1. Given moment of inertia of the original rod

For a rod of mass $M$ and length $L$, about an axis through its center and perpendicular to its length, $I_0=\frac{1}{12}ML^2$ This is given as $\alpha=\frac{1}{12}ML^2$

2. After cutting the rod into two equal parts

Each part has:

• mass $\dfrac{M}{2}$
• length $\dfrac{L}{2}$

These two pieces are joined symmetrically as a cross, with their centers coinciding.

The required axis passes through the common center and is normal to the plane of the cross.

3. Moment of inertia of one half-rod about this axis

For one half-rod, the axis passes through its own center and is perpendicular to its length. So, $I_1=\frac{1}{12}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2$

Now simplify: $I_1=\frac{1}{12}\cdot \frac{M}{2}\cdot \frac{L^2}{4}$ $I_1=\frac{ML^2}{96}$

4. Moment of inertia of the cross

Since there are two identical half-rods, $I_{\text{cross}}=2I_1=2\cdot \frac{ML^2}{96}$ $I_{\text{cross}}=\frac{ML^2}{48}$

5. Express in terms of $\alpha$

Since $\alpha=\frac{ML^2}{12}$ we get $I_{\text{cross}}=\frac{ML^2}{48}=\frac{1}{4}\cdot \frac{ML^2}{12}=\frac{\alpha}{4}$

6. Check options

• A: $\alpha/4$ ✅
• B: $\alpha/8$ ❌
• C: $\alpha$ ❌
• D: $\alpha/2$ ❌

Therefore, the correct answer is A.