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 $\frac{1}{3}$ rd of the force between $B$ and $C$. Then the distance of $C$ from $A$ is

Question

Dr. Abubacker · Accepted 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
[image:8_image73.emf]
Here, $\frac{M_{A}}{M_{B}} = \frac{4}{3}$, Force between $A$ and $C$ is
$F_{AC} = \frac{GMM_{A}}{x^{2}}$ …(i)
Force between $B$ and $C$ is
$F_{BC} = \frac{GMM_{B}}{(1 - x)^{2}}$ …(ii)
According to given problem $F_{AC} = \frac{1}{3}F_{BC}$
$	herefore\frac{GM_{A}M}{x^{2}} = \frac{1}{3}\left( \frac{GM_{B}M}{(1 - x)^{2}} ight)$ [Using (i) and (ii)]
$\frac{M_{A}}{x^{2}} = \frac{M_{B}}{3(1 - x)^{2}}\ or\ \frac{M_{A}}{M_{B}} = \frac{x^{2}}{3(1 - x)^{2}}$
$\Rightarrow \frac{4}{3} = \frac{x^{2}}{3(1 - x)^{2}}\ or\ 4 = \frac{x^{2}}{(1 - x)^{2}}$
Or $2 = \frac{x}{1 - x}$ or $2 - 2x = x$
$3x = 2$ or $x = \frac{2}{3}m$

Q. 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 $\frac{1}{3}$ rd of the force between $B$ and $C$. Then the distance of $C$ from $A$ is

Explanation

Reviewed by Dr. Abubacker

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