This question asks for the discharge ($Q$) in m³/s for a fluid undergoing laminar flow through a pipe. We are given the pipe's diameter and the maximum velocity along the centreline.

Understanding Laminar Pipe Flow

For laminar flow in a circular pipe, the velocity profile is parabolic. The velocity is zero at the pipe walls and maximum at the centreline. The relationship between the maximum velocity ($u_{max}$) and the average velocity ($u_{avg}$) across the pipe's cross-section is a key concept:

• Maximum Velocity ($u_{max}$): The highest velocity, found at the centreline of the pipe.
• Average Velocity ($u_{avg}$): The velocity that, when multiplied by the total flow area, gives the total discharge. For laminar flow, $u_{avg} = \frac{1}{2} u_{max}$.

Calculating Pipe Parameters

First, let's determine the radius and cross-sectional area of the pipe:

• Given Diameter, $D = 0.04 \text{ m}$.
• The radius, $R$, is half the diameter: $R = \frac{D}{2} = \frac{0.04 \text{ m}}{2} = 0.02 \text{ m}$.
• The cross-sectional area, $A$, is calculated using the formula for the area of a circle ($A = \pi R^2$): $A = \pi (0.02 \text{ m})^2 = \pi (0.0004) \text{ m}^2$.

Calculating Average Velocity

We are given the maximum velocity at the centreline:

• Maximum Velocity, $u_{max} = 1.5 \text{ m/s}$.

Using the relationship for laminar flow, we can find the average velocity:

• Average Velocity, $u_{avg} = \frac{1}{2} u_{max} = \frac{1}{2} \times 1.5 \text{ m/s} = 0.75 \text{ m/s}$.

Calculating Discharge

The discharge ($Q$) is the product of the cross-sectional area ($A$) and the average velocity ($u_{avg}$):

• Discharge, $Q = A \times u_{avg}$.
• Substituting the values we calculated: $Q = (0.0004\pi \text{ m}^2) \times (0.75 \text{ m/s})$.
• $Q = 0.0003 \pi \text{ m}^3/\text{s}$.

To match the options provided, we express the decimal $0.0003$ as a fraction:

• $0.0003 = \frac{3}{10000}$.
• Therefore, the discharge is: $Q = \frac{3\pi}{10000} \text{ m}^3/\text{s}$.

This result matches one of the options provided.