A larger spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M. The point masses are connected by rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only throght their mutual gravitational interaction. When the point mass nearer to M is at a distance r =3l form M, the tensin in the rod is zero for `m =k((M)/(288)).` The value of k is

Options

A. 5
B. 6
C. 7
D. 8

Answer

C

Explanation

( c) Both the point masses are connected by a light rod so they have same acceleration.
Suppose each point mass is moving with acceleration a towards larger mass M.
Using Newton's `2^(nd)` law of motion for point mass nearer to larger mass,
`F_(1)-F=ma`
`(GMm)/((3l)^(2))-(Gm^(2))/(l^(2))=ma` ....(i)

Again using `2^(nd)` law of motion for another mass `F_(2)+F=ma`
`(GMm)/((4l)^(2))-(Gm^(2))/(l^(2))=ma` ....(ii)
From eqn. (i) and (ii), we get
`(GM)/(9l^(2))-(Gm)/(l^(2))=(GM)/(16l^(2))+(Gm)/(l^(2))`
`(M)/(9)-(M)/(16)=m+m implies (7M)/(144)=2m`
`m=(7M)/(288)=k((M)/(288)) :. k=7`