JEE Physics 2025 Question with Solution

Given: A square lamina $OABC$ of side $10 \, \text{cm} = 0.1 \, \text{m}$ is pivoted at $O$ and remains stationary.

Find: The magnitude of $F$.

For rotational equilibrium, the sum of torques about the pivot $O$ must be zero.

Take counter-clockwise torque as positive and clockwise torque as negative.

From the figure and the extracted working:

• At $A$, the horizontal $10 \, \text{N}$ force has line of action through $O$, so its torque is zero.
• At $A$, the downward $10 \, \text{N}$ force gives clockwise torque:

• At $B$, the upward $10 \, \text{N}$ force gives counter-clockwise torque:

• At $B$, the rightward $10 \, \text{N}$ force gives clockwise torque:

• At $C$, the force $F$ acts leftward and gives counter-clockwise torque:

Applying equilibrium:

Therefore, the magnitude of the force is $10 \, \text{N}$. The correct option is C.

Given: Side of the square is $0.1 \, \text{m}$ and pivot is at $O$.

Find: Force $F$ such that the lamina remains in rotational equilibrium.

Use lever arm about $O$ for each force. Any force whose line of action passes through $O$ produces zero torque.

Clockwise torques:

Counter-clockwise torques:

Equating clockwise and counter-clockwise torques:

Hence, the required force is $10 \, \text{N}$.