A thin rod of length L is bent to form a semi | vedclass

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(D) The gravitational potential $V$ due to a point mass $m$ at a distance $r$ is given by $V = -\frac{Gm}{r}$.For a continuous distribution of mass,th

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$-\frac{\pi GM}{L}$

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(D) The gravitational potential $V$ due to a point mass $m$ at a distance $r$ is given by $V = -\frac{Gm}{r}$.For a continuous distribution of mass,the potential at the centre of a semicircle formed by a rod of length $L$ and mass $M$ is calculated by considering the distance of every mass element $dm$ from the centre.Since every point on the semicircle is at the same distance $r$ from the centre,the total potential is $V = \int -\frac{G dm}{r} = -\frac{G}{r} \int dm = -\frac{GM}{r}$.The length of the semicircle is $L = \pi r$,which implies $r = \frac{L}{\pi}$.Substituting $r$ into the potential formula: $V = -\frac{GM}{(L/\pi)} = -\frac{\pi GM}{L}$.

$A$ thin rod of length $L$ is bent to form a semicircle. The mass of the rod is $M$. What will be the gravitational potential at the centre of the circle?

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