$A$ thin metal rod of mass $M$ and length $L$ is cut into $4$ equal parts by cutting it perpendicular to its length. If the moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is $I$,then what is the moment of inertia of each part about a similar axis?

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)	$(iv) \frac{2ML^2}{3}$

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Two discs of same mass and same thickness have densities as $17 \, g/cm^3$ and $51 \, g/cm^3$. The ratio of their moment of inertia about their central axes is .........

Let $M$ be the mass and $L$ be the length of a thin uniform rod. In the first case,the axis of rotation passes through the center and is perpendicular to the length of the rod. In the second case,the axis of rotation passes through one end and is perpendicular to the length of the rod. The ratio of the radius of gyration in the first case to the second case is

$Assertion$ : $A$ judo fighter,in order to throw his opponent onto the mat,tries to initially bend his opponent and then rotate him around his hip.
$Reason$ : As the mass of the opponent is brought closer to the fighter's hip,the moment of inertia of the opponent about the axis of rotation decreases,which makes it easier to rotate the opponent.

The linear mass density of a thin rod $AB$ of length $L$ varies from $A$ to $B$ as $\lambda(x) = \lambda_{0}(1 + \frac{x}{L})$,where $x$ is the distance from $A$. If $M$ is the mass of the rod,then its moment of inertia about an axis passing through $A$ and perpendicular to the rod is $......ML^{2}$.