Assertion : A ring and a disc of same mass and radius begin to roll without slipping from the top of an inclined surface at `t=0`. The ring reaches the bottom of incline in time `t_(1)` while the disc reaches the bottom in time `t_(2)`, then `t_(1) lt t_(2)`
Reason : Disc will roll down tha plane with more acceleration because of its lesser value of moment of inertia.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is fasle
D. If Assertion is false but Reason is true

Explanation

To solve the problem, we need to analyze the assertion and reason provided in the question regarding the motion of a ring and a disc rolling down an inclined plane.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a ring and a disc of the same mass (M) and radius (R).
- Both objects start rolling down an inclined plane from rest at the same height.
- We need to determine the time taken by each object to reach the bottom of the incline, denoted as \( t_1 \) for the ring and \( t_2 \) for the disc.

2. **Moment of Inertia**:
- The moment of inertia (I) for the ring about its center is given by:
\[ I_{\text{ring}} = MR^2 \]
- The moment of inertia for the disc about its center is given by:
\[ I_{\text{disc}} = \frac{1}{2} MR^2 \]

3. **Calculating the K Factor**:
- The K factor is defined as:
\[ K = 1 + \frac{I}{MR^2} \]
- For the ring:
\[ K_{\text{ring}} = 1 + \frac{MR^2}{MR^2} = 2 \]
- For the disc:
\[ K_{\text{disc}} = 1 + \frac{\frac{1}{2} MR^2}{MR^2} = 1 + \frac{1}{2} = \frac{3}{2} \]

4. **Time to Reach the Bottom**:
- The time taken to reach the bottom of the incline is given by:
\[ t = \sqrt{\frac{2S}{g \sin \theta} \cdot K} \]
- Since the distance \( S \), gravitational acceleration \( g \), and angle \( \theta \) are the same for both objects, we can compare their times based on the K factor:
\[ t_1 \propto \sqrt{K_{\text{ring}}} = \sqrt{2} \]
\[ t_2 \propto \sqrt{K_{\text{disc}}} = \sqrt{\frac{3}{2}} \]

5. **Comparing Times**:
- Since \( K_{\text{ring}} = 2 \) and \( K_{\text{disc}} = \frac{3}{2} \), we find:
\[ t_1 = \sqrt{2} \quad \text{and} \quad t_2 = \sqrt{\frac{3}{2}} \]
- Since \( \sqrt{2} > \sqrt{\frac{3}{2}} \), we conclude that:
\[ t_1 > t_2 \]
- Therefore, the assertion \( t_1 < t_2 \) is **false**.

6. **Acceleration Comparison**:
- The acceleration \( a \) of a rolling body down the incline is given by:
\[ a = \frac{g \sin \theta}{K} \]
- For the ring:
\[ a_{\text{ring}} = \frac{g \sin \theta}{2} \]
- For the disc:
\[ a_{\text{disc}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]
- Since \( \frac{2g \sin \theta}{3} > \frac{g \sin \theta}{2} \), the disc has a greater acceleration than the ring.

7. **Conclusion**:
- The assertion is false, and the reason is true. Therefore, the correct answer is that the assertion is false and the reason is true.

### Final Answer:
- The assertion is false, and the reason is true. Hence, the correct option is D.