A : When a ring moves in pure rolling condition on ground it has 50% translational and 50% rotational energy.
R : `(KE_("trans"))/(KE_("rot"))=((1)/(2)MV^(2))/((1)/(2)l omega^(2))=((1)/(2)MV^(2))/((1)/(2)(MR^(2))(V^(2))/(R^(2)))=1:1`

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion,
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion,
C. If Assertion is true statement but Reason is false,
D. If both Assertion and Reason are false statements,

Answer

To solve the problem, we need to analyze the kinetic energy of a ring that is rolling without slipping. We will derive the translational and rotational kinetic energies and show that they are equal in a pure rolling condition.

### Step-by-Step Solution:

1. **Identify the Kinetic Energies**:
- The translational kinetic energy (KE_trans) of the ring is given by:
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \]
- The rotational kinetic energy (KE_rot) of the ring is given by:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
- For a ring, the moment of inertia \( I \) about its center is:
\[ I = m R^2 \]

2. **Relate Angular Velocity and Linear Velocity**:
- In pure rolling motion, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = R \omega \]
- Rearranging gives:
\[ \omega = \frac{v}{R} \]

3. **Substitute \( \omega \) in the Rotational Kinetic Energy**:
- Substitute \( \omega \) into the expression for \( KE_{\text{rot}} \):
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (m R^2) \left(\frac{v}{R}\right)^2 \]
- Simplifying this gives:
\[ KE_{\text{rot}} = \frac{1}{2} m R^2 \cdot \frac{v^2}{R^2} = \frac{1}{2} m v^2 \]

4. **Compare the Kinetic Energies**:
- Now we have:
\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \]
\[ KE_{\text{rot}} = \frac{1}{2} m v^2 \]
- Therefore, we can see that:
\[ KE_{\text{trans}} = KE_{\text{rot}} \]

5. **Calculate the Total Kinetic Energy**:
- The total kinetic energy (KE_total) is:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]

6. **Find the Ratios**:
- The ratio of translational kinetic energy to rotational kinetic energy is:
\[ \frac{KE_{\text{trans}}}{KE_{\text{rot}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} = 1 \]

### Conclusion:
- Since both the translational and rotational kinetic energies are equal, we conclude that in pure rolling conditions, a ring has 50% of its kinetic energy as translational and 50% as rotational. Thus, both statements A and R are correct.