Two bodies of masses 8 ,kg are placed at the | vedclass

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(B) The gravitational force between two masses is given by $F = \frac{G m_1 m_2}{r^2}$.Given $m_A = m_B = 8 \,kg$, $m_G = 2 \,kg$, and $AG = BG = 1 \,

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At a point $P$ on the line $CG$ such that $PG=\frac{1}{\sqrt{2}} \,m$

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(B) The gravitational force between two masses is given by $F = \frac{G m_1 m_2}{r^2}$.Given $m_A = m_B = 8 \,kg$, $m_G = 2 \,kg$, and $AG = BG = 1 \,m$.The force exerted by mass at $A$ on mass at $G$ is $F_A = \frac{G 	imes 8 	imes 2}{1^2} = 16G$.The force exerted by mass at $B$ on mass at $G$ is $F_B = \frac{G 	imes 8 	imes 2}{1^2} = 16G$.The angle between $F_A$ and $F_B$ is $120^{\circ}$.The resultant force $F_{AB}$ is given by $F_{AB} = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos 120^{\circ}}$.Since $F_A = F_B = 16G$, $F_{AB} = \sqrt{(16G)^2 + (16G)^2 + 2(16G)(16G)(-0.5)} = \sqrt{3(16G)^2 - (16G)^2} = 16G$.This resultant force $F_{AB}$ acts along the line $GC$ directed towards $C$.To make the net force on the $2 \,kg$ body zero, a fourth body of mass $m_C = 4 \,kg$ must be placed at a distance $x$ from $G$ along the line $GC$ such that the gravitational force $F_C$ exerted by it balances $F_{AB}$.$F_C = \frac{G 	imes 4 	imes 2}{x^2} = 16G$.$\frac{8G}{x^2} = 16G \Rightarrow x^2 = \frac{8}{16} = 0.5$.$x = \frac{1}{\sqrt{2}} \,m$.Thus, the fourth body should be placed at a point $P$ on the line $CG$ such that $PG = \frac{1}{\sqrt{2}} \,m$.

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