Two particles of equal mass ' m ' move in a circle of radius ' r ' under the action of their mutual gravitational attraction. The speed of each particle will be ( $\mathrm{G}=$ Universal gravitational constant)

Two particles of equal mass $m$ orbit each other in circular paths of radius $r$ under mutual gravitational attraction. The center of mass lies midway between them, so the separation distance is $d = 2r$.

The gravitational force providing the centripetal acceleration is $F_g = \frac{G m^2}{(2r)^2} = \frac{G m^2}{4r^2}$.

For circular motion with speed $v$, the required centripetal force is $F_c = \frac{m v^2}{r}$.

Equating these gives $\frac{G m^2}{4r^2} = \frac{m v^2}{r}$, which simplifies to $v^2 = \frac{G m}{4r}$.

Each particle's speed is therefore $v = \sqrt{\frac{G m}{4r}}$, corresponding to option A.

Final answer: $\boxed{\text{A}}$