Q.7The dimensions of coefficient of viscosity are _______.
(A) ML-1T-1
(B) ML-1T-2
(C) ML-2T-2
(D) ML-2T-1
The dimensions of the coefficient of viscosity are [M L^{-1} T^{-1}], making option (A) correct.
Question Solution
The coefficient of viscosity (η) arises from Newton’s law of viscosity: η = F / (A × (dv/dx)), where F is tangential force, A is area, and dv/dx is velocity gradient (dimensions [L T^{-1} / L] = [T^{-1}]).
Force F has dimensions [M L T^{-2}], area A has [L^2], and velocity gradient has [T^{-1}]. Substituting yields η = [M L T^{-2}] / ([L^2] [T^{-1}]) = [M L^{-1} T^{-1}].
Option Analysis
-
(A) [M L^{-1} T^{-1}]: Matches derived formula exactly, as confirmed across physics references.
-
(B) [M L^{-1} T^{-2}]: Resembles stress ([M L^{-1} T^{-2}]), but lacks the T^{-1} from velocity gradient inverse.
-
(C) [M L^{-2} T^{-2}]: Matches pressure dimensions, irrelevant to viscosity which involves velocity rate.
-
(D) [M L^{-2} T^{-1}]: Incorrect; L^{-2} might mimic surface tension per length, but viscosity requires L^{-1}.
In fluid mechanics, understanding the dimensions of coefficient of viscosity is essential for competitive exams like CSIR NET, JEE, and NEET. The coefficient of viscosity (η) quantifies a fluid’s resistance to flow, with dimensions of coefficient of viscosity given by [M¹ L⁻¹ T⁻¹].
Derivation Steps
Start with η = (F × dx) / (A × v), where dx is layer separation.
-
[F] = [M L T⁻²]
-
[dx] = [L]
-
[A] = [L²]
-
[v] = [L T⁻¹]
Thus, [η] = [M L T⁻² × L] / [L² × L T⁻¹] = [M L⁻¹ T⁻¹].
Common Exam MCQs
| Option | Dimensions | Why Incorrect? |
|---|---|---|
| A | [ML⁻¹T⁻¹] | Correct, matches derivation |
| B | [ML⁻¹T⁻²] | Stress dimensions, misses T factor |
| C | [ML⁻²T⁻²] | Pressure, no velocity gradient |
| D | [ML⁻²T⁻¹] | Unrelated to standard formula |
This dimensions of coefficient of viscosity [ML⁻¹T⁻¹] helps verify formulas like Poiseuille’s law in pipes.
