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.
If f is the frictional force between a solid sliding over another solid, and F is the viscous force when a liquid layer slides over another, then :

Options

A. f is independent of the area of the solid sliding over another solid.
B. f depends on the relative velocity of one solid with respect to the other
C. F depends on the area of the layer of the liquid.
D. F is independent of the relative velocity between adjacent layers.

Explanation

To solve the problem, we need to analyze the relationship between the viscous force in liquids and the frictional force in solids, as described by Newton's law of viscous flow.

### Step-by-Step Solution:

1. **Understanding Viscous Force**:
- The viscous force \( F \) acting on a layer of liquid is given by the equation:
\[ F = -\eta A \frac{dv}{dx} \]
- Here, \( \eta \) is the coefficient of viscosity, \( A \) is the area of the layer, and \( \frac{dv}{dx} \) is the velocity gradient (the change in velocity over the change in distance).

2. **Understanding Frictional Force**:
- The frictional force \( f \) between two solid surfaces is given by:
\[ f = \mu N \]
- Where \( \mu \) is the coefficient of friction and \( N \) is the normal force (which is dependent on the weight of the object).

3. **Independence of Area**:
- The frictional force \( f \) is independent of the area of contact between the two solid surfaces. This means that increasing the area does not increase the frictional force, as it only depends on the normal force and the coefficient of friction.

4. **Dependence on Relative Velocity**:
- For solids, the frictional force does not significantly depend on the relative velocity of the sliding surfaces until a certain limit is reached. Thus, it can be considered independent of relative velocity for small velocities.

5. **Dependence of Viscous Force on Area**:
- The viscous force \( F \) does depend on the area \( A \) between the layers of the liquid. This means that increasing the area will increase the viscous force.

6. **Dependence of Viscous Force on Relative Velocity**:
- The viscous force also depends on the velocity gradient \( \frac{dv}{dx} \), which is a measure of the relative velocity between adjacent layers of the liquid.

### Conclusion:
- Based on the analysis:
- The statement that the frictional force \( f \) is independent of the area of solid sliding over another solid is **true**.
- The statement that the frictional force \( f \) depends on the relative velocity of the solids is **incorrect** for small velocities.
- The statement that the viscous force \( F \) depends on the area of the layers of the liquid is **true**.
- The statement that the viscous force \( F \) is independent of the relative velocity between adjacent layers is **incorrect**.

Thus, the correct options are:
- \( f \) is independent of the area of solid sliding over another solid (True).
- \( F \) depends on the area of layers of the liquid (True).