Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) is:

Four identical uniform solid spheres,each of same mass '$M$' and radius '$R$',are placed touching each other as shown in the figure with centers $A, B, C, D$. If $I_{A}, I_{B}, I_{C}, I_{D}$ are the moments of inertia of these spheres respectively about an axis passing through their centers and perpendicular to the plane,then:

$A$ wire of mass $m$ and length $l$ is bent in the form of a circular ring. The moment of inertia of the ring about its axis is:

List-$I$	List-$II$
$(a)$ $MI$ of the rod (length $L$,mass $M$,about an axis $\perp$ to the rod passing through the midpoint)	$(i) \frac{8ML^2}{3}$
$(b)$ $MI$ of the rod (length $L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(ii) \frac{ML^2}{3}$
$(c)$ $MI$ of the rod (length $2L$,mass $M$,about an axis $\perp$ to the rod passing through its midpoint)	$(iii) \frac{ML^2}{12}$
$(d)$ $MI$ of the rod (length $2L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(iv) \frac{2ML^2}{3}$

Choose the correct answer from the options given below:

One circular ring and one circular disc,both having the same mass and radius. The ratio of their moments of inertia about the axis passing through their centers and perpendicular to their planes will be:

For a cube of side $a$ and mass $M$,the moment of inertia is minimum about which axis?