$A$ thin rod of length $L$ is bent in the form of a circle. Its mass is $M$. What force will act on a mass $m$ placed at the centre of this circle? $(G = \text{universal gravitational constant})$

Six objects are placed at the vertices of a regular hexagon. The geometric centre of the hexagon is at the origin with objects $1$ and $4$ on the $X$-axis (see figure). The mass of the $k$-th object is $m_k = k^i M |\cos \theta_k|$, where $i$ is an integer, $M$ is a constant with the dimension of mass, and $\theta_k$ is the angular position of the $k$-th vertex measured from the positive $X$-axis in the counter-clockwise sense. If the net gravitational force on a body at the centroid vanishes, the value of $i$ is

The mass of the moon is $\frac{M}{81}$,where $M$ is the mass of the earth. The distance between the earth and the moon is $60R$,where $R$ is the radius of the earth. At what distance from the center of the moon will the gravitational intensity be zero (in $R$)?

Gauss's law for gravitation is given by:

The gravitational field in a region is given by $\vec{E} = (5\,N/kg)\,\hat{i} + (12\,N/kg)\,\hat{j}$. If the potential at the origin is taken to be zero,then the ratio of the potential at the points $(12\,m, 0)$ and $(0, 5\,m)$ is:

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho(r) = \rho_{0} \left(1 - \frac{r^{2}}{R^{2}}\right)$. Then the gravitational field is maximum at: