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 centre of the square will be

The correct option is 4√2Gm²/a².

Here is the step-by-step derivation to find the gravitational force acting on the particle at the center of the square:

Step 1: Understand the Geometry of the System
Let the four corners of the square of side a be labeled in sequence as A, B, C, and D. The masses placed at these corners are:
- At A: $m_A=m$
- At B: $m_B=2m$
- At C: $m_C=3m$
- At D: $m_D=4m$

Let a particle of mass $m$ be placed at the center of the square, denoted as O.
The distance from the center O to any corner of the square is half the length of the diagonal.
Since the diagonal of a square of side a is $\sqrt{2}a$, the distance $r$ of each corner from the center is:
$r=\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$

Squaring both sides, we get:
$r^2=\frac{a^2}{2}$

Step 2: Calculate the Gravitational Forces due to each corner mass
Using Newton's law of gravitation, the force between the mass at the center and a corner mass $M$ is given by:
$F=\frac{GMm}{r^2}=\frac{GMm}{a^2/2}=\frac{2GMm}{a^2}$

Let us calculate the individual force magnitudes and their directions:
- Force due to corner A (directed towards A):
$F_A=\frac{2G\left(m\right)m}{a^2}=\frac{2G{m}^2}{a^2}$

- Force due to corner B (directed towards B):
$F_B=\frac{2G\left(2m\right)m}{a^2}=\frac{4G{m}^2}{a^2}$

- Force due to corner C (directed towards C):
$F_C=\frac{2G\left(3m\right)m}{a^2}=\frac{6G{m}^2}{a^2}$

- Force due to corner D (directed towards D):
$F_D=\frac{2G\left(4m\right)m}{a^2}=\frac{8G{m}^2}{a^2}$

Step 3: Resolve the Forces along the Diagonals
The forces acting along the diagonal AC are opposite in direction. The net force along diagonal AC is:
$F_{AC}=F_C-F_A=\frac{6G{m}^2}{a^2}-\frac{2G{m}^2}{a^2}=\frac{4G{m}^2}{a^2}$ (directed towards C)

Similarly, the forces acting along the diagonal BD are opposite in direction. The net force along diagonal BD is:
$F_{BD}=F_D-F_B=\frac{8G{m}^2}{a^2}-\frac{4G{m}^2}{a^2}=\frac{4G{m}^2}{a^2}$ (directed towards D)

Step 4: Find the Resultant Net Force
Since the diagonals of a square intersect at a right angle (90°), the vectors $F_{AC}$ and $F_{BD}$ are perpendicular to each other.
The magnitude of the resultant force $F_{\text{net}}$ is:
$F_{\text{net}}=\sqrt{F_{AC}^2+F_{BD}^2}$

Substituting the values:
$F_{\text{net}}=\sqrt{{\left(\frac{4G{m}^2}{a^2}\right)}^2+{\left(\frac{4G{m}^2}{a^2}\right)}^2}$

$F_{\text{net}}=\sqrt{2\cdot {\left(\frac{4G{m}^2}{a^2}\right)}^2}=\sqrt{2}\cdot \frac{4G{m}^2}{a^2}=\frac{4\sqrt{2}G{m}^2}{a^2}$

Thus, the magnitude of the net gravitational force on the particle at the center of the square is 4√2Gm²/a².