Question

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 ${\left(\frac{1}{3}\right)}^{rd}$ 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 $xm$ 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
$F_{AC}=\frac{GM{M}_A}{x^2}\dots \left(i\right)$
Force between $B$ and $C$ is
$F_{BC}=\frac{GM{M}_B}{{\left(1-x\right)}^2}\dots \left(ii\right)$
According to given problem
$F_{AC}=\frac{1}{3}{F}_{BC}$
$\therefore \frac{G{M}_{A}M}{x^2}=\frac{1}{3}\left(\frac{G{M}_{B}M}{{\left(1-x\right)}^2}\right)$ (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 $x=\frac{2}{3}m$