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.
A river is 5 m deep. The velocity of water on its surface is ` 2 ms^(-1) ` If the coefficient of viscosity of water is ` 10 ^(-3 ) Nsm ^(-2)` , the viscous force per unit area is :

Options

A. (a) ` 10 ^(-4) Nm^(-2) `
B. (b)` 2 xx 10 ^(-4) Nm^ (-2) `
C. (c)` 4 xx 10 ^(-4) Nm^(-2)`
D. (d)` 5 xx 10 ^(-4) Nm^(-2) `.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Depth of the river (h) = 5 m
- Velocity of water at the surface (V) = 2 m/s
- Coefficient of viscosity of water (η) = \(10^{-3}\) N·s/m²

### Step 2: Determine the velocity gradient (dv/dx)
In a river, the velocity of water decreases from the surface to the bottom. The velocity at the bottom layer (in contact with the riverbed) is zero because of the no-slip condition.

The velocity gradient (dv/dx) can be calculated as:
\[ \text{dv/dx} = \frac{V - 0}{h} = \frac{2 \, \text{m/s} - 0}{5 \, \text{m}} = \frac{2}{5} \, \text{s}^{-1} \]

### Step 3: Apply Newton's law of viscous flow
According to Newton's law of viscous flow, the viscous force per unit area (F/A) is given by:
\[ \frac{F}{A} = -\eta \frac{dv}{dx} \]
Substituting the values we have:
\[ \frac{F}{A} = -\left(10^{-3} \, \text{N·s/m}^2\right) \left(\frac{2}{5} \, \text{s}^{-1}\right) \]

### Step 4: Calculate the viscous force per unit area
Now calculate the above expression:
\[ \frac{F}{A} = -10^{-3} \times \frac{2}{5} = -\frac{2 \times 10^{-3}}{5} = -0.0004 \, \text{N/m}^2 \]
This can also be expressed as:
\[ \frac{F}{A} = -4 \times 10^{-4} \, \text{N/m}^2 \]

### Step 5: Interpret the result
The negative sign indicates that the viscous force opposes the motion. Therefore, the magnitude of the viscous force per unit area is:
\[ \frac{F}{A} = 4 \times 10^{-4} \, \text{N/m}^2 \]

### Final Answer
The viscous force per unit area is \(4 \times 10^{-4} \, \text{N/m}^2\).

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