We have two spheres,one of which is a hollow shell and the other solid. They have identical masses and moments of inertia about their respective diameters. The ratio of their radii is given by

Point masses $1, 2, 3$ and $4 \text{ kg}$ are lying at the points $(0,0,0), (2,0,0), (0,3,0)$ and $(-2,-2,0)$ respectively. The moment of inertia of this system about the $x$-axis will be:

$A$ 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?

Four point masses $m, 2m, 3m$ and $4m$ are kept at the corners $A, B, C$ and $D$ respectively of a square $ABCD$ of side '$b$'. The moment of inertia of the system about an axis perpendicular to the plane of the square and passing through the point $D$ is (in $mb^2$)

An annular ring has mass $10 \ kg$ and inner and outer radii are $5 \ m$ and $10 \ m$ respectively. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is