Falling vertically in glycerine, a metal sphere has a mass of 50 mg...

Given:
Mass of the sphere, $m = 50 \text{ mg} = 50 \times 10^{-6} \text{ kg}$
Radius of the sphere, $r = 1 \text{ mm} = 10^{-3} \text{ m}$
Coefficient of viscosity, $\eta = 8.0 \text{ poise} = 0.8 \text{ Pa}\cdot\text{s}$
Density of glycerine, $\rho = 1260 \text{ kg/m}^3$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

Volume of the sphere, $V = \dfrac{4}{3}\pi r^3$

Buoyant force, $F_B = V\rho g = \dfrac{4}{3}\pi r^3 \rho g$

Weight of the sphere, $W = mg$

At terminal velocity $v_t$, the net downward force is balanced by the viscous drag force (Stokes' law):

$6\pi\eta r v_t = W - F_B$

$v_t = \dfrac{mg - \dfrac{4}{3}\pi r^3 \rho g}{6\pi\eta r}$

Calculating the mass of displaced glycerine:
$m' = \dfrac{4}{3} \times \dfrac{22}{7} \times (10^{-3})^3 \times 1260 = 5.28 \times 10^{-6} \text{ kg}$

Net downward force:
$W - F_B = (50 \times 10^{-6} - 5.28 \times 10^{-6}) \times 9.8 = 44.72 \times 10^{-6} \times 9.8 = 4.38 \times 10^{-4} \text{ N}$

Viscous drag coefficient:
$6\pi\eta r = 6 \times \dfrac{22}{7} \times 0.8 \times 10^{-3} = 1.51 \times 10^{-2} \text{ kg/s}$

Terminal velocity:
$v_t = \dfrac{4.38 \times 10^{-4}}{1.51 \times 10^{-2}} = 2.9 \times 10^{-2} \text{ m/s} = 2.9 \text{ cm/s}$