Match List-I with List-II

List-I	List-II
(A) M.I. of solid sphere of radius R about any tangent.	(I) $$\frac{5}{3}MR^2$$
(B) M.I. of hollow sphere of radius R about any tangent.	(II) $$\frac{7}{5}MR^2$$
(C) M.I. of circular ring of radius R about its diameter.	(III) $$\frac{1}{4}MR^2$$
(D) M.I. of circular disc of radius R about any diameter.	(IV) $$\frac{1}{2}MR^2$$

We need to match the moment of inertia expressions using standard formulas and the parallel axis theorem.

(A) M.I. of solid sphere about a tangent:

M.I. about the center = $$\frac{2}{5}MR^2$$

By parallel axis theorem (shifting by distance $$R$$):

$$I = \frac{2}{5}MR^2 + MR^2 = \frac{2 + 5}{5}MR^2 = \frac{7}{5}MR^2$$

This matches (II).

(B) M.I. of hollow sphere about a tangent:

M.I. about the center = $$\frac{2}{3}MR^2$$

By parallel axis theorem:

$$I = \frac{2}{3}MR^2 + MR^2 = \frac{2 + 3}{3}MR^2 = \frac{5}{3}MR^2$$

This matches (I).

(C) M.I. of circular ring about its diameter:

By perpendicular axis theorem: $$I_z = I_x + I_y$$

For a ring, $$I_z = MR^2$$ (about axis through center, perpendicular to plane)

By symmetry, $$I_x = I_y$$, so: $$I_x = \frac{MR^2}{2} = \frac{1}{2}MR^2$$

This matches (IV).

(D) M.I. of circular disc about any diameter:

M.I. about axis perpendicular to disc through center = $$\frac{1}{2}MR^2$$

By perpendicular axis theorem and symmetry:

$$I_{\text{diameter}} = \frac{1}{2} \times \frac{1}{2}MR^2 = \frac{1}{4}MR^2$$

This matches (III).

The matching is: A-II, B-I, C-IV, D-III

The correct answer is Option A.