Consider a circular disc of mass M, radius R with surface mass density σ. The gravitational potential due to the disc at a point P lying on its axis at distance r from the centre is
Answer & explanation
Correct answer: option 3
$E=-2 \pi \sigma G \int\limits_0^\alpha \sin \theta d \theta$
$\Rightarrow E=-2 \pi \sigma G(1-\cos \alpha)$
$\Rightarrow E=-2 \pi \sigma G\left(1-\frac{r}{\sqrt{r^2+R^2}}\right)$
$V=-2 \pi \sigma r G\left(1-\frac{r}{\sqrt{r^2+R^2}}\right)$