v}}_{{\mathrm{t}}_2}}=\left(\frac{7.9{\rho }_2}{0.9{\rho }_2}\right){\left(\frac{1}{2}\right)}^2=\frac{79}{36}$$