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

Answer

7

Explanation

Due to gravitaional interaction connected masses have some acceleration
force eq. for mass nearer to larger mass
`(GMm)/((3l)^(2))-(Gm^(2))/(l^(2))=ma` …(i)
Force eq. for mass away from larger mass
`(GMm)/((4l)^(2))+(Gm^(2))/(l^(2))=ma` ...(ii)
From equation (i) & (ii)
`(GM)/(9l^(2))-(GM)/(l^(2))=(GM)/(16l^(2))+(GM)/(l^(2))implies(M)/(9)-(M)/(16)=m+m`
`(7M)/(144)=2mimpliesm=(7M)/(288)=k((M)/(288))`
`impliesK=7`