A thin wire of length ‘L’ and linear mass density ‘m’ is bent into a circular ring (in x-y plane) with centre ‘C’ as shown in figure. The moment of inertia of the ring about an axis yy will be: ____.

From the figure we see that the thin wire of length L is formed into a circular ring of radius R lying in the x‑y plane. The axis yy is a line parallel to the y‑axis that touches the ring at a point on its circumference (a tangent axis). The wire has a linear mass density m (mass per unit length).

First we express the radius R in terms of the given length L:

$L=2\pi R,R=\frac{L}{2\pi }$

The total mass of the ring is the linear density multiplied by its length:

$M=mL$

For a thin circular ring the moment of inertia about an axis through its centre and perpendicular to the plane (the z‑axis) is

$I$_z = MR2

Because the ring lies in a plane, the perpendicular‑axis theorem gives

$I$_z = I_x + I_y

The two diametral axes x and y through the centre are equivalent, so

$I$_x = I_y = I_z 2 = MR2 2

Now we need the moment of inertia about the tangent axis yy. Using the parallel‑axis theorem:

$I$_yy = I_y + Md2

Here d is the distance between the centre C and the tangent axis, which equals the radius R.

$I$_yy = MR2 2 + MR2 = 3 2 MR2

Substituting M = m L and R = L / (2π):

$I$_yy = 3 2 mL L 2π 2 = 3 2 mL L2 2π2 = 3mL3 8π2

Therefore, the moment of inertia of the thin circular wire about the tangent axis yy is

$\frac{3}{m}$

which matches the given correct option.