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

To find the value of \( \alpha \), we need to determine the net gravitational force at the center of the square when masses are placed at its corners.

Let's assume the side length of the square is \( a \). The center of the square will be equidistant from all four corners, and this distance can be calculated using the Pythagorean theorem:

\(d = \frac{a}{\sqrt{2}}\)

The gravitational force between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

\(F = \frac{G m_1 m_2}{r^2}\)

Here, the gravitational constant \( G \) can be considered uniform, and \(m_0\) is the mass at the center.

Case 1: Original Placement

The masses \( M, 2M, 3M, 4M \) are placed at the corners as shown:

Net force in the first case is the vector sum of all forces due to these masses:

\(F_1 = \frac{Gm_0}{(d^2)}(M + 2M + 3M + 4M) = \frac{Gm_0(10M)}{(\frac{a^2}{2})} = \frac{20Gm_0M}{a^2}\)

Case 2: 3M and 4M Interchanged

By interchanging \( 3M \) and \( 4M \), we have:

\(F_2 = \frac{Gm_0}{(d^2)}(M + 2M + 4M + 3M) = \frac{Gm_0(10M)}{(\frac{a^2}{2})} = \frac{20Gm_0M}{a^2}\)

Calculating the Ratio

Since the net forces \( F_1 \) and \( F_2 \) are equal, their ratio is:

\(\frac{F_1}{F_2} = \frac{20Gm_0M}{a^2} \div \frac{20Gm_0M}{a^2} = 1\)

This implies that:

\(\frac{F_1}{F_2} = \frac{\alpha}{\sqrt{5}}\)

Therefore, \( \alpha = \sqrt{5} \). The value rounded to an integer is:

Answer: The value of \( \alpha \) is 2.