$A$ flywheel rotates about its axis with a constant angular velocity. If a particle suddenly sticks to its rim,what will happen to its moment of inertia?

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:

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $a$. The moment of inertia of this system about an axis passing along one side of the triangle is:

$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Two rings of the same radius and mass are placed such that their centers are at the same point and their planes are mutually perpendicular. The moment of inertia of this system about an axis passing through the center and perpendicular to the plane of one of the rings is (mass of ring = $m$,radius = $r$):

If the radius of a solid sphere is doubled while keeping its mass constant,what is the ratio of its moment of inertia about any of its diameters?