A large 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 (see figure). The point masses are connected by a rigid massless rod of length ℓ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance $r = 3ℓ$ from M, the tension in the rod is zero for $m = k(M/288).$ The value of k is ______.

Step 1: Identify the positions of the masses.

The nearer point mass is at a distance of $3\ell$ from $M$, and the farther point mass is at a distance of $4\ell$ from $M$.

Step 2: Set up the condition for zero tension in the rod.

For the tension in the rod to be zero, the accelerations of both point masses along the line must be equal.

$$a_1 = a_2$$

Step 3: Write the acceleration of each point mass.

Taking the direction towards $M$ as negative, write the gravitational acceleration of each mass.

$$a_1 = -\frac{GM}{(3\ell)^2} + \frac{Gm}{\ell^2}$$ $$a_2 = -\frac{GM}{(4\ell)^2} - \frac{Gm}{\ell^2}$$

Step 4: Equate the accelerations and solve for $m$.

Set $a_1 = a_2$ to find the relation between $m$ and $M$.

$$-\frac{GM}{9\ell^2} + \frac{Gm}{\ell^2} = -\frac{GM}{16\ell^2} - \frac{Gm}{\ell^2}$$ $$\frac{2Gm}{\ell^2} = \frac{GM}{\ell^2}\left(\frac{1}{9} - \frac{1}{16}\right)$$ $$2m = M\left(\frac{7}{144}\right)$$ $$m = \frac{7M}{288}$$

Step 5: Determine the value of $k$.

Compare the expression for $m$ with the given relation $m = k\left(\frac{M}{288}\right)$.

$$k = 7$$

Answer: 7