Question

Four particles of masses $m,2m,3m$ and $4m$ are kept in sequence at the corners of a square of side $a$. The magnitude of gravitational force acting on a particle of mass $m$ placed at the center of the square will be

• A. $\frac{6m^{2}G}{a^2}$
• B. Zero
• C. $\frac{24m^{2}G}{a^2}$
• D. $\frac{4\sqrt{2}G{m}^2}{a^2}$

Answer 1

The correct option is D $\frac{4\sqrt{2}G{m}^2}{a^2}$

If two particles of mass $m$ are placed $x$ distance apart then force of attraction
$\frac{Gmm}{x^2}=F$ (Let)
Particle of mass $m$ is placed at the centre $\left(P\right)$ of square.
Force at point $P$ due to particle $A$,

$F_{PA}=\frac{Gmm}{x^2}=F$

similarly, $F_{PB}=\frac{G\left(2m\right)m}{x^2}=2F$

$F_{PC}=\frac{G\left(3m\right)m}{x^2}=3F$

And $F_{PD}=\frac{G\left(4m\right)m}{x^2}=4F$

Hence, the net force on $P$,
${\overrightarrow{F}}_{net}=F_{PA}+F_{PB}+F_{PC}+F_{PD}=2\sqrt{2}F$

${\overrightarrow{F}}_{\text{net}}=2\sqrt{2}\frac{Gmm}{x^2}$
Here, $x=\frac{a}{\sqrt{2}}$ (half of the diagonal of the square)
${\overrightarrow{F}}_{\text{net}}=2\sqrt{2}\frac{G{m}^2}{(a/\sqrt{2}{)}^2}$

$=\frac{4\sqrt{2}G{m}^2}{a^2}$