A particle is moving along x-axis under the action of a force, F which varies with its position (x) as `F prop x^(-1//4)`. The variation of power due to this force with x is

Options

A. `x^(1//3)`
B. `x^(1//8)`
C. `x^(1//2)`
D. `t^(1//4)`

Answer

To solve the problem step by step, we will analyze the relationship between force, acceleration, velocity, and power as given in the question.

### Step 1: Understanding the relationship between force and position
We are given that the force \( F \) varies with position \( x \) as:
\[ F \propto x^{-\frac{1}{4}} \]
This means we can express the force as:
\[ F = k x^{-\frac{1}{4}} \]
where \( k \) is a proportionality constant.

### Step 2: Relating force to acceleration
According to Newton's second law, force can also be expressed in terms of mass \( m \) and acceleration \( a \):
\[ F = m a \]
Since acceleration \( a \) can be expressed as \( a = \frac{dv}{dt} = v \frac{dv}{dx} \), we can write:
\[ F = m v \frac{dv}{dx} \]

### Step 3: Equating the two expressions for force
From the previous steps, we have:
\[ m v \frac{dv}{dx} = k x^{-\frac{1}{4}} \]
Rearranging gives:
\[ v \frac{dv}{dx} = \frac{k}{m} x^{-\frac{1}{4}} \]

### Step 4: Integrating to find the relationship between velocity and position
We can separate the variables and integrate:
\[ v \, dv = \frac{k}{m} x^{-\frac{1}{4}} \, dx \]
Integrating both sides:
\[ \int v \, dv = \int \frac{k}{m} x^{-\frac{1}{4}} \, dx \]
The left side integrates to:
\[ \frac{v^2}{2} \]
The right side integrates to:
\[ \frac{k}{m} \cdot \left( \frac{x^{3/4}}{3/4} \right) = \frac{4k}{3m} x^{3/4} \]
Thus, we have:
\[ \frac{v^2}{2} = \frac{4k}{3m} x^{3/4} + C \]
For simplicity, we can ignore the constant \( C \) for finding the relationship:
\[ v^2 \propto x^{3/4} \]
Taking the square root gives:
\[ v \propto x^{3/8} \]

### Step 5: Finding the expression for power
Power \( P \) is defined as the product of force and velocity:
\[ P = F \cdot v \]
Substituting the expressions we have:
\[ P \propto x^{-\frac{1}{4}} \cdot x^{\frac{3}{8}} = x^{-\frac{1}{4} + \frac{3}{8}} = x^{-\frac{2}{8} + \frac{3}{8}} = x^{\frac{1}{8}} \]

### Conclusion
Thus, the variation of power due to this force with \( x \) is:
\[ P \propto x^{\frac{1}{8}} \]