When a liquid flows in a tube, there is relative motion between adjacent layers of the liquid. This force is called the viscous force which tends to oppose the relative motion between the layers of the liquid. Newton was the first person to study the factors that govern the viscous force in a liquid. According to Newton’s law of viscous flow, the magnitude of the viscous force on a certain layer of a liquid is given by
` F = - eta A (dv)/(dx)`
where A is the area of the layer ` (dv)/(dx) ` is the velocity gradient at the layer and ` eta ` is the coefficient of viscosity of the liquid.
The dimensional formula for the coefficient of viscosity is :

Options

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

Explanation

To find the dimensional formula for the coefficient of viscosity (η), we start with the equation given in the question:

\[ F = -\eta A \frac{dv}{dx} \]

Where:
- \( F \) is the viscous force,
- \( A \) is the area,
- \( \frac{dv}{dx} \) is the velocity gradient,
- \( \eta \) is the coefficient of viscosity.

### Step 1: Rearranging the equation
We can rearrange the equation to solve for η:

\[ \eta = \frac{F}{A \frac{dv}{dx}} \]

### Step 2: Identify the dimensions of each term
1. **Force (F)**: The dimensional formula for force is given by Newton's second law, \( F = ma \):
- Mass (m) has the dimension [M].
- Acceleration (a) has the dimension [L T\(^{-2}\)].
- Therefore, the dimensional formula for force is:
\[ [F] = [M][L][T^{-2}] = [M L T^{-2}] \]

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

3. **Velocity gradient (\( \frac{dv}{dx} \))**:
- The velocity (v) has the dimension [L T\(^{-1}\)].
- The distance (x) has the dimension [L].
- Therefore, the velocity gradient is:
\[ \frac{dv}{dx} = \frac{[L T^{-1}]}{[L]} = [T^{-1}] \]

### Step 3: Substitute the dimensions back into the equation for η
Now substituting the dimensions into the equation for η:

\[ \eta = \frac{[M L T^{-2}]}{[L^2] \cdot [T^{-1}]} \]

### Step 4: Simplifying the expression
Now we simplify the expression:

\[ \eta = \frac{[M L T^{-2}]}{[L^2 T^{-1}]} = [M L T^{-2}] \cdot [L^{-2} T] \]

This simplifies to:

\[ \eta = [M L^{-1} T^{-1}] \]

### Conclusion
Thus, the dimensional formula for the coefficient of viscosity (η) is:

\[ \eta = [M^1 L^{-1} T^{-1}] \]

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

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