$A$ 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

In the diagram shown below,all three rods are of equal length $L$ and equal mass $M$. The system is rotated such that rod $B$ is the axis. What is the moment of inertia of the system?

Two rings of radius $R$ and $nR$ made of the same material have a ratio of moment of inertia about an axis passing through their centers and perpendicular to their planes as $1:8$. The value of $n$ is (mass per unit length $= \lambda$).

$A$ solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

Five particles of mass $2 \ kg$ each are attached to the rim of a circular disc of radius $0.1 \ m$ and negligible mass. The moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is ........ $kg \ m^2$.

$Assertion$ : Radius of gyration of a body is a constant quantity.
$Reason$ : The radius of gyration of a body about an axis of rotation may be defined as the root mean square distance of the particles from the axis of rotation.