A particle moves in the x-y plane under the action of a force,
` F = K [ (x)/((x^(2) + y^(2))^((3)/(2))) hat(i) + (y)/((x^(2) + y^(2))^((3)/(2))) hat(j) ]` where, K is a
constant. Work done by the force when the particel moves from (0,a) to (a, 0) along a circular path of radius a about the origin is

Options

A. `(2K pi)/(a)`
B. `(K pi)/(a)`
C. `(K pi)/(2a)`
D. 0

Answer

D

Explanation

According to the equestion,

Force on a particle moves in the x -y plane,
`F = k [ (x)/((x^(2) + y^(2))^(3//2)) hat(i) + (y)/((x^(2) + y^(2))^(3//2)) hat(j) ]`
Now , putting , x = r cos `theta`, y = r sin `theta` [ `because` From figure, ]
F = k `[ (r cos theta)/((r^(2) cos^(2) theta + r^(2) sin^(2) theta)^(3//2)) hat(i) + (r sin theta)/((r^(2) cos^(2) theta + r^(2) sin^(2) theta)^(3//2)) hat(j) ] `
`therefore " " sin^(2) theta + cos^(2) theta = 1`
F = k `[(r)/(r^(3)) (cos theta hat(i) + sin theta hat(j) ) ]`
F = `(k)/(r^(2)) [ cos theta hat(i) + sin theta hat(j)] `
`therefore` Work done , w = F `xx` s (`because s = 0` , for circle )
Now, we can say that the direction of force is along the radius of circle, Hence, the work done by this force will be zero.