1. Use energy conservation for rolling without slipping

For a body rolling down an incline from rest through the same vertical height $h$,

$mgh = K_{\text{total}}$

where total kinetic energy is the sum of translational and rotational kinetic energies:

$K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Since both bodies start from rest and roll without slipping down the same inclined plane, from the same height, each has:

$K_{\text{total}} = mgh$

So the total kinetic energies of the ring and the solid sphere are equal.

2. Therefore, find the ratio of total kinetic energies

Let

• $K_r$ = kinetic energy of ring
• $K_s$ = kinetic energy of solid sphere

Then

$K_r = mgh, \qquad K_s = mgh$

Hence,

$\frac{K_r}{K_s} = 1$

3. Match with the given form

Given:

$\frac{K_r}{K_s} = \frac{7}{x}$

But we found:

$\frac{K_r}{K_s} = 1$

So,

$\frac{7}{x} = 1 \implies x = 7$

4. Final answer

$x = 7$