To determine which pairs have equivalent dimensions, we examine the respective dimensional formulas.

For work (W) specified as $ \overrightarrow{F} \cdot \overrightarrow{s} $:

$ W = F s \cos \theta $

The dimensions are:

$ W = [MLT^{-2}][L] = [ML^2T^{-2}] $

For torque ($\overrightarrow{\tau}$) given as $ \overrightarrow{r} \times \overrightarrow{F} $:

$ \tau = rF \sin \theta $

The dimensions are:

$ \tau = [L] [MLT^{-2}] = [ML^2T^{-2}] $

Therefore, the dimensions of torque and work are identical.