Moment of inertia about an axis $$AB$$ for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass 10 kg and radius $$R$$ about an axis parallel to $$AB$$ axis with separation of 3 m as shown in figure. The value of $$R$$ is given as $$\sqrt{\frac{\alpha}{2}}$$. The value of $$\alpha$$ is _________.

For the rod (mass $$M = 40\ \text{kg}$$, length $$L = 3\ \text{m}$$) the axis $$AB$$ is perpendicular to the rod and passes through one of its ends.
Moment of inertia of a uniform rod about an axis perpendicular to the rod and through an end is

$$I_{\text{rod}} = \frac{1}{3} M L^{2}$$

Substituting the values:

$$I_{\text{rod}} = \frac{1}{3}\,(40)\,(3)^{2} = \frac{40 \times 9}{3} = 120\ \text{kg m}^{2}$$

For the solid sphere (mass $$m = 10\ \text{kg}$$, radius $$R$$) we need its moment of inertia about an axis parallel to its diameter but displaced by $$d = 3\ \text{m}$$ from the centre.
First write the moment of inertia about the central diameter:

$$I_{\text{cm}} = \frac{2}{5}\,m R^{2}$$

Apply the parallel-axis theorem, $$I = I_{\text{cm}} + m d^{2}$$, to shift the axis:

$$I_{\text{sphere}} = \frac{2}{5}\,m R^{2} + m d^{2}$$

Insert the numerical values $$m = 10\ \text{kg}$$ and $$d = 3\ \text{m}$$:

$$I_{\text{sphere}} = \frac{2}{5}\,(10) R^{2} + (10)(3)^{2} = 4 R^{2} + 90$$

The question states that the two moments of inertia are equal, so

$$I_{\text{rod}} = I_{\text{sphere}}$$

$$\Rightarrow 120 = 4 R^{2} + 90$$

Solve for $$R^{2}$$:

$$4 R^{2} = 120 - 90 = 30$$

$$\Rightarrow R^{2} = \frac{30}{4} = 7.5$$

The radius is expressed in the form $$R = \sqrt{\dfrac{\alpha}{2}}$$, so

$$R^{2} = \frac{\alpha}{2}$$

Compare with $$R^{2} = 7.5$$:

$$\frac{\alpha}{2} = 7.5 \quad\Longrightarrow\quad \alpha = 15$$

Hence the required value of $$\alpha$$ is 15.