$A$ light rod of length $l$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

$A$ solid sphere and a solid cylinder have the same mass and same radius. The ratio of the moment of inertia of the solid sphere about its diameter and the moment of inertia of the solid cylinder about its axis is

$A$ circular disc has radius $R_1$ and thickness $T_1$. Another circular disc made of the same material has radius $R_2$ and thickness $T_2$. If the moment of inertia of both discs are same and $\frac{R_1}{R_2}=2$, then $\frac{T_1}{T_2}=\frac{1}{\alpha}$. The value of $\alpha$ is . . . . . . .

Three point masses $m_1, m_2, m_3$ are placed at the vertices of an equilateral triangle of side $a$. The moment of inertia of the system about an axis passing through the vertex $m_1$ and perpendicular to the plane of the triangle is not what is asked; rather,the axis passes through $m_1$ and bisects the opposite side $m_2m_3$. What is the moment of inertia of the system about this axis?

The ratio of the masses of two circular rings is $1:2$ and the ratio of their diameters is $2:1$. The ratio of their moments of inertia is:

The moment of inertia of a ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is: