A square Lamina OABC of length 10 cm is pivoted at 'O'. Forces act on Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is:

For the lamina to remain stationary, the net torque about the pivot O must be zero:

$$\sum \tau_O = 0$$

$$\tau=\text{Force}\times\text{Perpendicular distance}$$

$$At\ Point\ A:The\ vertical\ 10\text{ N}\ (down)\ creates\ a\ clockwise\ torque:\ \tau_1=-10\cdot L$$

$$At\ Point\ B:The\ vertical\ 10\text{ N }(up)\ creates\ a\ counter\ clockwise\ torque:\ \tau_2=+10\cdot L$$

$$The\ horizontal\ 10\text{ N}\ (right)\ creates\ a\ clockwise\ torque:\ \tau_3=-10\cdot L$$

$$At\ Point\ C:The\ horizontal\ force\ F\ (left)\ creates\ a\ counter\ clockwise\ torque:\ \tau_4=+F\cdot L$$

$$The\ vertical\ force\ passes\ through\ O,\ so\ its\ torque\ is\ 0.$$

$$\sum \tau_O = (-10 \cdot L) + (10 \cdot L) + (-10 \cdot L) + (F \cdot L) = 0$$

$$-10L + FL = 0$$

$$F = 10 \text{ N}$$