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

Tension in the rod becomes zero when acceleration of both the blocks becomes equal just due to gravitational force and tension remains zero.

Let mass : `(GMm)/(9l^(2)) - (Gm^(2))/(l^(2)) = ma`
`rArr a = (GM)/(9l^(2)) - (Gm)/(l^(2))` ...(i)
Right mass : `(GMm)/(16l^(2)) + (Gm^(2))/(l^(2)) = ma`
`rArr a = (GM)/(16l^(2)) + (Gm)/(l^(2))` ...(ii)
Equating right hand sides of equations (i) and (ii) we can solve and then substitute given value of m to get `k = 7`.