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

Options

A. (a)`MV//A'`
B. (b)`MA'//V`
C. (c)`MVA'`
D. (d)`M//VA'`

Explanation

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

### Step 1: Understand the formula for viscosity
The coefficient of viscosity (η) is defined as the ratio of shear stress to shear rate. Mathematically, it can be expressed as:
\[ \eta = \frac{F}{A \cdot \frac{dv}{dx}} \]
where:
- \( F \) is the force applied,
- \( A \) is the area over which the force is applied,
- \( \frac{dv}{dx} \) is the velocity gradient.

### Step 2: Express force in terms of fundamental units
Force (F) can be expressed using Newton's second law:
\[ F = m \cdot a \]
where:
- \( m \) is mass,
- \( a \) is acceleration.

Acceleration can be expressed as:
\[ a = \frac{dv}{dt} \]
Thus, we can rewrite force as:
\[ F = m \cdot \frac{dv}{dt} \]

### Step 3: Substitute force into the viscosity formula
Substituting the expression for force into the viscosity formula gives:
\[ \eta = \frac{m \cdot \frac{dv}{dt}}{A \cdot \frac{dv}{dx}} \]

### Step 4: Simplify the expression
Rearranging the equation, we get:
\[ \eta = \frac{m \cdot dv}{A \cdot dt \cdot \frac{1}{dx}} = \frac{m \cdot dx}{A \cdot dt} \]

### Step 5: Identify dimensions of each term
Now, we need to express the dimensions of each term:
- Mass \( [M] \) has dimension \( M \).
- Area \( [A] \) has dimension \( L^2 \).
- Velocity \( [V] \) has dimension \( L/T \).

### Step 6: Substitute dimensions into the viscosity formula
Substituting the dimensions into the equation gives:
\[ \eta = \frac{M \cdot L}{L^2 \cdot T} = \frac{M}{L \cdot T} \]

### Step 7: Final dimension of viscosity
Thus, the dimension of the coefficient of viscosity (η) is:
\[ \eta = M \cdot L^{-1} \cdot T^{-1} \]

### Conclusion
The dimension of the coefficient of viscosity when mass, area, and velocity are chosen as fundamental units is:
\[ [M^1 L^{-1} T^{-1}] \]