The work done on a particle of mass $$m$$  by a force,
$$K\left[ {\frac{x}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}\hat i + \frac{y}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}\hat j} \right]$$
($$K$$  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-

Solution :
Let us consider a point on the circle

The equation of circle is $${x^2} + {y^2} = {a^2}$$
The force is
$$\eqalign{ & \vec F = K\left[ {\frac{{x\hat i}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}} + \frac{{y\hat j}}{{{{\left( {{x^2} + {y^2}} \right)}^{\frac{3}{2}}}}}} \right] \cr & \vec F = K\left[ {\frac{{x\hat i}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}} + \frac{{y\hat j}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}}} \right] \cr & \vec F = \frac{K}{{{a^3}}}\left[ {x\hat i + y\hat j} \right] \cr} $$
The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path. Therefore the angle between the force and displacement is $${90^ \circ }$$  and $$W=0$$
option (D) is correct.