A liquid whose coefficient of viscosity is ` eta ` flows on a horizontal surface. Let ` dx ` represent the vertical distance between two layers of liquid and dv represent the difference in the velocities of the two layers. Then the quantity ` eta (dv//dx )` has the same dimensions as:

Options

A. Acceleration
B. Force
C. Momentum
D. Pressure

Explanation

To solve the problem, we need to analyze the quantity \( \eta \left( \frac{dv}{dx} \right) \) and determine its dimensions.

### Step-by-Step Solution:

1. **Understanding the Terms**:
- \( \eta \) is the coefficient of viscosity, which is a measure of a fluid's resistance to flow.
- \( dv \) represents the difference in velocities between two layers of the liquid.
- \( dx \) represents the vertical distance between those two layers.

2. **Velocity Gradient**:
- The term \( \frac{dv}{dx} \) represents the velocity gradient, which is the change in velocity per unit distance.

3. **Dimensions of Velocity**:
- Velocity \( v \) has dimensions of \( [L T^{-1}] \), where \( L \) is length and \( T \) is time.

4. **Dimensions of Distance**:
- The distance \( dx \) has dimensions of \( [L] \).

5. **Calculating Dimensions of Velocity Gradient**:
- The dimensions of the velocity gradient \( \frac{dv}{dx} \) can be calculated as follows:
\[ \left[ \frac{dv}{dx} \right] = \frac{[L T^{-1}]}{[L]} = [T^{-1}] \]

6. **Dimensions of Coefficient of Viscosity**:
- The coefficient of viscosity \( \eta \) is defined as the ratio of shear stress to the velocity gradient. The dimensions of shear stress are:
\[ \text{Shear Stress} = \frac{\text{Force}}{\text{Area}} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]
- Therefore, the dimensions of viscosity \( \eta \) can be expressed as:
\[ [\eta] = \frac{[\text{Shear Stress}]}{[\text{Velocity Gradient}]} = \frac{[M L^{-1} T^{-2}]}{[T^{-1}]} = [M L^{-1} T^{-1}] \]

7. **Combining the Dimensions**:
- Now, we can find the dimensions of the quantity \( \eta \left( \frac{dv}{dx} \right) \):
\[ \left[ \eta \left( \frac{dv}{dx} \right) \right] = [M L^{-1} T^{-1}] \cdot [T^{-1}] = [M L^{-1} T^{-2}] \]

8. **Identifying the Result**:
- The dimensions \( [M L^{-1} T^{-2}] \) correspond to pressure, as pressure is defined as force per unit area.

### Conclusion:
Thus, the quantity \( \eta \left( \frac{dv}{dx} \right) \) has the same dimensions as pressure.

### Final Answer:
The quantity \( \eta \left( \frac{dv}{dx} \right) \) has the same dimensions as pressure.

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