Correct Answer - A::B::C::D
`M=Ni(OAxxAB)`
`=Ni(OAxxOC)`
`=(100)(1.2)[(0.4hatj)xx(0.03cos30^@hati+0.3sin30^@hatk)]`
`=(7.2hati-12.47hatk)A-m^2`
`tau=MxxB`
`=[(7.2hati-12.47hatk)xx(0.8hati)]`
`=(-9.98hatj)N-m`
`:. |tau|=9.98N-m`
Torque vector and expected direction of rotation is shown in figure.