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.You need to login to perform this action.
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