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(A) Let the mass of each point mass be $m$. The gravitational force between two masses $m$ separated by distance $r$ is $F = \frac{Gm^2}{r^2}$.For arr

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$C, B$ and $A$

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(A) Let the mass of each point mass be $m$. The gravitational force between two masses $m$ separated by distance $r$ is $F = \frac{Gm^2}{r^2}$.For arrangement $(A)$:The point mass $1$ is at one end. The other two masses are at distances $x$ and $y$ from it. Since both forces act in the same direction,the net force is $F_A = \frac{Gm^2}{x^2} + \frac{Gm^2}{y^2} = Gm^2 \left( \frac{1}{x^2} + \frac{1}{y^2} \right)$.For arrangement $(B)$:The point mass $1$ is at the corner of a right-angled triangle. The other two masses are at distances $x$ and $y$ along the perpendicular axes. The forces are $F_x = \frac{Gm^2}{x^2}$ and $F_y = \frac{Gm^2}{y^2}$. The net force is $F_B = \sqrt{F_x^2 + F_y^2} = Gm^2 \sqrt{\frac{1}{x^4} + \frac{1}{y^4}}$.For arrangement $(C)$:The point mass $1$ is in the middle. The other two masses are at distances $x$ and $y$ on opposite sides. The forces act in opposite directions,so the net force is $F_C = \left| \frac{Gm^2}{x^2} - \frac{Gm^2}{y^2} \right| = Gm^2 \left| \frac{1}{x^2} - \frac{1}{y^2} \right|$.Comparing the magnitudes,$F_C < F_B < F_A$ (assuming $x < y$). Thus,the increasing order is $C, B, A$.

There are three arrangements of three point masses (each of mass $m$). In which of the following three arrangements are they arranged in order of increasing magnitude of the net gravitational force on the point mass marked as $1$?

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