question_answer

Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length \[\ell \]is

A) \[\frac{Gm}{{{\ell }^{2}}}\operatorname{along}+x-axis\]          

B) \[\frac{Gm}{\pi {{\ell }^{2}}}\operatorname{along}+y-axis\]

C) \[\frac{2\pi Gm}{{{\ell }^{2}}}\operatorname{along}+x-axis\]

D) \[\frac{2\pi Gm}{{{\ell }^{2}}}\operatorname{along}+y-axis\]

Correct Answer: D

Solution :

[d] Let mass per unit light of wire, \[\lambda =\frac{M}{\ell }\] and \[\pi r=\ell ,r=\frac{\ell }{\pi }\]. mass of element, \[\operatorname{dm}=\lambda rd\theta  then dE =\frac{Gdm}{{{\operatorname{r}}^{2}}}\] \[\int\limits_{0}^{\pi }{dE\int\limits_{0}^{\pi }{\frac{G\lambda rd\theta }{{{r}^{2}}}\left( \hat{i}\cos \theta +\hat{j}\sin \theta  \right)}}\] \[E=\frac{G\lambda }{r}\left[ \int\limits_{0}^{\pi }{\hat{i}\cos \theta +\int\limits_{0}^{\pi }{\hat{j}\sin \theta }} \right]\] \[=\frac{2G\lambda }{r}\hat{j}=\frac{2GM}{\ell r}\hat{j}=\frac{2Gm\pi }{{{\ell }^{2}}}\hat{j}\](along y-axis