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

Three identical particles A, B, and C each of mass $$100 \text{ kg}$$ are placed in a straight line with $$AB = BC = 13 \text{ m}$$. Particle P of the same mass is placed at $$13 \text{ m}$$ from B on the perpendicular bisector of AC.

We place B at the origin, A at $$(-13, 0)$$, C at $$(13, 0)$$, and P at $$(0, 13)$$.

The distances from P to each particle are found to be $$BP = 13 \text{ m}$$, $$AP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$, and $$CP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$.

The gravitational force exerted on P by B is given by $$F_B = \frac{G \times 100 \times 100}{13^2} = \frac{10000G}{169}$$, which acts along BP (vertically upward in this setup).

The gravitational force exerted on P by A is $$F_A = \frac{G \times 100 \times 100}{(13\sqrt{2})^2} = \frac{10000G}{338}$$.

Since AP makes an angle of $$45°$$ with BP, the component of this force along BP is $$F_A \cos 45° = \frac{10000G}{338} \times \frac{1}{\sqrt{2}} = \frac{10000G}{338\sqrt{2}}$$.

By symmetry, the force from C has the same component along BP, and the horizontal components of forces from A and C cancel out, so the net force along BP is: $$F = F_B + 2F_A\cos 45°$$ $$F = \frac{10000G}{169} + 2 \times \frac{10000G}{338\sqrt{2}}$$ $$F = \frac{10000G}{169} + \frac{10000G}{169\sqrt{2}}$$ $$F = \frac{10000G}{169}\left(1 + \frac{1}{\sqrt{2}}\right)$$ $$F = \frac{10000G}{169} \times 1.707$$ $$F \approx 59.2 \times 1.707 \times G$$ $$F \approx 101G \approx 100G$$

Hence, the correct answer is Option B.