Answer:
[tex]2.5 \times 10^{11} N-m[/tex] upwards
Explanation:
Torque is the vector cross product of the force and radial distance.
[tex]\tau = rF sin \theta[/tex]
[tex]\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m[/tex]
The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.
