Terminal Velocity And Stokes' Law

Terminal velocity is reached when the downward weight of the sphere is balanced by the upward buoyant force and the viscous drag described by Stokes' law, $F=6\pi \eta rv$. Setting net force to zero gives $\frac{4}{3}\pi r^3(\rho -\sigma )g=6\pi \eta r{v}_t$, which rearranges to $v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$. For balls of the same material in the same oil, every factor except the radius is fixed, so $v_t\propto r^2$. Doubling the radius therefore multiplies the terminal velocity by $2^2=4$. The option 'same' ignores the radius dependence entirely. The option 'twice' assumes a linear rather than quadratic relation. The option 'half' even gives the wrong direction, since a larger ball falls faster. This quadratic dependence on radius is the central result of the NCERT treatment of viscous fall, and it holds only while the motion stays laminar so that Stokes' linear drag law applies. The weight of the sphere grows as $r^3$ whereas the viscous drag grows only as $r$, so their balance forces the steady speed to climb as $r^2$. As a check, larger raindrops and hailstones are known to reach higher steady speeds than tiny droplets, which agrees with the strong $r^2$ growth obtained here.