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 determine the dimensions of the quantity \( \eta \left( \frac{dv}{dx} \right) \) where \( \eta \) is the coefficient of viscosity, \( dv \) is the difference in velocities between two layers of liquid, and \( dx \) is the vertical distance between those layers.

### Step-by-Step Solution:

1. **Identify the dimensions of the coefficient of viscosity \( \eta \)**:
- The coefficient of viscosity \( \eta \) is defined as the ratio of shear stress to the shear rate.
- Shear stress has dimensions of pressure, which is given by:
\[ \text{Dimensions of pressure} = \frac{\text{Force}}{\text{Area}} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]
- Shear rate is defined as the velocity gradient, which has dimensions of:
\[ \text{Dimensions of shear rate} = \frac{\text{Velocity}}{\text{Distance}} = \frac{L T^{-1}}{L} = T^{-1} \]
- Therefore, the dimensions of viscosity \( \eta \) can be calculated as:
\[ \eta = \frac{\text{Shear stress}}{\text{Shear rate}} = \frac{M L^{-1} T^{-2}}{T^{-1}} = M L^{-1} T^{-1} \]

2. **Determine the dimensions of \( \frac{dv}{dx} \)**:
- Here, \( dv \) represents the difference in velocities, which has dimensions of:
\[ \text{Dimensions of velocity} = L T^{-1} \]
- The distance \( dx \) has dimensions of:
\[ \text{Dimensions of distance} = L \]
- Therefore, the dimensions of \( \frac{dv}{dx} \) are:
\[ \frac{dv}{dx} = \frac{L T^{-1}}{L} = T^{-1} \]

3. **Combine the dimensions of \( \eta \) and \( \frac{dv}{dx} \)**:
- Now we can find the dimensions of the product \( \eta \left( \frac{dv}{dx} \right) \):
\[ \text{Dimensions of } \eta \left( \frac{dv}{dx} \right) = \left( M L^{-1} T^{-1} \right) \left( T^{-1} \right) = M L^{-1} T^{-2} \]

4. **Compare with known dimensions**:
- The dimensions \( M L^{-1} T^{-2} \) correspond to the dimensions of pressure.

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