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. `3`
B. `6`
C. `7`
D. `9`

Answer

C

Explanation

Net force on the mass `m` at the left end of the rod
`= (GMm)/((3l)^(2)) - (Gm^(2))/(l^(2))`
Acceleration of the mass 'm' at the left end of the ros
`a_(1) = (1)/(m) [(GMm)/((3l)^(2)) - (Gm^(2))/(l^(2))]`
Similarly net force on the mass `m` at right end of the rod
`= (GM m)/((4l)^(2)) + (Gm^(2))/(l^(2))`
Acceleration of mass `m` at right end of the rod
`a_(2) = (1)/(m) [(GMm)/((4l)^(2)) - (Gm^(2))/(l^(2))]`
Since tension in the rod is zero. So, acceleration in the two masses should be same i.e. `a_(1) = a_(2)`
`(1)/(m) [(GMm)/((3l)^(2)) - (Gm^(2))/(l^(2))] = (1)/(m) [(GMm)/((4l)^(3)) - (Gm^(2))/(l^(2))]`
or `(GM m)/(l^(2)) [(1)/(9) - (1)/(16)] = (2Gm^(2))/(l^(2))`
or `(GM m)/(l^(2)) xx (7)/(9 xx 16) = (2Gm^(2))/(l^(2))`
or `(m)/(M) = (7)/(9 xx 16 xx 2) = (7)/(288)`
or `m = (7)/(288) M = (kM)/(288)` (Given)
so `k = 7`