Two lead balls of mass m and 2 m are placed at a separation d. A third ball of mass m is placed at an unknown location on the line joining the first two balls such that the net gravitational force experienced by the first ball is $\frac{6 {Gm}^{2}}{d^{2}} .$ What is the location of the third ball?

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$A t a d i s \tan c e \frac{2 d}{3} f r o m t h e b a l l o f m a s s m .$

$A t a d i s \tan c e \frac{d}{2 \sqrt{2}} f r o m t h e b a l l o f m a s s m .$

At a distance $\frac{d}{2} f r o m t h e b a l l o f m a s s m .$

At a distance $\frac{d}{4} f r o m t h e b a l l o f m a s s m .$

answer is A, D.

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Detailed Solution

There are two possibilities –
(1) the net force is towards right
(2) the net force is towards left (see figure)
When net force is towards right–
$\frac{Gm (2 m)}{d^{2}} + \frac{Gmm}{x^{2}} = \frac{6 {Gm}^{2}}{d^{2}} \Rightarrow \frac{1}{x^{2}} = \frac{4}{d^{2}} \Rightarrow x = \frac{d}{2}$

When the net force is to the left

$\begin{array}{l} \frac{Gmm}{x^{2}} - \frac{Gm (2 m)}{d^{2}} = \frac{6 {Gm}^{2}}{d^{2}} \\ \frac{1}{x^{2}} = \frac{8}{d} \\ x = \frac{d}{2 \sqrt{2}} \end{array}$