1. Write the expressions for kinetic energies

For a body rolling without slipping:

• Linear speed of centre of mass = $v$
• Angular speed = $\omega$
• Condition of rolling: $v = R\omega$

The linear kinetic energy of the centre of mass is $K_{\text{linear}} = \frac{1}{2}mv^2$

The rotational kinetic energy about the centre of mass is $K_{\text{rot}} = \frac{1}{2}I\omega^2$

2. Moment of inertia of a solid sphere about its centre

For a solid sphere, $I = \frac{2}{5}mR^2$

So, $K_{\text{rot}} = \frac{1}{2}\left(\frac{2}{5}mR^2\right)\omega^2$

Using $\omega = \frac{v}{R}$, $K_{\text{rot}} = \frac{1}{2}\cdot \frac{2}{5}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{5}mv^2$

3. Find the required ratio

$\frac{K_{\text{linear}}}{K_{\text{rot}}} = \frac{\frac{1}{2}mv^2}{\frac{1}{5}mv^2} = \frac{1/2}{1/5} = \frac{5}{2}$

4. Check options

• A: $\frac{3}{4}$ ❌
• B: $\frac{4}{3}$ ❌
• C: $\frac{5}{2}$ ✅
• D: $\frac{2}{5}$ ❌

Therefore, the correct answer is: $\boxed{\frac{5}{2}}$ which is Option C.