question_answer

The work done on 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 being a constant of appropriate dimension) when the particle is taken from the point (a, 0) to the point (3a,4a) along a circular path of radius a about the origin in the XY-plane is:

A)  \[\frac{2K\pi }{a}\]                               

B)  \[\frac{5K}{4\,a}\]

C)  \[\frac{K}{2\,a}\]                     

D)  \[\frac{4K}{5\,a}\]

Correct Answer: D

Solution :

A circle with radius r as shown below, such that the position of point P from centre O, i.e., OP = r is given as \[r=OP=x\hat{i}+y\hat{j}\]             \[F=\frac{k}{{{({{x}^{2}}+{{y}^{2}})}^{3/2}}}(x\hat{i}+y\hat{j})\,=\frac{k}{{{r}^{3}}}(\vec{r})=\frac{k}{{{r}^{2}}}(\hat{r})\] \[W=\vec{F}.d\vec{r}=\int\limits_{a}^{5a}{(r).d\vec{r}=\frac{4k}{5a}}\] Since F is along r or in radial direction. Hence the force is central. Therefore, work done do not depends on path but on initial and final position.