Question

A mass is at the centre of a square with four masses at the corners as shown.


Which of the following options hold true for the magnitude of gravitational force on the mass at centre of square is

• A. $F_A=F_B<F_C=F_D$
• B. $F_A<F_B<F_D<F_C$
• C. $F_A=F_B>F_C=F_D$
• D. None

Answer 1

The correct option is A $F_A=F_B<F_C=F_D$

The mass $M$ is placed at centre of the square, so its distance from all $4$ corners is equal.


In diagrams $\left(A\right)$ & $\left(B\right)$, the net force on central mass due to diagonal masses $5M$ and $2M$ gets cancelled respectively.


So, net force on central mass in $\left(A\right)$ and $\left(B\right)$ is given as

$F_{net}=\frac{G\left(3M\right)\left(M\right)}{r^2}-\frac{G\left(M\right)\left(M\right)}{r^2}$

$\Rightarrow F_{net}=\frac{2G{M}^2}{r^2}$

Similarly, in figures $\left(C\right)$ and $\left(D\right)$, the net force experienced due to diagonal masses $5M$ and $2M$ gets cancelled.


So, net force on central mass in $\left(c\right)$ and $\left(d\right)$ is given as

$F_{net}=\frac{F\left(3M\right)\left(2M\right)}{r^2}-\frac{G\left(2M\right)M}{r^2}$

$\Rightarrow F_{net}=\frac{4G{M}^2}{r^2}$

Thus,

$F_A=F_B<F_C=F_D$

Hence, option $\left(a\right)$ is correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{This problem intends to test your}\\ \text{application skill for superposition}\\ \text{principle along with observation of geometry.}\end{array}$$