\( 5.33 \times 10^{-10} \, \text{N} \)
\( 1.33 \times 10^{-9} \, \text{N} \)
The gravitational force is given by the formula:
\( F = \dfrac{G \cdot m_1 \cdot m_2}{r^2} \)
Substituting the values:
\( F = \dfrac{6.67 \times 10^{-11} \cdot 10 \cdot 20}{5^2} \)
\( F = \dfrac{6.67 \times 10^{-11} \cdot 200}{25} \)
\( F = \dfrac{1.334 \times 10^{-8}}{25} = 5.336 \times 10^{-10} \, \text{N} \)
Option 1: \( 5.33 \times 10^{-10} \, \text{N} \)
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 
