Question

Two masses $4\text{g}$ and $12\text{g}$, are placed horizontally at a distance of $12\text{cm}$ from each other. Another mass of $0.5\text{g}$ is placed in between them such that there is no net force acting on it. Find the distance (in $\text{cm}$) of the third mass from the mass of $4\text{g}$.

• A. $-6+6\sqrt{3}\text{cm}$
• B. $+6+6\sqrt{3}\text{cm}$
• C. $+6-6\sqrt{3}\text{cm}$
• D. $6\sqrt{3}\text{cm}$

Answer 1

The correct option is A $-6+6\sqrt{3}\text{cm}$

Given:
Mass of the body 1, ${\text{m}}_1=4\text{g}$
Mass of the body 2, ${\text{m}}_2=12\text{g}$
Mass of the body 3, ${\text{m}}_3=0.5\text{g}$
Let the third mass is placed at a distance $\text{x}\text{cm}$ from ${\text{m}}_1$ such that there is no net force acting on it.
If ${\text{r}}_{13}$ is the distance between ${\text{m}}_1$ and ${\text{m}}_3$, and ${\text{r}}_{23}$ is the distance between ${\text{m}}_2$ and ${\text{m}}_3$, then for the net force on ${\text{m}}_3$ to be zero:
${\text{F}}_{13}={\text{F}}_{23}$
$\frac{\text{G}{\text{m}}_1{\text{m}}_3}{{\text{r}}_{13}^2}=\frac{\text{G}{\text{m}}_2{\text{m}}_3}{{\text{r}}_{23}^2}$
$\frac{{\text{m}}_1}{{\text{r}}_{13}^2}=\frac{{\text{m}}_2}{{\text{r}}_{23}^2}$
$\frac{{\text{m}}_1}{{\text{m}}_2}=\frac{{\text{r}}_{13}^2}{{\text{r}}_{23}^2}$
$⟹\frac{4}{12}=\frac{{\text{x}}^2}{(12-\text{x}{)}^2}$
$3{\text{x}}^2={\text{x}}^2-24\text{x}+144$
$2{\text{x}}^2+24\text{x}-144=0$
${\text{x}}^2+12\text{x}-72=0$
$⟹\text{x}=-6\pm 6\sqrt{3}$
But, $\text{x}>0$. So,
$\text{x}=-6+6\sqrt{3}\text{cm}$