Inside a uniform spherical shell:
$(a)$ The gravitational field is zero.
$(b)$ The gravitational potential is zero.
$(c)$ The gravitational field is same everywhere.
$(d)$ The gravitational potential is same everywhere.
$(e)$ All of the above.
Choose the most appropriate answer from the options given below:

Three particles of equal mass $m$ are situated at the vertices of an equilateral triangle of side $l$. They are moving in a circle under the influence of their mutual gravitational interaction. Then their time period of revolution is directly proportional to

$A$ uniform sphere $A$ with radius $R$ exerts a force $F$ on a small particle $B$ situated at a distance $2R$ from the centre of the sphere. $A$ spherical portion of diameter $R$ is cut from the sphere $A$ as shown in the figure. If $F^{\prime}$ is the new gravitational force between the remaining part of the sphere $A$ and the particle $B$,then find the correct relation between $F$ and $F^{\prime}$.

The ratio of the $K.E.$ required to be given to a satellite to escape Earth's gravitational field to the $K.E.$ required to be given so that the satellite moves in a circular orbit just above Earth's atmosphere is:

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only,then the distance covered by the smaller body before collision is (in $R$)

$A$ small point mass $m$ is placed at a distance $2R$ from the centre $O$ of a big uniform solid sphere of mass $M$ and radius $R$. The gravitational force on $m$ due to $M$ is $F_1$. $A$ spherical part of radius $R/3$ is removed from the big sphere as shown in the figure and the gravitational force on $m$ due to the remaining part of $M$ is found to be $F_2$. The value of the ratio $F_1: F_2$ is