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 of \(10\) kg and radius \(R\) about an axis parallel to \(AB\) axis with separation of \(3\) m as shown in figure below. The value of \(R\) is given as \(\sqrt{\dfrac{\alpha}{2}}\). The value of \(\alpha\) is _______.

Moment of inertia of the rod about axis \(AB\) passing through one of its ends:
\(I_{\text{rod}} = \dfrac{ML^2}{3} = \dfrac{40 \times (3)^2}{3} = \dfrac{40 \times 9}{3} = 120\text{ kg}\cdot\text{m}^2\)

Moment of inertia of the solid sphere about the same axis \(AB\), using the parallel axis theorem \(I = I_{\text{cm}} + md^2\):
Mass \(m = 10\text{ kg}\), radius \(R\), distance from the centre of the sphere to axis \(AB\) is \(d = 3\text{ m}\).

\(I_{\text{sphere}} = \dfrac{2}{5}mR^2 + md^2 = \dfrac{2}{5}(10)R^2 + 10(3)^2 = 4R^2 + 90\)

Equating the two moments of inertia:
\(120 = 4R^2 + 90\)
\(4R^2 = 30\)
\(R^2 = \dfrac{15}{2}\)
\(R = \sqrt{\dfrac{15}{2}}\)

Comparing with \(R = \sqrt{\dfrac{\alpha}{2}}\):
\(\alpha = 15\)

Hence, \(\alpha = 15\).