Question

Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

• A. 1
• B. 3
• C. \( 2\sqrt{5} \)
• D. 2

Hint:

In gravitational force problems with symmetry, always resolve forces into perpendicular components and use vector addition.

Answer 1

The Correct Option is D

Each corner of the square is at the same distance from the center. Hence, gravitational force due to each mass is proportional to the mass itself.
Step 1: Resolve forces along perpendicular axes.
Let the side of the square be \( a \). Distance from center to each corner: \[ r = \frac{a}{\sqrt{2}}. \] Step 2: Case I — Original configuration.
Horizontal component: \[ (3M - M) \] Vertical component: \[ (4M - 2M) \] \[ F_1 \propto \sqrt{(2M)^2 + (2M)^2} = 2M\sqrt{2}. \] Step 3: Case II — After interchanging \(3M\) and \(4M\).
Horizontal component: \[ (4M - M) \] Vertical component: \[ (3M - 2M) \] \[ F_2 \propto \sqrt{(3M)^2 + (M)^2} = M\sqrt{10}. \] Step 4: Compute ratio.
\[ \frac{F_1}{F_2} = \frac{2M\sqrt{2}}{M\sqrt{10}} = \frac{2}{\sqrt{5}}. \] Thus, \[ \alpha = 2. \] Final Answer: \[ \boxed{2} \]