Three point masses,each of mass $m$,are kept at the corners of an equilateral triangle of side $L$. The system rotates about the center of the triangle. The period of rotation $T$ is directly proportional to:

Four particles,each of mass $M$ and equidistant from each other,move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

Three particles,each of mass $M$,are situated at the vertices of an equilateral triangle of side length $L$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation $L$. The initial speed that should be given to each particle is

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

$A$ star $2.5$ times the mass of the sun and collapsed to a size of $12 \;km$ rotates with a speed of $1.2 \;rev/s$. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun $= 2 \times 10^{30} \;kg$)

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 $R$)