The coefficient of viscosity has the dimensional formula

Options

A. a. `[ML T^(-2)]`
B. b. `[M L^(2) T^(-2)]`
C. c. `[ML^(0) T^(-2)]`
D. d. `[ML ^(-1) T^(-1)]`

Answer

To find the dimensional formula for the coefficient of viscosity, we can follow these steps:

### Step 1: Understand the definition of viscosity
The coefficient of viscosity (η) is defined as the ratio of shear stress to the shear rate (velocity gradient). Mathematically, it can be expressed as:
\[ \eta = \frac{\text{Shear Stress}}{\text{Shear Rate}} \]

### Step 2: Define shear stress
Shear stress (τ) is defined as the force (F) applied per unit area (A):
\[ \tau = \frac{F}{A} \]
The dimensional formula for force (F) is:
\[ [F] = MLT^{-2} \]
The dimensional formula for area (A) is:
\[ [A] = L^2 \]
Thus, the dimensional formula for shear stress (τ) becomes:
\[ [\tau] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

### Step 3: Define shear rate
Shear rate is defined as the velocity gradient, which is the change in velocity (v) with respect to distance (x):
\[ \text{Shear Rate} = \frac{dv}{dx} \]
The dimensional formula for velocity (v) is:
\[ [v] = LT^{-1} \]
The dimensional formula for distance (x) is:
\[ [x] = L \]
Thus, the dimensional formula for shear rate becomes:
\[ \left[\frac{dv}{dx}\right] = \frac{[v]}{[x]} = \frac{LT^{-1}}{L} = T^{-1} \]

### Step 4: Combine the dimensions to find viscosity
Now, substituting the dimensions of shear stress and shear rate into the formula for viscosity:
\[ [\eta] = \frac{[\tau]}{\left[\frac{dv}{dx}\right]} = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1} \]

### Conclusion
Thus, the dimensional formula for the coefficient of viscosity is:
\[ \eta = ML^{-1}T^{-1} \]

### Final Answer
The correct option for the dimensional formula of the coefficient of viscosity is \( ML^{-1}T^{-1} \).

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