$A$ thin metal wire of length $L$ and uniform linear mass density $Q$ is bent into a circular coil with $O$ as the center. The moment of inertia of the coil about the axis $XX'$ is:

The moment of inertia of a circular ring of mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is

$(a)$ Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point $(x, y)$ in the $x-y$ plane from an axis through the origin and perpendicular to the plane is $x^{2}+y^{2}$)
$(b)$ Prove the theorem of parallel axes. (Hint: If the centre of mass of a system of $n$ particles is chosen to be the origin,$\sum m_{i} r_{i}=0$)

The moment of inertia of a circular ring about an axis perpendicular to its plane and passing through its center is $200 \, gm \cdot cm^2$. The moment of inertia about its diameter is ....... $gm \cdot cm^2$.

$A$ thin uniform rod has mass $M$ and length $L$. The moment of inertia about an axis perpendicular to it and passing through the point at a distance $\frac{L}{3}$ from one of its ends,will be

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.