Question

If initially the force on \(m_0\) is \(F_0\). When the positions of \(4m\) and \(3m\) are interchanged, the force becomes \(F'\). If \[ \frac{F_0}{F'}=\frac{\alpha}{\sqrt{5}}, \] find \(\alpha\).

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For force problems involving symmetry:
Resolve forces into perpendicular components
Use proportionality to avoid unnecessary constants
Only differences in opposite forces contribute to the net force

Answer 1

The Correct Option is B

Concept:
The gravitational force on a mass due to several surrounding masses is the vector sum of individual forces. Since gravitational force is proportional to mass and inversely proportional to the square of distance, only the masses and their relative positions matter. \[ F = G\frac{m_0 m}{r^2} \]
Step 1: Identify forces acting on \(m_0\). From the figure, four masses are placed at the corners of a square of side \(L\): \[ 4m,\; 3m,\; m,\; 2m \] The distance from the center \(m_0\) to each corner is: \[ r=\frac{L}{\sqrt{2}} \] Hence force due to a mass \(km\) is: \[ F_k = G\frac{m_0 (km)}{(L/\sqrt{2})^2} = \frac{2Gm_0 m}{L^2}k \]
Step 2: Initial force \(F_0\). Resolve forces along \(x\) and \(y\) axes. Initially, opposite corners contain \(4m\) and \(2m\) (vertical pair), and \(3m\) and \(m\) (horizontal pair). Net components: \[ F_{x0} \propto (3m - m) = 2m \] \[ F_{y0} \propto (4m - 2m) = 2m \] Thus: \[ F_0 \propto \sqrt{(2m)^2+(2m)^2}=2m\sqrt{2} \]
Step 3: Final force \(F'\) after interchange of \(4m\) and \(3m\). Now vertical pair is \(3m\) and \(2m\), horizontal pair is \(4m\) and \(m\). Net components: \[ F_{x'} \propto (4m - m)=3m \] \[ F_{y'} \propto (3m - 2m)=m \] Thus: \[ F' \propto \sqrt{(3m)^2+(m)^2}=m\sqrt{10} \]
Step 4: Compute the ratio. \[ \frac{F_0}{F'}=\frac{2m\sqrt{2}}{m\sqrt{10}} =\frac{2\sqrt{2}}{\sqrt{10}} =\frac{2}{\sqrt{5}} \] Comparing with: \[ \frac{F_0}{F'}=\frac{\alpha}{\sqrt{5}} \] \[ \alpha = 2 \] \[ \boxed{\alpha=2} \]