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 $r = 3 l$ from $M$ , the tension in the rod is zero for $m = k (\frac{M}{288})$ . The value of $k$ is

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Detailed Solution

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 second law of motion for point mass nearer to larger mass, $F_{1} - F = m a \Rightarrow \frac{G M m}{{(3 l)}^{2}} - \frac{G m^{2}}{l^{2}} = m a . . (i)$

Again using second law of motion for the other mass, $F_{2} + F = m a$

$\frac{G M m}{{(4 l)}^{2}} + \frac{G m^{2}}{l^{2}} = m a . . . . . (i i)$

From eqn. $(i)$ and $(i i)$ , we get

$\frac{G M}{9 l^{2}} - \frac{G m}{l^{2}} = \frac{G M}{16 l^{2}} + \frac{G m}{l^{2}} \frac{M}{9} - \frac{M}{16} = m + m \Rightarrow \frac{7 M}{144} = 2 m m = \frac{7 M}{288} = k (\frac{M}{288})$