Dimensions [ML⁻¹T⁻¹] are related to

The correct option is coefficient of viscosity.

Let's derive the dimensional formulas for each of the options step-by-step to see which one matches the given dimensions ${\mathrm{ML}}^{-1}{T}^{-1}$.

1. Coefficient of Viscosity ($\eta$):
According to Newton's formula for viscous force:
$F=\eta A\frac{dv}{dx}$
where $F$ is the force, $A$ is the area, and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging the formula for the coefficient of viscosity ($\eta$):
$\eta =\frac{F}{A\cdot \frac{dv}{dx}}$
Now, substituting the dimensional formulas for the individual quantities:
• Force ($F$) = $\left[{\mathrm{MLT}}^{-2}\right]$
• Area ($A$) = $\left[L^2\right]$
• Velocity gradient ($\frac{dv}{dx}$) = $\frac{\left[{\mathrm{LT}}^{-1}\right]}{\left[L\right]}=\left[T^{-1}\right]$
Putting these into the equation for $\eta$:
$\left[\eta \right]=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[L^2\right]\left[T^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{T}^{-1}\right]$
This matches the dimensions given in the question.

2. Work:
Work = Force × Displacement
$\left[\text{Work}\right]=\left[{\mathrm{MLT}}^{-2}\right]\times \left[L\right]=\left[{\mathrm{ML}}^2{T}^{-2}\right]$

3. Torque:
Torque = Force × perpendicular distance
$\left[\text{Torque}\right]=\left[{\mathrm{MLT}}^{-2}\right]\times \left[L\right]=\left[{\mathrm{ML}}^2{T}^{-2}\right]$

4. Energy:
Energy has the same dimensions as work:
$\left[\text{Energy}\right]=\left[{\mathrm{ML}}^2{T}^{-2}\right]$

Thus, the dimensions $\left[{\mathrm{ML}}^{-1}{T}^{-1}\right]$ belong uniquely to the coefficient of viscosity.