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

Infinite number of bodies, each of mass $2\, kg$ are situated on $x-$ axis at distances $1\, m, 2\, m, 4 \,m, 8\, m, ... ,$ respectively, from the origin. The resulting gravitational potential due to this system at the origin will be

- [AIPMT 2013]

A particle of mass $M$ is situated at the centre of a spherical shell of same mass and radius $a$. The gravitational potential at a point situated at $\frac{a}{2}$ distance from the centre, will be

Taking the gravitational potential at a point infinte distance away as zero, the gravitational potential at a point $A$ is $-5\, unit$. If the gravitational potential at point infinite distance away is taken as $+ 10\, units$, the potential at point $A$ is ......... $unit$

Two particles of identical mass are moving in circular orbits under a potential given by $V(r)=K r^{-n}$, where $K$ is a constant. If the radii of their orbits are $r_1 \cdot r_2$ and their speeds are $v_1 \cdot v_2$, respectively. Then,

- [KVPY 2016]

Gravitational potential at the centre of curvature of a hemispherical bowl of radius $R$ and mass $M$ is $V.$