An air bubble of radius 2 mm is rising at a steady speed of 2.5 cm...

Given:
Radius of the bubble, $r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$
Steady velocity, $v = 2.5 \text{ cm s}^{-1} = 2.5 \times 10^{-2} \text{ m s}^{-1}$
Coefficient of viscosity, $\eta = 0.5 \text{ kg m}^{-1} \text{ s}^{-1}$

According to Stokes' law, the magnitude of the viscous drag force is:
$F = 6\pi\eta r v$
$F = 6\pi \times 0.5 \times (2 \times 10^{-3}) \times (2.5 \times 10^{-2})$
$F = 3\pi \times 5 \times 10^{-5}$
$F = 15\pi \times 10^{-5} \text{ N} = 1.5\pi \times 10^{-4} \text{ N}$

Viscous drag always opposes the relative motion of the object through the fluid. Since the air bubble is rising (moving vertically upwards), the viscous drag force must act vertically downwards.

Answer: $1.5\pi \times 10^{-4} \text{ N}$, vertically downwards