A

Step 1: Identify the key parameters. The problem mentions that the velocity profile is parabolic with a moving plate speed of $v_0$ and stationary plate below it. The thickness of the oil layer is $d$.
Step 2: For a parabolic velocity profile in a Newtonian fluid, shear stress ($\tau$) can be expressed as:
$$\tau = \mu \frac{du}{dy}$$
where $\mu$ is the dynamic viscosity and $\frac{du}{dy}$ is the velocity gradient. For a parabolic profile, the velocity at a distance $y$ from the moving plate can be expressed as:
$$u(y) = \frac{4v_0}{d^2}(d - y)y$$
Step 3: Calculate the velocity gradient for $y = 0$:
$$\frac{du}{dy} = \frac{4v_0}{d^2}(d - 2y)$$
At $y = 0$, this becomes:
$$\frac{du}{dy} = \frac{4v_0}{d^2}(d) = \frac{4v_0}{d}$$
Step 4: Substitute this into the shear stress formula:
$$\tau = \mu \frac{4v_0}{d}$$
This indicates that the shear stress on the moving plate when the flow profile is parabolic depends linearly on the velocity $v_0$ and inversely on $d$.
Step 5: For a linear velocity profile (assuming a Newtonian fluid), the velocity profile is given by:
$$u(y) = v_0 \frac{y}{d}$$
Step 6: The velocity gradient at the moving plate is:
$$\frac{du}{dy} = \frac{v_0}{d}$$
Step 7: Substitute this into the shear stress formula:
$$\tau_{linear} = \mu \frac{v_0}{d}$$
Conclusion: Therefore, the shear stress on the upper plate increases with the speed of the plate, and the specific expressions for shear stress in both scenarios have been derived.