Which of the following is correct dimension for dynamics viscosity(μ)?
- AMLT⁻¹
- BML⁻¹T⁻¹Correct
- CML²T⁻¹
- DML⁻¹T⁻²
Solution
Concept
Dimensional formula of dynamic viscosity
Formula
NONE
Step-by-step solution
1. Dynamic viscosity (μ) is defined by Newton's law of viscosity as shear stress divided by velocity gradient: μ = τ / (du/dy). 2. The dimensions of shear stress (Force/Area) are M L⁻¹ T⁻², and the velocity gradient (du/dy) has dimensions T⁻¹. 3. Dividing the dimensions gives [μ] = (M L⁻¹ T⁻²) / (T⁻¹) = M L⁻¹ T⁻¹. Answer: B
Explanation
Dynamic viscosity is defined as the ratio of shear stress to the velocity gradient in a flowing fluid. Since shear stress has units of force per unit area and velocity gradient has units of inverse time, dynamic viscosity evaluates to units of mass divided by length and time. In dimensional terms, this converts to M L inverse T inverse.
Why the other options are wrong
A) MLT⁻¹ represents linear momentum. C) ML²T⁻¹ represents angular momentum or action. D) ML⁻¹T⁻² represents stress, pressure, or elastic modulus.
Shortcut
Dynamic viscosity unit is Pa·s or kg/(m·s) which gives dimensions ML⁻¹T⁻¹.
Common mistake
Confusing the dimensions of dynamic viscosity (ML⁻¹T⁻¹) with kinematic viscosity (L²T⁻¹) or pressure (ML⁻¹T⁻²).
Correct answer: B) ML⁻¹T⁻¹