Question

\(M, L, T\) correspond to dimensions representing mass, length and time, respectively. What is the dimension of viscosity?

• A. \(M^1 L^{-2} T^{-1}\)
• B. \(M^1 L^{-1} T^{-1}\)
• C. \(M^1 L^{-1} T^{1}\)
• D. \(M^1 L^{-2} T^{-2}\)

Hint:

- Dynamic viscosity always has the dimension \(ML^{-1}T^{-1}\).
- Kinematic viscosity has dimension \(L^2 T^{-1}\).

Answer 1

The Correct Option is B

Step 1: Viscosity (dynamic viscosity \(\eta\)) is defined from Newton’s law of viscosity: \[ \tau = \eta \frac{du}{dy} \] where \(\tau\) is shear stress, \(du/dy\) is velocity gradient.
Step 2: Shear stress is force per area: \[ \tau = \frac{F}{A} \] Dimension of force \(F = MLT^{-2}\).
Area \(A = L^2\).
So, shear stress dimension = \(\frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}\).
Step 3: Velocity gradient = \(\frac{du}{dy}\).
Velocity \(u = LT^{-1}\). Dividing by length \(L\): \[ \frac{du}{dy} = T^{-1} \] Step 4: From \(\tau = \eta \cdot (du/dy)\): \[ \eta = \frac{\tau}{du/dy} = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1} \] \[ \boxed{ML^{-1}T^{-1}} \]