The correct option is D 0.94
As we can see here, 20 N force which is passing through the center O will not produce any torque about point O
For 30 N force which is acting at 45∘ angle with x−axis, its x− component will pass through the origin, hence its horizontal component will also not cause any torque about O.
Let ⨂ represent vector going into the plane and
⨀ out of the plane
Consider ⨀ direction as positive
Hence, torque because of 20 N force which is acting in tangential direction :
τ1=r⊥F=0.1×30=3 N-m ⨂
Torque produced by y component of 30 N force acting at angle 45∘:
τ2=r⊥F=0.05×30sin45∘
=0.05×30×1√2=1.5√2 Nm ⨀
Torque produced by 20 N force acting in negatve y direction:
τ3=r⊥×F=0.05×20
τ=1 Nm⨀
Hence, Net torque produced on O
τnet=τ1+τ2+τ3
τnet=−3+1.5√2+1
τnet=−0.94 Nm
or |τnet|=0.94 Nm