QJEEAdv15Ph1707: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph1707 Solution: A large spherical mass M is fixed at one position and two…

The acceleration $\vec{a}$ of the point masses are equal because they are connected by a massless rigid rod.


Consider the situation when tension in the rod is zero. The gravitational forces on the two point masses are shown in the figure. The forces $f_1=\frac{G M m}{r^2}$ and $f_3=\frac{G M m}{(r+l)^2}$ are due to the attraction by the larger mass $M$ . The force $f_2=\frac{G m m}{l^2}$ is due to mutual attraction between the two point masses. Apply Newton's second law on the two point masses to get

$$\frac{GMm}{r^2}-\frac{Gmm}{l^2}=ma$$ ....... (1)

$$\frac{GMm}{(r+l{)}^2}+\frac{Gmm}{l^2}=ma$$ ......... (2)

From eqn. (1) and (2), we get

$$\begin{array}{rl}& \frac{GM}{9l^2}-\frac{Gm}{l^2}=\frac{GM}{16l^2}+\frac{Gm}{l^2}\\ \\ & \frac{M}{9}-\frac{M}{16}=m+m\Rightarrow \frac{7M}{144}=2m\end{array}$$

$$m=\frac{7M}{288}=k\left(\frac{M}{288}\right)$$

$$\therefore$$ k = 7