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Three identical particle A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately :

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42 G

21 G

100 G

59 G

answer is B.

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Detailed Solution

$\begin{array}{l} F = \frac{GMM}{r^{2}} + \sqrt{2} \frac{GMM}{(\sqrt{2} r)^{2}} \\ = \frac{GMM}{r^{2}} (1 + \frac{1}{\sqrt{2}}) \\ = \frac{G \times 10^{4}}{13^{2}} (1 + \frac{1}{\sqrt{2}}) \\ F ≃ 100 G \end{array}$

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