The work done on a particle of mass $m$ by a force, $K\left[\frac{x}{(x^2+y^2)^{3/2}}\hat{i} + \frac{y}{(x^2+y^2)^{3/2}}\hat{j}\right]$ ($K$ being a constant of appropriate dimensions), when the particle is taken from the point $(a,0)$ to the point $(0,a)$ along a circular path of radius $a$ about the origin in the $x$-$y$ plane is

Step 1: Express the force vector in terms of the position vector.

Define the position vector $\vec{r}$ and its magnitude $r$.

$$\vec{r} = x\hat{i} + y\hat{j}$$ $$r = \sqrt{x^2 + y^2}$$

Rewrite the given force $\vec{F}$ using $\vec{r}$ and $r$.

$$\vec{F} = K\left[\frac{x}{(x^{2}+y^{2})^{3/2}}\hat{i}+\frac{y}{(x^{2}+y^{2})^{3/2}}\hat{j}\right]$$ $$\vec{F} = \frac{K}{r^3}\left(x\hat{i} + y\hat{j}\right)$$ $$\vec{F} = \frac{K}{r^3}\vec{r}$$

Step 2: Analyze the direction of the force and displacement.

The force is directed radially outward along the position vector $\vec{r}$.

$$\vec{F} \parallel \vec{r}$$

For a circular path of radius $a$ centered at the origin, the infinitesimal displacement vector $d\vec{r}$ is always tangential to the path.

$$d\vec{r} \perp \vec{r}$$

Step 3: Calculate the work done.

Since the force is radial and the displacement is tangential, they are perpendicular at every point along the path.

$$\vec{F} \cdot d\vec{r} = 0$$

Integrate the work done from the starting point $A(a,0)$ to the ending point $B(0,a)$.

$$W = \int_{A}^{B} \vec{F} \cdot d\vec{r}$$ $$W = 0$$

Step 4: Verify using the potential energy of the conservative force.

The force is conservative and can be expressed as the negative gradient of a potential energy function $U(r)$.

$$\vec{F} = -\nabla U$$ $$U(r) = \frac{K}{r}$$

Calculate the work done as the negative change in potential energy between the initial point $A(a,0)$ and final point $B(0,a)$.

$$W = -\left(U(a) - U(a)\right)$$ $$W = -\left(\frac{K}{a} - \frac{K}{a}\right)$$ $$W = 0$$

Answer: $0$