The moment of inertia of a thin uniform rod of mass ' $M$ ' and length ' $L$ ', about an axis perpendicular…

Moment of inertia via parallel axis theorem:

The moment of inertia about the center of mass axis is $I_{\text{CM}} = \frac{ML^2}{12}$.

The distance from the given axis to the center of mass is $d = |\frac{L}{2} - \frac{L}{4}| = \frac{L}{4}$.

Applying the parallel axis theorem $I = I_{\text{CM}} + Md^2$:

$I = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{16}$

Combining terms with common denominator 48:

$I = \frac{4ML^2}{48} + \frac{3ML^2}{48} = \frac{7ML^2}{48}$

Final moment of inertia: $\frac{7ML^2}{48}$