JEE Main 2026 — Work Power Energy Question with Solution

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:

$N+mg=\frac{m{v}^2}{R}$

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:

$3mg+mg=\frac{m{v}^2}{R}$

$4mg=\frac{m{v}^2}{R}\Rightarrow v^2=4gR$

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):

$mgh=mg(2R)+\frac{1}{2}m{v}^2$

Substituting $v^2=4gR$ into the energy equation:

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

$mgh=2mgR+2mgR=4mgR$

Therefore, $h=4R$.

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

Answer: $4$