# Q. A uniform chain of mass M and length L is hanging over a rough horizontal table as shown in the figure. Find the torque of the gravitational force above the corner of the table. Assume the chain...

Q. A uniform chain of mass M and length L is hanging over a rough horizontal table as shown in the figure. Find the torque of the gravitational force above the corner of the table. Assume the chain to be thin.

A) zero

B) `(MgL)/16`

C) `(MgL)/8`

D) `(MgL)/4`

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Torque refers to the rotation due to a force. It is the cross product of the displacement from the pivot and the force.

If the angle between force and the displacement vector is `theta` , the resulting torque is equal to `|F|*|d|*sin theta` .

In the given figure, half the chain lies above the table and the rest is hanging down. The torque due to gravitational force acting on the portion that is hanging down is 0. This is due to the fact that the force and the displacement vector are in the same direction.

The center of gravity of the portion lying above the table is at a distance `L/4` from the corner of the table. The force acting downwards is equal to `(M/2)*g` . This gives the torque as `(L/4)(M/2)*g*sin 90` = `(M*L*g)/8`

**The correct answer is option C.**