Question

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

• A. $\frac{2}{3}\,m$
• B. $\frac{1}{3}\,m$
• C. $\frac{1}{4}\,m$
• D. $\frac{2}{7}\,m$

Answer 1

Answer: (A)

Let a point mass $C$ is placed at a distance of $x\, m$ from the point mass $A$ as shown in the figure.

Here, $\frac{M_{A}}{M_{B}} = \frac{4}{3}$
Force between $A$ and $C$ is
$F_{AC} = \frac{GMM_{A}}{x^{2}}\quad\ldots\left(i\right)$
Force between $B$ and $C$ is
$F_{BC} = \frac{GMM_{B}}{\left(1-x\right)^{2}}\quad \ldots \left(ii\right)$
According to given problem
$F_{AC} = \frac{1}{3}\,F_{BC}$
$\therefore \frac{GM_{A}M}{x^{2}} = \frac{1}{3}\left(\frac{GM_{B}M}{\left(1-x\right)^{2}}\right)$ (Using$\left(i\right)$ and $\left(ii\right)$)
$\frac{M_{A}}{x^{2}} = \frac{M_{B}}{3\left(1-x\right)^{2}} $
or $\frac{M_{A}}{M_{B}} = \frac{x^{2}}{3\left(1-x\right)^{2}}$
$\frac{4}{3} = \frac{x^{2}}{3\left(1-x\right)^{2}}$ or $4 = \frac{x^{2}}{\left(1-x\right)^{2}}$
or $2 = \frac{x}{1-x}$ or $2-2x = x$
$3x = 2$ or $x = \frac{2}{3} m$