1. Set up the square

Let the side of the square be $a$.

Choose one corner mass, say at $A$. The other three masses are at:

• adjacent corner $B$
• adjacent corner $D$
• opposite corner $C$

We need the net gravitational force on the mass at $A$ due to the other three masses.

2. Forces due to adjacent corner masses

Each adjacent mass is at distance $a$, so each exerts force of magnitude $F_1 = \frac{Gm^2}{a^2}.$

These two forces are perpendicular to each other. Their resultant is $F_{BD} = \sqrt{F_1^2+F_1^2} = \sqrt{2}\,F_1 = \sqrt{2}\,\frac{Gm^2}{a^2}.$

This resultant acts along the diagonal toward the center/opposite corner.

3. Force due to opposite corner mass

The opposite corner is at distance $a\sqrt{2}.$

So the force due to that mass is $F_C = \frac{Gm^2}{(a\sqrt{2})^2} = \frac{Gm^2}{2a^2}.$

This force is also along the same diagonal.

4. Net force on the chosen mass

Since both resultants are along the same diagonal, add them directly: $F_{\text{net}} = \sqrt{2}\,\frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2}.$

So, $F_{\text{net}} = \left(\sqrt{2}+\frac12\right)\frac{Gm^2}{a^2}.$

Given that $F_{\text{net}} = \left(\frac{2\sqrt{2}+1}{32}\right)\frac{Gm^2}{L^2}.$

Note that $\sqrt{2}+\frac12 = \frac{2\sqrt{2}+1}{2}.$

Thus, $\frac{2\sqrt{2}+1}{2a^2} = \frac{2\sqrt{2}+1}{32L^2}.$

Cancel $(2\sqrt{2}+1)$ from both sides: $\frac{1}{2a^2} = \frac{1}{32L^2}.$

Therefore, $2a^2 = 32L^2$ $a^2 = 16L^2$ $a = 4L.$

5. Check options

• A: $4L$ ✅
• B: $3L$ ❌
• C: $2L$ ❌
• D: $\frac{L}{2}$ ❌

So the correct answer is A.