Q58 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$.
- A$2 \sqrt{5}$
- B3
- C1
- Correct answerD2
Solution
Let $k = \frac{G}{r^2\sqrt{2}}$ where $r = a/\sqrt{2}$.
Net field at center has components along $x$ and $y$.
Config 1 (original):
$g_x = k(-M+2M+3M-4M) = 0$,
$g_y = k(-M-2M+3M+4M) = 4kM$.
$F_1 = 4kM$.
Config 2 (3M and 4M swapped):
$g_x = k(-M+2M+4M-3M) = 2kM$,
$g_y = k(-M-2M+4M+3M) = 4kM$.
$F_2 = kM\sqrt{4+16} = 2\sqrt{5}\,kM$.
$\frac{F_1}{F_2} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \Rightarrow \alpha = 2$.
Net field at center has components along $x$ and $y$.
Config 1 (original):
$g_x = k(-M+2M+3M-4M) = 0$,
$g_y = k(-M-2M+3M+4M) = 4kM$.
$F_1 = 4kM$.
Config 2 (3M and 4M swapped):
$g_x = k(-M+2M+4M-3M) = 2kM$,
$g_y = k(-M-2M+4M+3M) = 4kM$.
$F_2 = kM\sqrt{4+16} = 2\sqrt{5}\,kM$.
$\frac{F_1}{F_2} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \Rightarrow \alpha = 2$.
Hence, the correct answer is option: 'D'.