tau$$ in a fluid is given by:

$$ \tau = \eta \frac{du}{dy} $$

where $$\eta$$ is the dynamic viscosity of the fluid, and $$\frac{du}{dy}$$ is the velocity gradient in the direction perpendicular to the flow.

Given the velocity profile:

$$ v = k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) $$

First, we need to calculate the velocity gradient $$\frac{dv}{dy}$$:

$$ \frac{dv}{dy} = \frac{d}{dy} \left( k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) \right) $$

Applying the differentiation, we get:

$$ \frac{dv}{dy} = k \left( \frac{d}{dy} \left( \frac{2 y^2}{a^2} \right) - \frac{d}{dy} \left( \frac{y^3}{a^3} \right) \right) $$

Differentiating term by term, we have:

$$ \frac{dv}{dy} = k \left( \frac{2 \cdot 2y}{a^2} - \frac{3y^2}{a^3} \right) = k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} \right) $$

We need to evaluate this gradient at $$y = a$$:

$$ \left. \frac{dv}{dy} \right|_{y=a} = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4}{a} - \frac{3}{a} \right) = k \left( \frac{4-3}{a} \right) = \frac{k}{a} $$

Now, using the given viscosity of the water, $$2 \eta$$, the shear stress $$\tau$$ is given by:

$$ \tau = 2 \eta \left( \frac{k}{a} \right) = \frac{2 \eta k}{a} $$

Therefore, the shear stress between layers of water at $$y=a$$ is:

$$ \tau = \frac{2 \eta k}{a} $$

The correct option is Option A:

$$ \frac{2 \eta k}{a} $$

Fluid Mechanics

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