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7.Gravitation
normal
A thin rod of length $L$ is bent to form a semicircle. The mass of rod is $M.$ What will be the gravitational potential at the centre of the circle?
A
$-\frac {GM}{L}$
B
$ - \frac{{GM}}{{2\pi L}}$
C
$ - \frac{{\pi GM}}{{2L}}$
D
$ - \frac{{\pi GM}}{{L}}$
Solution
Gravitational potential of the centre $=\frac{-\mathrm{GM}}{\mathrm{r}}$
since, $\pi r=L \Rightarrow r=\frac{L}{\pi}$
Gravitational potential $=\frac{-\mathrm{GM}}{(\mathrm{L} / \pi)}$
$=\frac{-\pi G M}{L}$
Standard 11
Physics