Two small particles of mass m each are placed at the vertices A and B of a right angle isoceles triangle right angled at C. If AB=l, find the gravitational field strength at C

Gravitational field strength $\left(E_G\right)$:
It is the force acting on a unit mass at the given point in a gravitational field (of another mass).
           $E_G=\frac{Gm}{d^2}$
Here, G is universal gravitational constant, m is mass of the source of gravitational field, 
         d is distance between the given point and source of the gravitational field.

Length of the sides of right angled isosceles triangle:


From Pythagoras theorem,

       $$\begin{gathered} d^2+d^2=L^2 \\ 2d^2=L^2 \\ d^2=\frac{L^2}{2} \end{gathered}$$

Gravitational field strength due to mass at point C due to mass at point A:

                   $$\begin{gathered} E_{G1}=\frac{Gm}{d^2} \\ =\frac{2Gm}{L^2} \end{gathered}$$

Direction: Towards A (Upward)

Gravitational field strength due to mass at point C due to mass at point B:

                   $$\begin{gathered} E_{G2}=\frac{Gm}{d^2} \\ =\frac{2Gm}{L^2} \end{gathered}$$

Direction: Towards B (Right side)

Net gravitational field strength at point C:

The angle between $E_{G1}$ and $E_{G2}$ is 90 degrees.

Hence, net gravitational field strength is,  

           $$\begin{gathered} E_G=\sqrt{{\left(E_{G1}\right)}^2+{\left(E_{G2}\right)}^2} \\ =\sqrt{2}\left(\frac{2Gm}{L^2}\right) \\ =\frac{2\sqrt{2}Gm}{L^2} \end{gathered}$$

Answer:
Gravitational field strength at point C is, ${\mathit{E}}_{\mathbf{G}}\mathbf{=}\frac{\mathbf{2}\sqrt{\mathbf{2}}\mathbf{G}\mathbf{m}}{{\mathbf{L}}^{\mathbf{2}}}$