If $V$ is the gravitational potential due to a sphere of uniform density on its surface,then its value at the center of the sphere will be:

An infinite number of objects,each of mass $1 \ kg$,are placed on the $x$-axis on both sides of $x=0$ at positions $x = \pm 1 \ m, \pm 2 \ m, \pm 4 \ m, \pm 8 \ m, \ldots$ and so on. The magnitude of the resultant gravitational potential (in $SI$ units) at $x=0$ is ($G$ is the universal gravitational constant).

Four particles,each of mass $M$,are located at the vertices of a square with side $L$. The gravitational potential due to this system at the centre of the square is:

$A$ particle of mass $M$ is placed at the centre of a uniform spherical shell of mass $2M$ and radius $R$. The gravitational potential on the surface of the shell is

From a solid sphere of mass $M$ and radius $R$,a spherical portion of radius $R/2$ is removed,as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$,the potential at the centre of the cavity thus formed is
($G =$ gravitational constant)

$A$ particle of mass $m$ is released from infinity and it moves towards a large solid sphere of mass $M$ and radius $R$ made of a jelly-like material. Neglecting the resistance offered by the material of the sphere,find the velocity of the particle when it crosses the centre of the sphere.