A

Step 1: Understand the Velocity Profile
The velocity profile is given as:
$$v = \frac{\beta}{4\mu}\left(\frac{d^2}{4} - r^2\right)$$
where \(v\) is the velocity at a radial distance \(r\) from the centerline, \(\beta\) is a constant, \(\mu\) is the viscosity of water, \(d\) is the diameter of the pipe.

Step 2: Shear Stress at the Wall
The shear stress \(\tau\) at any point in a fluid can be calculated using the formula:
$$\tau = \mu \frac{dv}{dr}$$
We first need to calculate \(\frac{dv}{dr}\).

Step 3: Differentiate the Velocity Profile
Differentiate the velocity equation with respect to \(r\):
$$\frac{dv}{dr} = \frac{\beta}{4\mu} \cdot (-2r) = -\frac{\beta r}{2\mu}$$

Step 4: Calculate Shear Stress at the Wall (r = d/2)
At the wall of the pipe, \(r = \frac{d}{2}\):
$$\frac{dv}{dr}\bigg|_{r=\frac{d}{2}} = -\frac{\beta (\frac{d}{2})}{2\mu} = -\frac{\beta d}{4\mu}$$
Now substituting into the shear stress equation:
$$\tau_w = \mu \left(-\frac{\beta d}{4\mu}\right) = -\frac{\beta d}{4}$$

Step 5: Calculate Shear Stress at r = d/4
At \(r = \frac{d}{4}\):
$$\frac{dv}{dr}\bigg|_{r=\frac{d}{4}} = -\frac{\beta (\frac{d}{4})}{2\mu} = -\frac{\beta d}{8\mu}$$
Now substituting into the shear stress equation:
$$\tau_{d/4} = \mu \left(-\frac{\beta d}{8\mu}\right) = -\frac{\beta d}{8}$$

Step 6: Calculate the Drag on the Pipe
The drag force is given by:
$$F_d = \tau_w \cdot A$$
where \(A\) is the surface area of the pipe segment. The surface area of the pipe for length \(L\) is:
$$A = \pi d L$$
Substituting the shear stress of the wall:
$$F_d = \left(-\frac{\beta d}{4}\right) \cdot \pi d L$$
Therefore the drag induced on the pipe by the water in the direction of flow over distance \(L\) is:
$$F_d = -\frac{\beta \pi d^2 L}{4}$$
Since drag is a force in the direction of flow, we can present the answer in magnitude.

Final Summary
Shear Stress at the wall: \(\tau_w = -\frac{\beta d}{4}\), Shear Stress at \(r = \frac{d}{4}\): \(\tau_{d/4} = -\frac{\beta d}{8}\), and Drag:
$$F_d = -\frac{\beta \pi d^2 L}{4}$$
Thus, the answers are provided for the shear stresses and the drag.