$A$ thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $XX'$ is

$I_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of the disc. $I_2$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to the axis $CM$ at a distance $\frac{2R}{3}$ from the centre. The ratio of $I_2$ to $I_1$ is $\frac{I_2}{I_1} = \frac{x}{9}$. The value of $x$ is ($R =$ radius of the disc).

Match Column-$I$ with Column-$II$:

Column-$I$	Column-$II$
$(1)$ Perpendicular Axis Theorem	$(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem	$(b)$ $I_z = I_x + I_y$

Where, $d =$ distance between two parallel axes.

$A$ circular disc of mass $9M$ and radius $R$ has a smaller disc of radius $R/3$ cut from it. Calculate the moment of inertia of the remaining portion about an axis passing through the center of the original disc and perpendicular to its plane. (in $MR^2$)

The moment of inertia of a sphere about its diameter is $40 \ kg \cdot m^2$. Find the moment of inertia about any tangent.

The moment of inertia of a uniform circular ring,having a mass $M$ and a radius $R$,about an axis tangential to the ring and perpendicular to its plane,is