Which of the following represents the correct dimensions of the coefficient of viscosity ?

The correct option is [ML⁻¹T⁻¹].

To find the dimensions of the coefficient of viscosity ($\eta$), we can use Newton's formula for viscous force:
$F=-\eta A\frac{dv}{dx}$

Where:
• $F$ is the viscous force
• $A$ is the area of the contact surface
• $\frac{dv}{dx}$ is the velocity gradient (change in velocity per unit distance)

Rearranging the formula to solve for the coefficient of viscosity ($\eta$), we get:
$\eta =\frac{F}{A\left(\frac{dv}{dx}\right)}$

Now, let's substitute the dimensions of each quantity:
• Dimensions of Force ($F$) = $\left[{\text{MLT}}^{-2}\right]$
• Dimensions of Area ($A$) = $\left[{\text{L}}^2\right]$
• Dimensions of Velocity Gradient ($\frac{dv}{dx}$) = $\frac{\left[{\text{LT}}^{-1}\right]}{\left[\text{L}\right]}=\left[{\text{T}}^{-1}\right]$

Substituting these dimensional formulas back into the rearranged equation:
$\left[\eta \right]=\frac{\left[{\text{MLT}}^{-2}\right]}{\left[{\text{L}}^2\right]\left[{\text{T}}^{-1}\right]}$

Simplifying the expression:
$\left[\eta \right]=\left[\text{M}\right]·\left[{\text{L}}^{1-2}\right]·\left[{\text{T}}^{-2-(-1)}\right]$
$\left[\eta \right]=\left[{\text{ML}}^{-1}{\text{T}}^{-1}\right]$

Therefore, the dimensional formula of the coefficient of viscosity is $\left[{\text{ML}}^{-1}{\text{T}}^{-1}\right]$.