Question

The dimension of dynamic viscosity is:

• A. \( M^1 L^{-1} T^{-2} \)
• B. \( M^1 L^{-1} T^{-1} \)
• C. \( M^0 L^2 T^{-1} \)
• D. \( M^0 L^0 T^0 \)

Hint:

Dynamic viscosity always has the dimension of stress divided by rate of strain, i.e. \( M L^{-1} T^{-1} \).

Answer 1

The Correct Option is B

Step 1: Recall formula of dynamic viscosity. Dynamic viscosity (\(\eta\)) is defined as: \[ \eta = \frac{\text{Shear Stress}}{\text{Velocity Gradient}} \]

Step 2: Dimension of shear stress. \[ \text{Shear stress} = \frac{\text{Force}}{\text{Area}} \] Force has dimension \( MLT^{-2} \). Area has dimension \( L^2 \). So, \[ [\text{Shear stress}] = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2} \]

Step 3: Dimension of velocity gradient. \[ \text{Velocity Gradient} = \frac{\text{Velocity}}{\text{Length}} \] Velocity has dimension \( L T^{-1} \). So, \[ [\text{Velocity Gradient}] = \frac{LT^{-1}}{L} = T^{-1} \]

Step 4: Dimension of viscosity. \[ [\eta] = \frac{M L^{-1} T^{-2}}{T^{-1}} = M L^{-1} T^{-1} \]

Final Answer: \[ \boxed{M^1 L^{-1} T^{-1}} \]