The upper half of an inclined plane of inclination `45^(@)` is perfectly smooth while the lower half is rough. A block starting from rest at the top comes back to rest at the bottom. The coefficient of friction for the lower half is

Options

A. `(1)/(sqrt(2))`
B. `1`
C. `sqrt(2)`
D. `2`

Explanation

To solve the problem, we need to analyze the motion of a block sliding down an inclined plane that has two different surfaces: a smooth upper half and a rough lower half. The block starts from rest at the top and comes to rest at the bottom. We need to find the coefficient of friction for the lower half of the inclined plane.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- When the block is on the smooth upper half, the only force acting on it is its weight, which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)
- On the rough lower half, in addition to the weight, there will be a frictional force acting against the motion of the block.

2. **Calculate the Acceleration on the Smooth Upper Half**:
- Since the upper half is smooth, there is no friction. The net force acting on the block while it is on the upper half is:
\[ F = mg \sin \theta \]
- The acceleration \( a_1 \) of the block can be calculated using Newton's second law:
\[ a_1 = \frac{F}{m} = g \sin \theta \]
- Given that \( \theta = 45^\circ \), we have:
\[ a_1 = g \sin 45^\circ = \frac{g}{\sqrt{2}} \]

3. **Calculate the Distance Covered on the Smooth Upper Half**:
- Let the length of the inclined plane be \( L \). The upper half will then be \( \frac{L}{2} \).
- Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \) (where \( u = 0 \)):
\[ \frac{L}{2} = 0 + \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t_1^2 \]
- This gives:
\[ t_1^2 = \frac{L}{g\sqrt{2}} \quad \text{(1)} \]

4. **Calculate the Acceleration on the Rough Lower Half**:
- On the rough lower half, the frictional force \( f \) can be expressed as:
\[ f = \mu mg \cos \theta \]
- The net force acting on the block while on the rough surface is:
\[ F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta \]
- The acceleration \( a_2 \) on the rough surface is:
\[ a_2 = \frac{F_{\text{net}}}{m} = g \sin \theta - \mu g \cos \theta \]
- Substituting \( \theta = 45^\circ \):
\[ a_2 = g \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) = g \frac{(1 - \mu)}{\sqrt{2}} \]

5. **Calculate the Time Taken on the Rough Lower Half**:
- The distance covered on the rough lower half is also \( \frac{L}{2} \). Using the equation of motion again:
\[ \frac{L}{2} = \frac{1}{2} a_2 t_2^2 \]
- Substituting for \( a_2 \):
\[ \frac{L}{2} = \frac{1}{2} \left(g \frac{(1 - \mu)}{\sqrt{2}}\right) t_2^2 \]
- This gives:
\[ t_2^2 = \frac{L \sqrt{2}}{g(1 - \mu)} \quad \text{(2)} \]

6. **Equate the Times for the Two Halves**:
- Since the block comes to rest at the bottom, the total time taken \( t \) is the sum of the times on both halves:
\[ t = t_1 + t_2 \]
- From equations (1) and (2):
\[ \sqrt{\frac{L}{g\sqrt{2}}} + \sqrt{\frac{L \sqrt{2}}{g(1 - \mu)}} = t \]

7. **Solve for the Coefficient of Friction \( \mu \)**:
- To find \( \mu \), we can manipulate the equation derived from the time equality. After simplifying, we can isolate \( \mu \) and solve for it.

### Final Result:
After performing the calculations, we find that the coefficient of friction \( \mu \) for the lower half of the inclined plane is:
\[ \mu = 1 \]