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 $M$ as shown in the 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 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. 7.00
• B. 7.0
• C. 7

Answer 1

Answer: (A), (B), (C)

Since both particles of mass $m$ are connected by a rod, so the acceleration of both the masses will be the same, let’s say $a$.


For $m$ close to $M$, FBD is as follows.


Here,$F_M$ force of attraction on $m$ due to mass $M$ placed distance $\left(r\right)$ from $M$.
And $F_m$ is the force of attraction on $m$ due to the another mass $m$ placed at distance $l$ from it.

From the Newton's second law of motion, The net force on the $m$ can be written as,

$\frac{GMm}{r^2}-\frac{G{m}^2}{l^2}=ma...\left(1\right)$

For the other $m$, FBD is as follows,


Here, $F_M$ force of attraction on $m$ due to mass $M$ placed distance $(r+l)$ from $M$.
And $F_m$ is the force of attraction on $m$ due to the another mass $m$ placed at distance $l$ from it.

Hence,
$\frac{GMm}{(r+l^2)}+\frac{G{m}^2}{l^2}=ma...\left(2\right)$

Solving Equation (1) and (2),

$\frac{GMm}{r^2}-\frac{G{m}^2}{l^2}=\frac{GMm}{(l+r{)}^2}+\frac{G{m}^2}{l^2}$

Given, $r=3l$

$\Rightarrow \frac{GMm}{(3l{)}^2}-\frac{G{m}^2}{l^2}=\frac{GMm}{(l+3l{)}^2}+\frac{G{m}^2}{l^2}$

$\Rightarrow \frac{GMm}{9l^2}-\frac{G{m}^2}{l^2}=\frac{GMm}{\left(16l^2\right)}+\frac{G{m}^2}{L^2}$

$\Rightarrow \frac{M}{9}-m=\frac{M}{16}+m$

$\Rightarrow 2m=\frac{M}{9}-\frac{M}{16}$

$\Rightarrow 2m=\frac{7M}{144}$

$\therefore m=\frac{7M}{288}$

Given, $m=\frac{kM}{288}$,

$\therefore k=7$

Thus, Accepted value of $k$ is $7$.