Use Newton's law of viscosity

The viscous force is given by $F = \eta \, A \, \frac{dv}{dx}$ where:

• $F$ = force
• $\eta$ = coefficient of viscosity
• $A$ = area
• $\dfrac{dv}{dx}$ = velocity gradient

Write dimensions of each quantity

• Force: $[F] = [MLT^{-2}]$
• Area: $[A] = [L^2]$
• Velocity gradient: $\left[\frac{dv}{dx}\right] = \frac{[LT^{-1}]}{[L]} = [T^{-1}]$

Find dimensions of $\eta$

From $F = \eta A \frac{dv}{dx}$ we get $[\eta] = \frac{[F]}{[A]\,[dv/dx]}$

Substituting: $[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]}$

$[\eta] = [ML^{-1}T^{-1}]$

Match with options

• A: $ML^{-1}T^{-1}$ ✅
• B: $MLT^{-1}$
• C: $ML^{-1}T^{-2}$
• D: $ML^{-2}T^{-2}$

Final answer

The correct dimensions of coefficient of viscosity are $\boxed{ML^{-1}T^{-1}}$ So, the correct option is A.