A uniform straight wire of mass M and length L is bent to form a...

Let $R$ be the radius of the circular arc.

The length of the arc is given as $L$ and the angle subtended at the centre is $\theta = 120^\circ = \dfrac{2\pi}{3}$ radians.

Using the relation between arc length, radius, and angle, we have:

$L = R\theta = R\left(\dfrac{2\pi}{3}\right)$

$\Rightarrow R = \dfrac{3L}{2\pi}$

The moment of inertia of a circular arc of mass $M$ and radius $R$ about an axis passing through its centre of curvature and perpendicular to its plane is:

$I = \int R^2 dm = MR^2$

Substituting the value of $R$, we get:

$I = M\left(\dfrac{3L}{2\pi}\right)^2$

$I = \dfrac{9ML^2}{4\pi^2}$

Answer: $\dfrac{9ML^2}{4\pi^2}$