The dimensional formula for coefficient of viscosity is

Options

A. `ML^(2)T^(-2)`
B. `ML^(-1)T^(-1)`
C. `MLT^(-2)`
D. `M^(2)L^(-2)T^(-1)`

Answer

To find the dimensional formula for the coefficient of viscosity, we can start from its definition. The coefficient of viscosity (η) can be expressed using the formula:

\[ \eta = \frac{F \cdot T}{A \cdot V} \]

Where:
- \( F \) is the force,
- \( T \) is the time,
- \( A \) is the area,
- \( V \) is the velocity.

### Step 1: Identify the dimensions of each variable
1. **Force (F)**: The dimensional formula for force is given by Newton's second law:
\[ F = m \cdot a \]
Where \( m \) is mass (M) and \( a \) is acceleration (L/T²). Therefore, the dimension of force is:
\[ [F] = M \cdot \left(\frac{L}{T^2}\right) = M L T^{-2} \]

2. **Time (T)**: The dimension of time is:
\[ [T] = T \]

3. **Area (A)**: The area is given by length squared:
\[ [A] = L^2 \]

4. **Velocity (V)**: The velocity is defined as length per time:
\[ [V] = \frac{L}{T} \]

### Step 2: Substitute the dimensions into the viscosity formula
Now we substitute these dimensions into the formula for the coefficient of viscosity:

\[ \eta = \frac{F \cdot T}{A \cdot V} = \frac{(M L T^{-2}) \cdot T}{L^2 \cdot \left(\frac{L}{T}\right)} \]

### Step 3: Simplify the expression
Now, we simplify the expression step by step:

1. Substitute the dimensions:
\[ \eta = \frac{M L T^{-2} \cdot T}{L^2 \cdot \frac{L}{T}} = \frac{M L T^{-1}}{L^2 \cdot \frac{L}{T}} \]

2. Simplify the denominator:
\[ L^2 \cdot \frac{L}{T} = \frac{L^3}{T} \]

3. Now, substitute this back into the equation:
\[ \eta = \frac{M L T^{-1}}{\frac{L^3}{T}} = \frac{M L T^{-1} \cdot T}{L^3} = \frac{M L^2}{L^3} \cdot T^{-1} \]

4. Finally, simplify:
\[ \eta = M L^{-1} T^{-1} \]

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