A small spherical ball of radius $$r$$, falling through a viscous medium of negligible density has terminal velocity $$v$$. Another ball of the same mass but of radius $$2r$$, falling through the same viscous medium will have terminal velocity:

We need to find the terminal velocity of a ball of radius $$2r$$ with the same mass as a ball of radius $$r$$.

According to Stokes' law, the terminal velocity is given by $$v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$$. Since the medium has negligible density ($$\rho_l \approx 0$$), this reduces to $$v_t = \frac{2r^2\rho_s g}{9\eta}$$.

Because the spheres have the same mass, we have $$\frac{4}{3}\pi r_1^3 \rho_1 = \frac{4}{3}\pi r_2^3 \rho_2$$ with $r_1 = r$ and $r_2 = 2r$. Substituting these radii gives $$\rho_2 = \rho_1 \left(\frac{r_1}{r_2}\right)^3 = \rho_1 \left(\frac{r}{2r}\right)^3 = \frac{\rho_1}{8}$$.

Then the ratio of their terminal velocities is $$\frac{v_2}{v_1} = \frac{r_2^2 \rho_2}{r_1^2 \rho_1} = \frac{(2r)^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4r^2 \times \rho_1/8}{r^2 \times \rho_1} = \frac{4}{8} = \frac{1}{2}$$. Hence $$v_2 = \frac{v}{2}$$.

The correct answer is Option 1: $$\frac{v}{2}$$.