Q. Two point masses $A$ and $B$ having masses in the ratio $4:3$ are separated by a distance of $1m.$ When another point mass $C$ of mass $M$ is placed in between $A$ and $B,$ the force between $A$ and $C$ is $\left(\frac{1}{3}\right)^{r d}$ of the force between $B$ and $C.$ Then the distance of $C$ from $A$ is:

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A

$\frac{2}{3}m$

B

$\frac{1}{3}m$

C

$\frac{1}{4}m$

D

$\frac{2}{7}m$

Solution:

$F_{C A}=\frac{1}{3}F_{C B}$
$\Rightarrow \frac{G 4 M m}{r^{2}}=\frac{G 3 M m}{\left(\right. 1 - r \left(\left.\right)^{2}}\times \frac{1}{3}$
On solving $r=\frac{2}{3}m$

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