A particle is projected along a horizontal field whose coefficient of friction varies as `mu=A//r^2`, where r is the distance from the origin in meters and A is a positive constant. The initial distance of the particle is `1m` from the origin and its velocity is radially outwards. The minimum initial velocity at this point so the particle never stops is

Options

A. (a) `oo`
B. (b) `2sqrt(gA)`
C. (c) `sqrt(2gA)`
D. (d) `4sqrt(gA)`

Explanation

To solve the problem, we will use the work-energy theorem, which states that the work done by all forces acting on a particle is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
The only force acting on the particle in this scenario is the frictional force, which opposes the motion. The coefficient of friction is given as \( \mu = \frac{A}{r^2} \).

2. **Set Up the Work-Energy Theorem**:
According to the work-energy theorem:
\[ W_{\text{total}} = \Delta KE \]
where \( W_{\text{total}} \) is the work done by friction and \( \Delta KE \) is the change in kinetic energy.

3. **Calculate the Work Done by Friction**:
The work done by friction can be expressed as:
\[ W_{\text{friction}} = -\int_{1}^{\infty} \mu \cdot mg \, dx \]
Substituting \( \mu \):
\[ W_{\text{friction}} = -\int_{1}^{\infty} \frac{A}{x^2} mg \, dx \]

4. **Evaluate the Integral**:
The integral simplifies to:
\[ W_{\text{friction}} = -mgA \int_{1}^{\infty} \frac{1}{x^2} \, dx \]
The integral \( \int \frac{1}{x^2} \, dx \) evaluates to \( -\frac{1}{x} \), so:
\[ W_{\text{friction}} = -mgA \left[-\frac{1}{x}\right]_{1}^{\infty} = -mgA \left(0 + 1\right) = -mgA \]

5. **Determine the Change in Kinetic Energy**:
The initial kinetic energy \( KE_i \) is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
The final kinetic energy \( KE_f \) is 0 since the particle stops at infinity. Thus:
\[ \Delta KE = KE_f - KE_i = 0 - \frac{1}{2} m u^2 = -\frac{1}{2} m u^2 \]

6. **Apply the Work-Energy Theorem**:
Setting the work done by friction equal to the change in kinetic energy:
\[ -mgA = -\frac{1}{2} m u^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gA = \frac{1}{2} u^2 \]

7. **Solve for the Initial Velocity \( u \)**:
Rearranging gives:
\[ u^2 = 2gA \]
Taking the square root:
\[ u = \sqrt{2gA} \]

### Final Answer:
The minimum initial velocity \( u \) at the point where the particle is 1 meter from the origin so that it never stops is:
\[ u = \sqrt{2gA} \]