1. Use standard moments of inertia and parallel axis theorem

We match each item in List-I.

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2. (A) Solid sphere of radius $R$ about any tangent

For a solid sphere, moment of inertia about any diameter is: $I_{\text{diameter}}=\frac{2}{5}MR^2$

A tangent axis is parallel to a diameter and at distance $R$ from the center, so by parallel axis theorem: $I_{\text{tangent}}=I_{\text{diameter}}+MR^2$ $I_{\text{tangent}}=\frac{2}{5}MR^2+MR^2=\frac{7}{5}MR^2$

So, $A \to (II)$

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3. (B) Hollow sphere of radius $R$ about any tangent

For a hollow sphere (spherical shell), moment of inertia about any diameter is: $I_{\text{diameter}}=\frac{2}{3}MR^2$

Again using parallel axis theorem for tangent axis: $I_{\text{tangent}}=\frac{2}{3}MR^2+MR^2=\frac{5}{3}MR^2$

So, $B \to (I)$

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4. (C) Circular ring of radius $R$ about its diameter

For a ring, moment of inertia about axis perpendicular to its plane through center is: $I_z=MR^2$

By perpendicular axis theorem for a planar body: $I_z=I_x+I_y$

Since for a ring, both diameters are equivalent, $I_x=I_y$

Hence, $MR^2=2I_{\text{diameter}}$ $I_{\text{diameter}}=\frac{1}{2}MR^2$

So, $C \to (IV)$

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5. (D) Circular disc of radius $R$ about any diameter

For a disc, moment of inertia about axis perpendicular to plane through center is: $I_z=\frac{1}{2}MR^2$

Using perpendicular axis theorem, $I_z=I_x+I_y$

By symmetry, $I_x=I_y=I_{\text{diameter}}$

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

Thus, $D \to (III)$

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6. Final matching

$A \to II, \quad B \to I, \quad C \to IV, \quad D \to III$

This corresponds to Option A.

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7. Comparison with stored answer

Stored correct answer = A.

Our derived answer = A.

So the answer agrees with the stored answer.