The ratios of the masses and radii of two circular rings are $1:2$ and $2:1$ respectively. What is the ratio of their moments of inertia?

Two discs $A$ and $B$ of same material and thickness have radii $R$ and $3R$ respectively. Their moments of inertia about their axis will be in the ratio

Three particles of mass $m$ each are placed at the vertices of an equilateral triangle $ABC$ of side length $\ell$. What is the moment of inertia of the system about the axis $AX$ passing through vertex $A$ and perpendicular to side $AB$ in the plane of the triangle?

$A$ rigid body can be hinged about any point on the $x$-axis. When it is hinged such that the hinge is at $x$,the moment of inertia is given by $I = 2x^2 - 12x + 27$. The $x$-coordinate of the centre of mass is:

Two loops $A$ and $B$ of radii $R_1$ and $R_2$ are made from a uniform wire. If the moment of inertia of $A$ is $I_A$ and that of $B$ is $I_B$,then find the ratio $R_2 / R_1$ given that $I_A / I_B = 27$.

$A$ circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$,and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $\frac{t}{4}$. The ratio of the moment of inertia $\frac{I_Y}{I_X}$ is: