A particle moves in the x-y plane under the a | vedclass

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(D) The given force is $F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$.In polar coordinates,$x = r \cos \

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(D) The given force is $F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} ight]$.In polar coordinates,$x = r \cos 	heta$ and $y = r \sin 	heta$,where $r = \sqrt{x^2+y^2}$.Substituting these into the force expression:$F = K \left[ \frac{r \cos 	heta}{r^3} \hat{i} + \frac{r \sin 	heta}{r^3} \hat{j} ight] = \frac{K}{r^2} (\cos 	heta \hat{i} + \sin 	heta \hat{j})$.This force is a central force directed radially outward,i.e.,$F = \frac{K}{r^2} \hat{r}$.The work done by a central force is given by $W = \int F \cdot dr = \int \frac{K}{r^2} dr$.Since the particle moves along a circular path of radius $a$,the distance $r$ from the origin is constant $(r = a)$.For a circular path,the displacement vector $d\vec{l}$ is always perpendicular to the radial force vector $\vec{F}$.Therefore,the dot product $\vec{F} \cdot d\vec{l} = 0$ at every point on the path.Thus,the total work done $W = \int \vec{F} \cdot d\vec{l} = 0$.

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