Q.

The work done on a particle of mass m by a force, K[x(x2+y2)3/2i^+y(x2+y2)3/2j^]  (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 xy plane is               [2013]

1 2Kπa  
2 Kπa  
3 Kπ2a  
4 0  

Ans.

(4)

Radius of circular path =a

The equation of circle is x2+y2=a2

Given : force

F=K[xi^(x2+y2)3/2+yj^(x2+y2)3/2]

F=K[xi^(a2)3/2+yj^(a2)3/2]

F=Ka3[xi^+yj^]

The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path.

Here the angle between the force which acts radially outwards and displacement which is tangential to the circular path is 90°

 Work done, W=FScosθ=0