JEE Advanced 2013 · Paper 1 · Physics Q8

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PhysicsSingle correct+2Question 8 of 60

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

\dfrac{2K\pi}a

\dfrac{K\pi}a

\dfrac{K\pi}{2a}

Answer and worked solution

Official answer: D

Radial force. $\vec F=\dfrac{K\vec r}{r^3}$ points along the radius, and on a circle about the origin the displacement is always tangential. So $\vec F\cdot d\vec r=0$ and the work is zero. (Equivalently $F$ is conservative with potential depending only on $r$ .)

(D)

Topics:Potential energy and conservative forces

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JEE Advanced 2013 · Paper 1, Physics question 8