Q 11-07-123JEE MainJEE Main 2022 (25 Jul, Shift 1)Medium
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
Answer: (B) $100G$
Force due to $B$ (distance 13 m): $\dfrac{G(100)^2}{169} = 59.2G$, along $PB$.
$A$ and $C$ are at $13\sqrt2\ \text{m}$ at $45^\circ$ on either side of $PB$; their perpendicular components cancel:
$$2\times\frac{G(100)^2}{338}\cos45^\circ = 2\times29.6G\times0.707 = 41.8G$$
$$F = 59.2G + 41.8G\approx100G$$
Solution by Sreeraj P, M.Sc Physics