Moment of inertia of a body about two perpendicular axes $X$ and $Y$ in the plane of the lamina are $20 \text{ kg m}^2$ and $25 \text{ kg m}^2$ respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of $X$ and $Y$ axes is:

$A$ uniform cylinder has a radius $R$ and length $L$. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length,then

$ABC$ is a plane lamina in the shape of an equilateral triangle. $D$ and $E$ are the midpoints of $AB$ and $AC$,respectively,and $G$ is the centroid of the lamina. The moment of inertia of the lamina about an axis passing through $G$ and perpendicular to the plane $ABC$ is $I_{0}$. If the part $ADE$ is removed,the moment of inertia of the remaining part about the same axis is $\frac{NI_{0}}{16}$,where $N$ is an integer. The value of $N$ is

$(a)$ Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point $(x, y)$ in the $x-y$ plane from an axis through the origin and perpendicular to the plane is $x^{2}+y^{2}$)
$(b)$ Prove the theorem of parallel axes. (Hint: If the centre of mass of a system of $n$ particles is chosen to be the origin,$\sum m_{i} r_{i}=0$)

$A$ square lamina of side $b$ has the same mass as a disc of radius $R$. The moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio $\frac{b}{R}$ is

Four identical solid spheres each of mass $m$ and radius $a$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about one side of the square is: