NEET UG 2019 — Physics Mechanics

The terminal velocity achieved by ball in a viscous fluid is $$v_{\mathrm{t}}=\frac{2(\rho-\sigma) \mathrm{r}^2 \mathrm{~g}}{9 \eta}$$vt​=9η2(ρ−σ)r2 g​ where, $\rho=$ρ= density of metal of ball, $\sigma=$σ= density of viscous medium, $\mathrm{r}=$r= radius of ball and $\eta=$η= coefficient of viscosity of medium Terminal velocity of first ball, $$\begin{aligned} \mathrm{v}_{\mathrm{t}_1} & =\frac{2\left(\rho_1-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta} \\ & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta} \end{aligned}$$vt1​​​=9η2(ρ1​−σ)r12​ g​=9η2(8ρ2​−σ)r12​ g​​ ...(i) $\left[\because \rho_1=8 \rho_2\right]$[∵ρ1​=8ρ2​] Similarly, for second ball $$\mathrm{v}_{\mathrm{t}_2}=\frac{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}}{9 \eta}$$vt2​​=9η2(ρ2​−σ)r22​ g​ From Eq. (i) and (ii), we get $$\begin{aligned} \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}} & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}} \times \frac{9 \eta}{9 \eta} \\ & =\left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^2 \end{aligned}$$vt2​​vt1​​​​=2(ρ2​−σ)r22​ g2(8ρ2​−σ)r12​ g​×9η9η​=(ρ2​−0.1ρ2​8ρ2​−0.1ρ2​​)(r2​r1​​)2​ ..(iii) $\left[\because \sigma=0.1 \rho_2\right]$[∵σ=0.1ρ2​] Here, $\mathrm{r}_1=1 \mathrm{~mm}$r1​=1 mm and $\mathrm{r}_2=2 \mathrm{~mm}$r2​=2 mm Substituting these values in Eq. (iii), we get $$\Rightarrow \quad \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}=\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\right)^2=\frac{79}{36}$$⇒vt2​​vt1​​​=(0.9ρ2​7.9ρ2​​)(21​)2=3679​