The great downhill race

As the objects will roll down without slipping on the inclined plane, assuming any particular body with radius $r$ , mass $m$ and radius of gyration equal to $r\sqrt{k}$ and using the principle of conservation of mechanical energy (with reference for potential set at the base of the plane) we get

$mgh = \dfrac{mv^2}{2} + \dfrac{I {\omega}^2}{2}$

As the motion is a pure rolling motion we have : $\omega = \dfrac vr$ and so

$\begin{aligned} & mgh = \dfrac{mv^2}{2} + \dfrac{\left( kmr^2 \right) \left( \frac {v^2}{r^2} \right)}{2} \\ \implies & gh = \dfrac{v^2}{2} + \dfrac{kv^2}{2} \\ \implies & v = \sqrt{\dfrac{2gh}{1+k}} \end{aligned}$

Now with the help of the equation : $v^2 = u^2 + 2as$ , since $u$ and $s$ remain the same for all bodies, therefore $a \propto v^2$ which also means that $a \propto |v|$ and the greater the acceleration, the less time it takes to cover the required distance $s$ .

Thus we have

$t \propto \dfrac{1}{a} \propto \dfrac{1}{v} \propto \sqrt{1+k}$

So the spheres will take less time to reach the bottom as their $\sqrt{1+k}$ values are less than that of the discs. Also further note that the order is irrespective of the size of the body and only depends on its shape.

Thus, the correct order is : $\boxed{C=D < A=B}$ .

**Sorry, that's an error in the explanation. I meant to say that the spheres will reach the bottom first, followed by the cylinders. Ignore the last statement