KCET 2021 Physics Solved Paper

According to the question, the arrangement of the masses is as shown below, Gravitational force between two masses is given as $F=\frac{G m_{1} m_{2}}{r^{2}}$ where, $G=$ gravitational constant and $r=$ distance between them. From the given values, we can say that force between masses at A and G. $=$ Force between masses at B and G ⇒ $F_{A}=\frac{G m_{A} m_{G}}{AG^{2}}$ and $F_{B}=\frac{G m_{B} m_{G}}{BG^{2}}$ or $F_{A}=F_{B}=\frac{G \times 8 \times 2}{1^{2}}=16 G$ ...(i) $[\because$ Given, $m_{1}=m_{A}=m_{B}=8 \text{ kg}, m_{2}=m_{G}=2 \text{ kg},$,$AG=BG=r=1 \text{ m}]$ From the figure, resultant of $F_{A}$ and $F_{B}$ is given as $F_{AB}=\sqrt{F_{A}^{2}+F_{B}^{2}+2 F_{A} F_{B} \cos \theta}$ $=\sqrt{2 F_{A}^{2}+2 F_{A}^{2} \cos 120^{\circ}}$ $=\sqrt{F_{A}^{2}}=\sqrt{(26 G)^{2}}=16 G$ [from Eq. (i)] ⇒$F_{AB}=F_{A}=F_{B}=16 G$ For resultant force on $2 \text{ kg}$ body to be zero, $4 \text{ kg}$ body should be placed at a certain distance from G such that $F_{CG}=-F_{AE}$ ⇒ $|F_{CG}|=|F_{AB}|$ or $\frac{G \times m_{C} m_{G}}{x^{2}}=G(16)$ where, $x=$ distance between $4 \text{ kg}$ and $2 \text{ kg}$ body. Here, $m_{C}=4 \text{ kg}$ ⇒ $\frac{G \times 4 \times 2}{x^{2}}=G(16)$ ⇒ $x^{2}=\frac{1}{2}$ ⇒ $x=\frac{1}{\sqrt{2}} \text{ m}$ This means, $4 \text{ kg}$ body should be placed at point P on line CG such that $PG=\frac{1}{\sqrt{2}} \text{ m}$.