Net gravitational force at the centre of a square is found to be \( F_{1} \) when four particles having mass 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 \( \frac{F_{1}}{F_{2}} \) is \( \frac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is __________ .

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Answer & Solution

Correct Answer

(A) 2

Step-by-step Solution

Detailed explanation

Initial configuration

\( F=2\sqrt{2}\frac{Gmm_{0}}{r^{2}} \)
New configuration

\( F'=\sqrt{10}\frac{Gmm_{0}}{r^{2}}\Rightarrow\frac{F}{F'}=2\sqrt{2}.\frac{1}{\sqrt{10}}=\frac{2}{\sqrt{5}} \)
\( \alpha=2 \)

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