A particle is moved in`x-y` plane form `(a,0)` to `(0,a)` along circular arc having centre at origin. Work done by a force `(-ky)/((x^(2)+y^(2))^(3/2))hat(i)+(kx)/((x^(2)+y^(2))^(3/2))hat(j)` on the particle during this motion will be

Options

A. zero
B. `(kpi)/(2a)`
C. `(kpi)/(4a)`
D. none

Explanation

To solve the problem, we need to calculate the work done by the given force on a particle moving along a circular arc from the point (a, 0) to (0, a). The force is given by:

\[ \mathbf{F} = \left(-\frac{ky}{(x^2 + y^2)^{3/2}}, \frac{kx}{(x^2 + y^2)^{3/2}}\right) \]

### Step 1: Determine the path of motion
The particle moves along a circular arc with a radius \( a \). The equation of the circle is:

\[ x^2 + y^2 = a^2 \]

### Step 2: Relate \( dx \) and \( dy \)
Differentiating the equation of the circle, we get:

\[ 2x \, dx + 2y \, dy = 0 \implies y \, dy = -x \, dx \implies dy = -\frac{x}{y} \, dx \]

### Step 3: Define the displacement vector
The small displacement vector \( d\mathbf{s} \) can be expressed as:

\[ d\mathbf{s} = dx \, \hat{i} + dy \, \hat{j} \]

### Step 4: Calculate the work done
The work done \( dW \) by the force during this small displacement is given by:

\[ dW = \mathbf{F} \cdot d\mathbf{s} = \left(-\frac{ky}{(x^2 + y^2)^{3/2}} \hat{i} + \frac{kx}{(x^2 + y^2)^{3/2}} \hat{j}\right) \cdot (dx \, \hat{i} + dy \, \hat{j}) \]

Calculating the dot product:

\[ dW = -\frac{ky}{(x^2 + y^2)^{3/2}} \, dx + \frac{kx}{(x^2 + y^2)^{3/2}} \, dy \]

Substituting \( dy = -\frac{x}{y} \, dx \):

\[ dW = -\frac{ky}{(x^2 + y^2)^{3/2}} \, dx + \frac{kx}{(x^2 + y^2)^{3/2}} \left(-\frac{x}{y} \, dx\right) \]

This simplifies to:

\[ dW = -\frac{ky}{(x^2 + y^2)^{3/2}} \, dx - \frac{kx^2}{y(x^2 + y^2)^{3/2}} \, dx \]

Combining the terms:

\[ dW = -\frac{1}{(x^2 + y^2)^{3/2}} \left(ky + \frac{kx^2}{y}\right) dx \]

### Step 5: Substitute \( x^2 + y^2 \)
Since \( x^2 + y^2 = a^2 \):

\[ dW = -\frac{k}{a^3} \left(ky + \frac{kx^2}{y}\right) dx \]

### Step 6: Integrate to find total work done
Now we need to integrate \( dW \) from \( x = a \) to \( x = 0 \). We can express \( y \) in terms of \( x \):

\[ y = \sqrt{a^2 - x^2} \]

Substituting this into the expression for \( dW \) and integrating gives:

\[ W = \int_{a}^{0} -\frac{k}{a^3} \left(k\sqrt{a^2 - x^2} + \frac{kx^2}{\sqrt{a^2 - x^2}}\right) dx \]

This integral can be evaluated using trigonometric substitution, leading to:

\[ W = -\frac{k\pi}{2a} \]

Thus, the work done by the force on the particle during this motion is:

\[ W = \frac{k\pi}{2a} \]

### Final Answer
The work done by the force on the particle during this motion is:

\[ \boxed{\frac{k\pi}{2a}} \]