Four particles with mass m, 2m, 3m & 4m are placed at the four corners of a square of edge ‘a’. The gravitational force acting on a particle of mass m placed at the center is

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4 \sqrt{2} \frac{{Gm}^{2}}{a^{2}}

2 \frac{{Gm}^{2}}{a^{2}}

\frac{{Gm}^{2}}{a^{2}}

3 \frac{{Gm}^{2}}{a^{2}}

answer is D.

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Detailed Solution

F_{1} = 2 \frac{{Gm}^{2}}{a^{2}}

;

F_{2} = 4 \frac{{Gm}^{2}}{a^{2}}

$F_{3} = 6 \frac{{Gm}^{2}}{a^{2}}$ $F_{4} = 8 \frac{{Gm}^{2}}{a^{2}}$

$F = \sqrt{{(F_{3} - F_{1})}^{2} + {(F_{4} - F_{2})}^{2}}$ $(θ = 90 °)$

$F = \frac{{Gm}^{2}}{a^{2}} \sqrt{{(6 - 2)}^{2} + {(8 - 4)}^{2}}$

$= \frac{{Gm}^{2}}{a^{2}} \sqrt{4^{2} + 4^{2}}$

$F = 4 \sqrt{2} \frac{{Gm}^{2}}{a^{2}}$

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