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Q.
Two particles each of mass 'm' are placed at $A$ and $C$ are such $AB=BC$ . The gravitational force on the third particle placed at $D$ at a - distance $L$ on the perpendicular bisector of the line $AC$ is
Since $AB = BC = DB = L$, we have $AD =\sqrt{2} L$
Now force between particles placed at A and D is: $F = Gmm /(\sqrt{2} L )^{2}=$ $Gm ^{2} / 2 L ^{2}$
Similar force particle at C will exert on particle at D.
So total force on particle $D$ will be ${ }_{2} F$.
Now the resultant component of these two forces along DB will be: $2 Fcos 45^{\circ}$
Since $AB = BC = BD = L$ and $\angle DBC =\angle DBA =90^{\circ}$, we get $\angle ADB =$ $\angle BDC =45^{\circ}$
So, the resultant force on mass $m$ at $D$ will be $\frac{ Gm ^{2}}{\sqrt{2} L ^{2}}$