If mass M, area A and velocity V are chosen as fundamental units, then the dimension of coefficient of viscosity will be:

Options

A. `MV//A`
B. `MA//V`
C. `MVA`
D. `M//VA`

Explanation

To find the dimension of the coefficient of viscosity when mass (M), area (A), and velocity (V) are chosen as fundamental units, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Coefficient of Viscosity (η)**:
The coefficient of viscosity is defined as the ratio of shear stress to shear rate. Mathematically, it can be expressed as:
\[ \eta = \frac{\text{Force}}{\text{Area} \times \text{Velocity Gradient}} \]

2. **Express Shear Stress**:
Shear stress is given by:
\[ \text{Shear Stress} = \frac{\text{Force}}{\text{Area}} \]
Therefore, we can rewrite the coefficient of viscosity as:
\[ \eta = \frac{\text{Force}}{\text{Area} \times \frac{dv}{dx}} \]

3. **Substitute Force**:
Force (F) can be expressed in terms of mass (M) and acceleration (a):
\[ F = M \cdot a \]
Acceleration (a) can be expressed as the change in velocity (dv) over time (dt):
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ F = M \cdot \frac{dv}{dt} \]

4. **Substituting into the Viscosity Equation**:
Now substituting the expression for force into the viscosity equation:
\[ \eta = \frac{M \cdot \frac{dv}{dt}}{\text{Area} \times \frac{dv}{dx}} \]

5. **Simplifying the Expression**:
The velocity gradient \(\frac{dv}{dx}\) can be rearranged:
\[ \eta = \frac{M \cdot \frac{dv}{dt}}{A \cdot \frac{dv}{dx}} = \frac{M}{A} \cdot \frac{dx}{dt} \]
Here, \(\frac{dx}{dt}\) is the velocity (V).

6. **Final Expression for Viscosity**:
Therefore, we can express the coefficient of viscosity as:
\[ \eta = \frac{M}{A} \cdot V \]

7. **Writing Dimensions**:
Now, we can express the dimensions of the coefficient of viscosity:
\[ [\eta] = M \cdot A^{-1} \cdot V \]

### Conclusion:
The dimension of the coefficient of viscosity when mass (M), area (A), and velocity (V) are chosen as fundamental units is:
\[ [\eta] = M \cdot A^{-1} \cdot V \]