Gujarati
Hindi
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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.