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=2tantheta`
B. `mu=tantheta`
C. `mu=2//(tantheta)`
D. `mu=1//tantheta`

Answer

To solve the problem, we need to analyze the motion of a body sliding down an inclined plane that has two different surfaces: a smooth upper half and a rough lower half.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The inclined plane has an angle of inclination \( \phi \).
- The upper half of the plane is smooth (frictionless), while the lower half is rough.

2. **Identifying Forces**:
- When the body is at the top, it starts from rest and will slide down due to gravity.
- On the smooth surface, there is no friction, so the only force acting on the body is its weight component along the incline.

3. **Calculating Work Done by Gravity**:
- The work done by gravity as the body moves down the entire length of the incline (let's denote the total length as \( 2L \)) can be calculated.
- The vertical height \( h \) corresponding to the length \( 2L \) is \( h = 2L \sin \phi \).
- The work done by gravity \( W_g \) is given by:
\[ W_g = mgh = mg(2L \sin \phi) = 2mgL \sin \phi \]

4. **Calculating Work Done by Friction**:
- On the rough surface, the frictional force \( F_f \) acting against the motion is given by \( F_f = \mu mg \cos \phi \), where \( \mu \) is the coefficient of friction.
- The work done by friction \( W_f \) over the distance \( L \) (the length of the rough part) is:
\[ W_f = -F_f \cdot L = -\mu mg \cos \phi \cdot L \]

5. **Setting Up the Equation**:
- Since the body comes to rest at the bottom, the total work done on the body must equal zero (the work done by gravity must equal the work done against friction):
\[ W_g + W_f = 0 \]
- Substituting the expressions for work done:
\[ 2mgL \sin \phi - \mu mg \cos \phi \cdot L = 0 \]

6. **Simplifying the Equation**:
- We can cancel \( mgL \) from both sides (assuming \( m \neq 0 \) and \( L \neq 0 \)):
\[ 2 \sin \phi - \mu \cos \phi = 0 \]
- Rearranging gives:
\[ \mu \cos \phi = 2 \sin \phi \]
- Therefore, we can express \( \mu \) as:
\[ \mu = \frac{2 \sin \phi}{\cos \phi} = 2 \tan \phi \]

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