A wheel of diameter 2 m can be rotated about an axis passing through its centre by a moment of force equal to 2.0 N m. What minimum force must be applied on its rim?

Given:
Moment of force = 2.0 N m
Diameter = 2 m so radius = 1 m
Therefore perpendicular distance d = radius = 1 m
Force f = ?

$\text{Moment of force} = \text{force f} \times \text{distance d} \\[1em] \text{Force f}= \dfrac{\text{Moment of force}}{\text{distance d}} \\[1em] = \dfrac{\text{2.0}}{\text{1}} \\[1em] = 2 \text{ N}$

So, minimum force required is 2 N.