A body which was released from rest, slides down on an inclined plane of inclination `theta`. The coefficient of friction dowh the plane varies as `mu = kx` where k is a spositive constant and x is the distance moved by the body down the plane. The speed of the body 'v' versus distance 'x' graph will be as :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of a body sliding down an inclined plane with a varying coefficient of friction. Let's break it down step by step.

### Step 1: Identify the forces acting on the body
When the body is on the inclined plane, the following forces act on it:
- Gravitational force (weight) \( mg \)
- Normal force \( N \)
- Frictional force \( f \)

The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

The frictional force, which opposes the motion, is given by:
\[ f = \mu N \]
Given that \( \mu = kx \), where \( k \) is a positive constant and \( x \) is the distance moved down the plane, we have:
\[ f = kx N \]

### Step 2: Determine the normal force
The normal force \( N \) acting on the body is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 3: Write the expression for frictional force
Substituting the expression for \( N \) into the frictional force equation:
\[ f = kx (mg \cos \theta) \]

### Step 4: Apply Newton's second law
The net force acting on the body along the incline is:
\[ F_{\text{net}} = mg \sin \theta - f \]
Substituting the expression for \( f \):
\[ F_{\text{net}} = mg \sin \theta - kx (mg \cos \theta) \]

Using Newton's second law \( F = ma \), we can write:
\[ ma = mg \sin \theta - kx (mg \cos \theta) \]
Dividing through by \( m \):
\[ a = g \sin \theta - kx (g \cos \theta) \]

### Step 5: Relate acceleration to velocity and distance
Acceleration \( a \) can also be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]

Substituting this into our equation gives:
\[ v \frac{dv}{dx} = g \sin \theta - kx (g \cos \theta) \]

### Step 6: Integrate to find the relationship between \( v \) and \( x \)
Rearranging the equation:
\[ v dv = \left( g \sin \theta - kx g \cos \theta \right) dx \]

Integrating both sides:
\[ \int v \, dv = \int \left( g \sin \theta - kx g \cos \theta \right) dx \]
This results in:
\[ \frac{v^2}{2} = g \sin \theta \cdot x - \frac{k g \cos \theta}{2} x^2 + C \]

### Step 7: Determine the constant of integration
Since the body is released from rest at \( x = 0 \) (where \( v = 0 \)):
\[ 0 = g \sin \theta \cdot 0 - \frac{k g \cos \theta}{2} \cdot 0^2 + C \implies C = 0 \]

### Step 8: Final expression for velocity
Thus, we have:
\[ v^2 = 2g \sin \theta \cdot x - k g \cos \theta \cdot x^2 \]

### Step 9: Analyze the velocity versus distance graph
The equation \( v^2 = 2g \sin \theta \cdot x - k g \cos \theta \cdot x^2 \) is a quadratic equation in terms of \( x \). The graph of \( v \) versus \( x \) will be a parabola that opens downwards (since the coefficient of \( x^2 \) is negative).

### Conclusion
The speed \( v \) versus distance \( x \) graph will be a downward-opening parabola, indicating that the speed increases to a maximum value and then decreases as the distance \( x \) increases.