The upper half of an inclined plane with inclination `phi` is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Options

A. `mu=(1)/( tan theta)`
B. `mu=(2)/( tan theta)`
C. `mu= 2 tan theta`
D. `mu= tan theta`

Explanation

To solve the problem step by step, we will analyze the motion of the body on the inclined plane, considering both the smooth and rough sections.

### Step 1: Analyze the motion from the top to the bottom of the smooth section
1. **Identify the parameters**: The inclined plane has an angle of inclination \( \phi \). The upper half (length \( L/2 \)) is smooth, and the lower half (also length \( L/2 \)) is rough with a coefficient of friction \( \mu \).
2. **Initial conditions**: The body starts from rest, so the initial velocity \( U = 0 \).
3. **Acceleration on the smooth section**: The acceleration \( a \) of the body while sliding down the smooth section is given by:
\[ a = g \sin \phi \]
4. **Using the kinematic equation**: We can find the maximum velocity \( V_{\text{max}} \) at the bottom of the smooth section (point B) using the equation:
\[ V^2 = U^2 + 2aS \]
Substituting the values:
\[ V_{\text{max}}^2 = 0 + 2(g \sin \phi)(L/2) \]
Simplifying this gives:
\[ V_{\text{max}}^2 = gL \sin \phi \]

### Step 2: Analyze the motion from the bottom of the smooth section to the bottom of the rough section
1. **For the rough section**: The body starts with a velocity \( V_{\text{max}} \) and will come to rest at the bottom (point C).
2. **Net acceleration on the rough section**: The net acceleration while moving down the rough section is:
\[ a_{\text{net}} = g \sin \phi - \mu g \cos \phi \]
3. **Using the kinematic equation again**: Since the final velocity at point C is zero, we can write:
\[ 0 = V_{\text{max}}^2 + 2a_{\text{net}}S \]
Substituting \( S = L/2 \) and \( a_{\text{net}} \):
\[ 0 = gL \sin \phi + 2\left(g \sin \phi - \mu g \cos \phi\right)\left(\frac{L}{2}\right) \]
Simplifying gives:
\[ 0 = gL \sin \phi + gL \sin \phi - \mu gL \cos \phi \]
\[ 0 = 2gL \sin \phi - \mu gL \cos \phi \]

### Step 3: Solve for the coefficient of friction \( \mu \)
1. **Rearranging the equation**:
\[ \mu gL \cos \phi = 2gL \sin \phi \]
Dividing both sides by \( gL \) (assuming \( gL \neq 0 \)):
\[ \mu \cos \phi = 2 \sin \phi \]
2. **Solving for \( \mu \)**:
\[ \mu = \frac{2 \sin \phi}{\cos \phi} = 2 \tan \phi \]

### Final Answer
The coefficient of friction \( \mu \) for the lower half of the inclined plane is:
\[ \mu = 2 \tan \phi \]