(A) : The time of ascent for a body projected to move up a rough inclined plane is less than the time of descent.
(R ) : The retardation for upward motion is more than the acceleration for down motion on rough inclined plane.

Options

A. Both (A) and (R) are true and (R) is the correct explanation of (A)
B. Both (A) and (R) are true and (R) is not the
C. (A) is true but ( R) is false
D. Both (A) and (R ) are false

Answer

To solve the question, we need to analyze the assertion and reason provided.

### Assertion (A):
The time of ascent for a body projected to move up a rough inclined plane is less than the time of descent.

### Reason (R):
The retardation for upward motion is more than the acceleration for downward motion on a rough inclined plane.

### Step-by-Step Solution:

1. **Understanding Forces on the Inclined Plane**:
- When a body is projected upwards on a rough inclined plane, it experiences two main forces acting against its motion:
- The gravitational force component acting down the incline, which is \( mg \sin \theta \).
- The frictional force acting down the incline, which is \( \mu mg \cos \theta \) (where \( \mu \) is the coefficient of friction).

2. **Calculating Retardation During Ascent**:
- The total retarding force when moving upwards is:
\[ F_{\text{retard}} = mg \sin \theta + \mu mg \cos \theta \]
- The retarding acceleration \( a_r \) during ascent can be calculated as:
\[ a_r = g \sin \theta + \mu g \cos \theta \]

3. **Time of Ascent**:
- Using the kinematic equation for distance \( s = \frac{1}{2} a t^2 \), where \( s \) is the distance traveled (length of the incline \( L \)), we can express the time of ascent \( t_a \):
\[ L = \frac{1}{2} a_r t_a^2 \implies t_a = \sqrt{\frac{2L}{a_r}} = \sqrt{\frac{2L}{g \sin \theta + \mu g \cos \theta}} \]

4. **Calculating Acceleration During Descent**:
- When the body is moving downwards, the frictional force acts in the opposite direction:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- The acceleration \( a_d \) during descent is:
\[ a_d = g \sin \theta - \mu g \cos \theta \]

5. **Time of Descent**:
- Similarly, the time of descent \( t_d \) can be expressed as:
\[ L = \frac{1}{2} a_d t_d^2 \implies t_d = \sqrt{\frac{2L}{a_d}} = \sqrt{\frac{2L}{g \sin \theta - \mu g \cos \theta}} \]

6. **Comparing Times of Ascent and Descent**:
- To compare \( t_a \) and \( t_d \), we observe that:
- The denominator for \( t_a \) (which is \( g \sin \theta + \mu g \cos \theta \)) is greater than the denominator for \( t_d \) (which is \( g \sin \theta - \mu g \cos \theta \)).
- Since \( t_a \) has a larger denominator, it follows that:
\[ t_a < t_d \]
- Therefore, the assertion (A) is true.

7. **Verifying the Reason**:
- The reason states that the retardation for upward motion is more than the acceleration for downward motion.
- We can see that:
\[ a_r = g \sin \theta + \mu g \cos \theta > g \sin \theta - \mu g \cos \theta = a_d \]
- Thus, the reason (R) is also true.

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