Question

Two particles each of mass $m$ are placed at A and C are such $AB=BC=L$. The gravitational force on the third particle placed at D at a distance L on the perpendicular bisector of the line AC is :

• A. $\frac{G{m}^2}{\sqrt{2}{L}^2}$ along BD
• B. $\frac{G{m}^2}{\sqrt{2}{L}^2}$ along DB
• C. $\frac{G{m}^2}{L^2}$ along AC
• D. none of these

Answer 1

The correct option is B $\frac{G{m}^2}{\sqrt{2}{L}^2}$ along DB

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$$=G{m}^2/2L^2$
Similar force particle at C will exert on particle at D.
So total force on particle D will be 2F.
Now the resultant component of these two forces along DB will be: 2Fcos ${45}^{\circ }$
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 $\frac{G{m}^2}{\sqrt{2}{L}^2}$