The work done an a particle of mass `m` by a force
`K[(x)/((x^(2) + y^(2))^(3//2)) hati +(y)/((x^(2) + y^(2^(3//2))) hatj)]`
(K being a constant of appropriate dimensions), when the partical is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in x - y plane is

Options

A. (a) `(2Kx)/(a)`
B. (b) `(Kx)/(a)`
C. (c) `(Kx)/(2a)`
D. (d) `0`

Explanation

To solve the problem of finding the work done on a particle of mass \( m \) by the given force as it moves along a circular path from the point \( (a, 0) \) to the point \( (0, a) \), we can follow these steps:

### Step 1: Identify the Force Acting on the Particle
The force acting on the particle is given by:
\[ \mathbf{F} = K \left[ \frac{x}{(x^2 + y^2)^{3/2}} \hat{i} + \frac{y}{(x^2 + y^2)^{3/2}} \hat{j} \right] \]

### Step 2: Define the Path of Motion
The particle moves along a circular path of radius \( a \) about the origin. The equation of the circular path is:
\[ x^2 + y^2 = a^2 \]
As the particle moves from \( (a, 0) \) to \( (0, a) \), it traces a quarter of the circle.

### Step 3: Substitute the Circular Path into the Force Equation
Since \( x^2 + y^2 = a^2 \), we can express the force in terms of \( a \):
\[ \mathbf{F} = K \left[ \frac{x}{a^3} \hat{i} + \frac{y}{a^3} \hat{j} \right] \]
This simplifies to:
\[ \mathbf{F} = \frac{K}{a^3} (x \hat{i} + y \hat{j}) \]

### Step 4: Analyze the Direction of Force and Displacement
The displacement \( d\mathbf{s} \) of the particle along the circular path is always tangential to the circle. The force \( \mathbf{F} \) is directed radially outward from the origin, while the displacement is tangential.

### Step 5: Determine the Angle Between Force and Displacement
Since the force is radial and the displacement is tangential, the angle \( \theta \) between the force and the displacement is \( 90^\circ \).

### Step 6: Calculate the Work Done
The work done \( W \) by a force is given by the dot product of the force and the displacement:
\[ W = \int \mathbf{F} \cdot d\mathbf{s} \]
However, since \( \theta = 90^\circ \), we have:
\[ \cos(90^\circ) = 0 \]
Thus, the work done is:
\[ W = |\mathbf{F}| |\mathbf{s}| \cos(90^\circ) = 0 \]

### Conclusion
The work done on the particle as it moves from \( (a, 0) \) to \( (0, a) \) along the circular path is:
\[ \boxed{0 \text{ joules}} \]