$A$ ring of mass $m$ and radius $r$ is melted and recast into a sphere. The moment of inertia of the sphere will be:

Consider the following statements:
Assertion $(A)$: The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass.
Reason $(R)$: The weight of a rigid body always acts through its centre of mass in a uniform gravitational field.
Of these statements:

$A$ massless equilateral triangle $EFG$ of side $'a'$ (as shown in the figure) has three particles of mass $m$ situated at its vertices. The moment of inertia of the system about the line $EX$ perpendicular to $EG$ in the plane of $EFG$ is $\frac{N}{20} ma^{2}$,where $N$ is an integer. The value of $N$ is

$A$ disc $X$ of radius $R$ and thickness $t$ is made of an iron plate. Another disc $Y$ of radius $4R$ and thickness $t/4$ is also made of the same iron plate. What is the relationship between their moments of inertia $I_x$ and $I_y$?

Three bodies have equal masses $m$. Body $A$ is a solid cylinder of radius $R$,body $B$ is a square lamina of side $R$,and body $C$ is a solid sphere of radius $R$. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina)?

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: