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

Choose the correct answer from the options given below:

We recall the standard formula for the moment of inertia of a uniform slender rod of length $$l$$ and mass $$m$$ about an axis perpendicular to the rod:

1. When the axis passes through the mid-point of the rod, the formula is

$$I_{\text{mid}}=\frac{1}{12}\,m\,l^{2}.$$

2. When the axis passes through one end of the rod, the formula is

$$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$

With these two relations ready, we now evaluate each item in List-I one by one and match with List-II.

Term (a): The rod has length $$L$$ and mass $$M$$. The axis is perpendicular to the rod and passes through its mid-point, so we use the mid-point formula.

$$I_{(a)} \;=\;\frac{1}{12}\,M\,L^{2}.$$

Checking List-II, we see that $$\frac{ML^{2}}{12}$$ is entry (iii). So, (a) corresponds to (iii).

Term (b): The rod has the same length $$L$$ but double the mass, i.e. $$2M$$. The axis is at one end, so we use the end-point formula.

First write the formula: $$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$ Now substitute $$m=2M$$ and $$l=L$$:

$$I_{(b)} \;=\;\frac{1}{3}\,(2M)\,L^{2} =\;\frac{2\,M\,L^{2}}{3}.$$

Looking at List-II, $$\dfrac{2ML^{2}}{3}$$ is entry (iv). Therefore, (b) matches (iv).

Term (c): The rod now has length $$2L$$ while the mass is $$M$$. The axis again passes through the mid-point, so we use the mid-point formula.

Formula: $$I_{\text{mid}}=\frac{1}{12}\,m\,l^{2}.$$ Substituting $$m=M$$ and $$l=2L$$ gives

$$I_{(c)} \;=\;\frac{1}{12}\,M\,(2L)^{2} =\;\frac{1}{12}\,M\,(4L^{2}) =\;\frac{4}{12}\,M\,L^{2} =\;\frac{1}{3}\,M\,L^{2}.$$

In List-II, $$\dfrac{ML^{2}}{3}$$ is entry (ii). Thus, (c) corresponds to (ii).

Term (d): The rod now has length $$2L$$ and mass $$2M$$, with the axis at one end. Hence we again use the end-point formula.

Formula: $$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$ Substituting $$m=2M$$ and $$l=2L$$ yields

$$I_{(d)} \;=\;\frac{1}{3}\,(2M)\,(2L)^{2} =\;\frac{2}{3}\,M\,(4L^{2}) =\;\frac{8}{3}\,M\,L^{2}.$$

The expression $$\dfrac{8ML^{2}}{3}$$ appears in List-II as entry (i). Consequently, (d) matches with (i).

Gathering all the matches we have obtained:

(a) → (iii),  (b) → (iv),  (c) → (ii),  (d) → (i).

When we inspect the four options, we see that this exact sequence is given in Option C.

Hence, the correct answer is Option C.