Gravitation Question 341
Question: Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length $ [\ell ] $ is
Options:
A) $ [\frac{Gm}{{{\ell }^{2}}}along+x-axis] $
B) $ [\frac{Gm}{\pi {{\ell }^{2}}}along+y-axis] $
C) $ [\frac{2\pi Gm}{{{\ell }^{2}}}along+x-axis] $
D) $ [\frac{2\pi Gm}{{{\ell }^{2}}}along+y-axis] $
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Answer:
Correct Answer: D
Solution:
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Let mass per unit light of wire, $ [\lambda =\frac{M}{\ell }] $
and $ [\pi r=\ell ,r=\frac{\ell }{\pi }] $ .
mass of element,
$ [dm=\lambda rd\theta $ then $ dE =\frac{Gdm}{{{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
