Ray Optics and Optical Instruments Part 3

When a spherical ball falls through a viscous fluid, it eventually reaches a constant speed called terminal velocity. This occurs when the net force acting on the ball is zero.The forces acting on the ball are :Weight (W) which acts downwards. $\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{~g}$ (where $\sigma$ is the density of the ball).Up-thrust $\left(\mathrm{F}_{\mathrm{B}}\right)$ is Acting upwards. $F_B=\frac{4}{3} \pi r^3 \rho g$ (where $\rho$ is the density of the fluid).Viscous drag $\left(\mathrm{F}_{\mathrm{v}}\right)$ is acting upwards. According to Stokes' Law, $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r} \mathrm{v}$, where $\eta$ is the viscosity of fluid and v is the speed of the ball.At equilibrium :$$\mathrm{W}=\mathrm{F}_{\mathrm{B}}+\mathrm{F}_{\mathrm{v}}$$$$\Rightarrow$$ $\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 \rho g+6 \pi \eta r v$$$\Rightarrow$$ $6 \pi \eta r v=\frac{4}{3} \pi r^3 g(\sigma-\rho)$$$\Rightarrow$$ $$\mathrm{v}=\frac{2 \mathrm{r}^2 \mathrm{~g}(\sigma-\rho)}{9 \eta}$$For balls of the same material in the same fluid :$$\mathrm{v} \propto \mathrm{r}^2$$Therefore :$$\frac{\mathrm{v}_2}{\mathrm{v}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2$$Initial radius of the ball is $r_1=6 \mathrm{~mm}$ and the initial terminal velocity is $v_1=20 \mathrm{~cm} / \mathrm{s}$Final radius of the ball is $r_2=3 \mathrm{~mm}$Substituting the values :$\frac{v_2}{20}=\left(\frac{3}{6}\right)^2$$$\Rightarrow$$ $$\frac{v_2}{20}=\frac{1}{4} \Rightarrow v_2=\frac{20}{4}=5 \mathrm{~cm} / \mathrm{s}$$Therefore, the terminal velocity of the second ball will be $5 \mathrm{~cm} / \mathrm{s}$.Hence, the correct answer is 5.