Rods AB and CD, each of length 2 m, are joined together to form a cross. Four force acts on the ends A, B, C and D as shown in the figure. The magnitude of torque about the origin O due to these forces in N-m is equal to

The torque due to couple $F_A$  of and is $F_B$   clockwise with magnitude
 ${\tau }_{AB}=F{l}_{AB}=\left(2\right)\left(2\right)=4Nm.$
The torque due to couple of  $F_C$ and $F_D$  is also clockwise with magnitude
 ${\tau }_{CD}=F{l}_{CD}=\left(1\right)\left(2\right)=Nm.$
Thus, net torque about the origin O is  ${\tau }_O={\tau }_{AB}+{\tau }_{CD}=6Nm$.