A point mass 'm' is located at a distance 'r' | vedclass

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(C) Let the rod lie along the $x$-axis with its near end at $x = r$ and its far end at $x = r + L$ from the point mass '$m$' located at the origin.Con

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$\frac{G M m}{r(r+L)}$

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(C) Let the rod lie along the $x$-axis with its near end at $x = r$ and its far end at $x = r + L$ from the point mass '$m$' located at the origin.Consider a small element of the rod of length '$dx$' at a distance '$x$' from the point mass '$m$'.The mass of this element is $dm = \lambda dx = \frac{M}{L} dx$.The gravitational force '$dF$' exerted by this element on the point mass '$m$' is given by:$dF = \frac{G m dm}{x^2} = \frac{G m (M/L) dx}{x^2} = \frac{G M m}{L} \frac{dx}{x^2}$.To find the total force '$F$',we integrate '$dF$' from $x = r$ to $x = r + L$:$F = \int_{r}^{r+L} \frac{G M m}{L} \frac{dx}{x^2} = \frac{G M m}{L} \left[ -\frac{1}{x} \right]_{r}^{r+L}$.$F = \frac{G M m}{L} \left( -\frac{1}{r+L} - (-\frac{1}{r}) \right) = \frac{G M m}{L} \left( \frac{1}{r} - \frac{1}{r+L} \right)$.$F = \frac{G M m}{L} \left( \frac{r+L-r}{r(r+L)} \right) = \frac{G M m}{L} \left( \frac{L}{r(r+L)} \right) = \frac{G M m}{r(r+L)}$.

$A$ point mass '$m$' is located at a distance '$r$' from the end of a uniform thin rod of mass '$M$' and length '$L$' as shown in the figure. The magnitude of the gravitational force of attraction is

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