Three identical particles A, B and C of mass 100 kg each are placed… | JEE Main 2022 (25 Jul, Shift 1)

Q13JEE Main 2022 · 25 JUL · SHIFT 1Difficult

Newton's Law of Gravitation

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:

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Answer: (b)

B pulls $P$ from $13$ m: $F_B = \frac{G(100)^2}{169} \approx 59.2G$ .
A and C are at $13\sqrt{2}$ m: each $\frac{G(100)^2}{338} \approx 29.6G$ , at $45^\circ$ to $PB$ .
Sideways parts cancel; along $PB$ : $2 \times 29.6G \cos45^\circ \approx 41.8G$ .
$F \approx 59.2G + 41.8G \approx 101G$
So the answer is (b) $100G$ .

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