## College Physics (4th Edition)

We can write the expression for the magnitude of the drag force for a sphere falling through a viscous fluid: $F_d = 6\pi \eta ~r~v$ A falling object reaches terminal velocity when the opposing drag forces are equal in magnitude to the gravitational force: $mg = F_d$ $mg = 6\pi \eta ~r~v$ $v = \frac{mg}{6\pi \eta ~r}$ We can use the density $\rho$ to find an expression for the mass of a sphere with radius $r$: $m = (\rho)(\frac{4}{3}\pi~r^3)$ We can replace this expression for the mass $m$ in the terminal velocity equation: $v = \frac{mg}{6\pi \eta ~r}$ $v = \frac{(\frac{4}{3}\rho \pi~r^3)g}{6\pi \eta ~r}$ $v = \frac{2~\rho \pi~r^2~g}{9\pi \eta}$ From this equation, we can see that for spheres with the same density $\rho$, the sphere with a larger radius will have a larger terminal velocity. Since A has a larger radius, A has a larger terminal velocity. The correct answer is: (a) A has the larger terminal velocity.