The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through a point at a distance $L/4$ from one of its ends and perpendicular to the length of the rod is

The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is $6$ times its radius $R$, is:

$A$ solid sphere of mass $5 \text{ kg}$ and a disc of mass $4 \text{ kg}$ have the same radius. The ratio of the moment of inertia of the sphere about its tangent to the moment of inertia of the disc about a tangent in its plane will be $x : y$. The value of $x$ and $y$ respectively is

$A$ 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 the figure. The moment of inertia of the ring about an axis $yy'$ will be:

$A$ thin uniform square lamina of side $a$ is placed in the $xy$-plane with its sides parallel to $X$ and $Y$-axes and with its centre coinciding with the origin. Its moment of inertia about an axis passing through a point on the $Y$-axis at a distance $y = 2a$ and parallel to the $X$-axis is equal to its moment of inertia about an axis passing through a point on the $X$-axis at a distance $x = d$ and perpendicular to the $xy$-plane. Then the value of $d$ is:

From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed concentrically. If the mass of the removed disc is $M$,what is the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre?