A liquid is flowing through a horizontal channel. The speed of flow `(v)` depends on height `(y)` from the floor as `v = v_(0)[2((y)/(h))-((y)/(h))^(2)]`. Where `h` is the height of liquid in the channel and `v_(0)` is the speed of the top layer. Coefficient of viscosity is `eta`. Then the shear stress that the liquid exerts on the floor is.
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Options

A. `(2etav_(0))/(h)`
B. `(3etav_(0))/(h)`
C. `(4etav_(0))/(h)`
D. `(2etav_(0))/(3h)`

Explanation

To find the shear stress that the liquid exerts on the floor of a horizontal channel, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Velocity Profile**:
The speed of flow \( v \) is given by the equation:
\[ v = v_0 \left( 2 \frac{y}{h} - \left( \frac{y}{h} \right)^2 \right) \]
where \( v_0 \) is the speed of the top layer, \( h \) is the height of the liquid in the channel, and \( y \) is the height from the floor.

2. **Calculate the Velocity Gradient**:
The velocity gradient (which is the change in velocity with respect to height) is given by:
\[ \frac{dv}{dy} \]
To find this, we differentiate \( v \) with respect to \( y \):
\[ \frac{dv}{dy} = \frac{d}{dy} \left( v_0 \left( 2 \frac{y}{h} - \left( \frac{y}{h} \right)^2 \right) \right) \]
Using the product and chain rules, we get:
\[ \frac{dv}{dy} = v_0 \left( \frac{2}{h} - \frac{2y}{h^2} \right) \]

3. **Evaluate the Velocity Gradient at the Floor**:
At the floor, where \( y = 0 \):
\[ \frac{dv}{dy} \bigg|_{y=0} = v_0 \left( \frac{2}{h} - 0 \right) = \frac{2v_0}{h} \]

4. **Use the Shear Stress Formula**:
The shear stress \( \tau \) exerted by the liquid on the floor can be calculated using the formula:
\[ \tau = \eta \frac{dv}{dy} \]
Substituting the value of the velocity gradient we found:
\[ \tau = \eta \left( \frac{2v_0}{h} \right) \]

5. **Final Expression for Shear Stress**:
Thus, the shear stress that the liquid exerts on the floor is:
\[ \tau = \frac{2 \eta v_0}{h} \]

### Conclusion:
The shear stress exerted by the liquid on the floor of the channel is given by:
\[ \tau = \frac{2 \eta v_0}{h} \]