Question

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 $I$ 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 $m = k\left(\frac{M}{288}\right)$. The value of $k$ is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (C)

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 Newtons $2^{nd}$ law of motion for point mass nearer to larger mass,
$F_1- F = ma$
$\frac{GMm}{\left(3l\right)^{2} }-\frac{Gm^{2}}{l^{2}} = ma\quad\ldots\left(i\right)$

Again using $2^{nd}$ law of motion for the other mass
$F_{2} + F = ma$
$\frac{GMm}{\left(4l\right)^{2} }-\frac{Gm^{2}}{l^{2}} = ma\quad \ldots \left(ii\right)$
From eqn. $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{GM}{9l^{2}}-\frac{Gm}{l^{2}} = \frac{GM}{16l^{2}}+\frac{Gm}{l^{2}}$
$\frac{M}{9}-\frac{M}{16} = m+m$
$\Rightarrow \frac{7\,M}{144} = 2m$
$m = \frac{7\,M}{288} = k\left(\frac{M}{288}\right)$
$\therefore k = 7$