A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height \(h\) as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height \(h = \alpha R\). The value of \(\alpha\) is _______.

Let \(v\) be the velocity of the body at the highest point of the circular loop.

At the highest point, the forces acting on the body are its weight \(mg\) (downwards) and the normal reaction \(N\) (downwards). The net downward force provides the necessary centripetal acceleration:

\(N + mg = \dfrac{mv^2}{R}\)

Given that the body exerts a force equal to three times its weight on the track at the highest point, we have \(N = 3mg\). Substituting this into the equation:

\(3mg + mg = \dfrac{mv^2}{R}\)

\(4mg = \dfrac{mv^2}{R} \Rightarrow v^2 = 4gR\)

Applying the principle of conservation of mechanical energy between the point of release and the highest point of the loop (which is at a height of \(2R\) from the bottom):

\(mgh = mg(2R) + \dfrac{1}{2}mv^2\)

Substituting \(v^2 = 4gR\) into the energy equation:

\(mgh = 2mgR + \dfrac{1}{2}m(4gR)\)

\(mgh = 2mgR + 2mgR = 4mgR\)

Therefore, \(h = 4R\).

Comparing this with the given relation \(h = \alpha R\), we get \(\alpha = 4\).

Answer: \(4\)