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 given assertion and reason question, we will analyze the kinetic energies involved in the rolling motion of a ring.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a ring rolls without slipping on the ground, it exhibits both translational and rotational motion. The translational motion refers to the movement of the center of mass of the ring, while the rotational motion refers to the spinning of the ring around its center.

2. **Kinetic Energy Formulas**:
- The translational kinetic energy (KE_trans) of the ring is given by:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the ring and \( v \) is the linear velocity of the center of mass.
- The rotational kinetic energy (KE_rot) of the ring is given by:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the ring and \( \omega \) is the angular velocity.

3. **Moment of Inertia of the Ring**:
- The moment of inertia \( I \) for a ring about its central axis is:
\[ I = mR^2 \]
where \( R \) is the radius of the ring.

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

5. **Substituting into the Rotational Kinetic Energy**:
- Now substituting \( \omega \) into the rotational kinetic energy formula:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (mR^2) \left(\frac{v}{R}\right)^2 = \frac{1}{2} mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{2} mv^2 \]

6. **Finding the Ratio of Kinetic Energies**:
- Now, we can find the ratio of translational kinetic energy to rotational kinetic energy:
\[ \frac{KE_{\text{trans}}}{KE_{\text{rot}}} = \frac{\frac{1}{2} mv^2}{\frac{1}{2} mv^2} = 1 \]

7. **Conclusion**:
- Since the ratio of translational kinetic energy to rotational kinetic energy is 1:1, this means that both energies are equal when the ring is rolling without slipping. Hence, the assertion that the ring has 50% translational and 50% rotational energy is true.

### Final Answer:
Both the assertion (A) and reason (R) are true, and the reason correctly explains the assertion.

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