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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 is 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 

Answer:

Explanation:

$r=OP=x\hat{i}+y\hat{j}$

   $F=\frac{k}{(x^{2}+y^{2})^{3/2}}(x\hat{i}+y\hat{j})$

$=\frac{k}{r^{3}}(r)$

  Since F is along with r or in radial direction. Therefore , work done is zero