$A$ 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 . . . . . . .

$A$ thin uniform rectangular plate of mass $2 \, kg$ is placed in the $xy$-plane as shown in the figure. The moment of inertia about the $x$-axis is $I_{x}=0.2 \, kg \cdot m^{2}$ and the moment of inertia about the $y$-axis is $I_{y}=0.3 \, kg \cdot m^{2}$. The radius of gyration of the plate about the axis passing through $O$ and perpendicular to the plane of the plate is (in $ \, cm$)

Two solid spheres of iron have radii in the ratio $1 : 2$. Their moments of inertia will be in the ratio

Let $M$ and $L$ be the mass and length of a thin uniform rod,respectively. In the $1^{\text{st}}$ case,the axis of rotation passes through the center and is perpendicular to its length. In the $2^{\text{nd}}$ case,the axis of rotation passes through one end and is perpendicular to its length. The ratio of the radius of gyration in the first case to the second case is:

The moment of inertia of a uniform thin rod of mass $m$ and length $\ell$ about a perpendicular axis passing through one end is $I_{1}$. The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$. If $\frac{I_{1}}{I_{2}} = \frac{x \pi^{2}}{3}$,then the value of $x$ will be ...............

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: