Interpret the figure and geometry

Let the square lamina have side length $a=10\text{ cm}=0.1\text{ m}.$

It is pivoted at the corner $O$. Since the lamina remains stationary, the net torque about the pivot $O$ must be zero.

Key idea: torque of a force about the pivot

Torque magnitude is $\tau = F \times d_\perp,$ where $d_\perp$ is the perpendicular distance from the pivot to the line of action of the force.

Use the given force configuration

From the figure, the shown forces produce opposite torques about $O$.

• One force has a moment arm of side length $10\text{ cm}$.
• The force $F$ acts along a direction such that its effective perpendicular distance from $O$ is also obtained from the square geometry.

Balancing torques about $O$ gives $\tau_{\text{clockwise}}=\tau_{\text{anticlockwise}}.$

On evaluating the moments from the given arrangement, we get $F\times 10 = 100$ in consistent units, hence $F=10\text{ N}.$

Check options

• A: $10\text{ N}$ ✅
• B: $0$ ❌
• C: $10\sqrt2\text{ N}$ ❌
• D: $20\text{ N}$ ❌

Final answer

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