To solve this problem, let's break it down step-by-step using fluid dynamics principles, specifically Bernoulli's equation and the continuity equation.

Given Data and Observations:

1. Fluid: Water, with density $\rho$.
2. Point 1 (top of the cistern): Diameter is much larger than the pipe, implying $v_1 \approx 0$.
3. Height ($h$) from point 1 to point 2: 1.25 m.
4. Point 2: Open to the atmosphere, so $p_2 = p_{atm}$.
5. Diameters:
   • Section before point 2: $8 \text{ cm}$ (radius $r_2 = 0.04 \text{ m}$)
   • Section at point 3: $6 \text{ cm}$ (radius $r_3 = 0.03 \text{ m}$)

Calculations:

Let's approach each part as stated:

(a) Velocity at Point 2, $v_2$

Using Bernoulli’s equation between points 1 and 2: $p_1 + \frac{1}{2} \rho v_1^2 + \rho g h = p_2 + \frac{1}{2} \rho v_2^2$ Since $v_1 \approx 0$ and $p_1 = p_2 = p_{atm}$, the equation simplifies to: $\rho g h = \frac{1}{2} \rho v_2^2$ Solving for $v_2$: $v_2 = \sqrt{2gh} = \sqrt{2 \cdot 9.81 \, \text{m/s}^2 \cdot 1.25 \, \text{m}}$ $v_2 \approx 4.95 \, \text{m/s}$

(b) Flow Rate $Q$

The flow rate $Q$ is given by: $Q = A_2 v_2$ where $A_2$ is the cross-sectional area at point 2: $A_2 = \pi r_2^2 = \pi (0.04 \, \text{m})^2$ $A_2 \approx 0.005024 \, \text{m}^2$ Now, calculate $Q$: $Q = 0.005024 \, \text{m}^2 \times 4.95 \, \text{m/s}$ $Q \approx 0.0249 \, \text{m}^3/\text{s} \approx 0.025 \, \text{m}^3/\text{s}$

(c) Velocity at Point 3, $v_3$

Using the continuity equation $A_2 v_2 = A_3 v_3$: $v_3 = \frac{A_2 v_2}{A_3}$ where $A_3 = \pi r_3^2 = \pi (0.03 \, \text{m})^2$: $A_3 \approx 0.002827 \, \text{m}^2$ $v_3 = \frac{0.005024 \, \text{m}^2 \times 4.95 \, \text{m/s}}{0.002827 \, \text{m}^2}$ $v_3 \approx 8.8 \, \text{m/s}$

(d) Pressure at Point 3, $p_3$

Applying Bernoulli’s equation between points 2 and 3: $p_2 + \frac{1}{2} \rho v_2^2 = p_3 + \frac{1}{2} \rho v_3^2$ Since $p_2 = p_{atm}$, solve for $p_3$: $p_3 = p_2 + \frac{1}{2} \rho (v_2^2 - v_3^2)$ Substitute $v_2 = 4.95 \, \text{m/s}$ and $v_3 = 8.8 \, \text{m/s}$: $p_3 = p_{atm} + \frac{1}{2} \rho (4.95^2 - 8.8^2)$ Assuming $\rho = 1000 \, \text{kg/m}^3$ for water: $p_3 = p_{atm} + \frac{1}{2} \cdot 1000 \, \text{kg/m}^3 \cdot (-58.9475 \, \text{m}^2/\text{s}^2)$ $p_3 = p_{atm} - 29473.75 \, \text{Pa}$ Convert to atm ($1 \, \text{atm} \approx 101325 \, \text{Pa}$): $p_3 \approx 0.75 \, \text{atm}$

Summary of Results:

1. Velocity at Point 2, $v_2$: $4.95 \, \text{m/s}$
2. Flow Rate $Q$: $0.025 \, \text{m}^3/\text{s}$
3. Velocity at Point 3, $v_3$: $8.8 \, \text{m/s}$
4. Pressure at Point 3, $p_3$: $0.75 \, \text{atm}$

Would you like more details on any of the steps or have any questions?

Here are five related questions for further exploration:

1. How would the results change if the height $h$ was increased?
2. What would be the effect on $v_3$ if the radius at point 3 was altered?
3. How does Bernoulli’s equation apply if the fluid was not incompressible?
4. What would happen to the flow rate if the fluid density changed?
5. How would the pressure at point 3 change if it was not open to the atmosphere?

Tip: Remember that in fluid dynamics, using Bernoulli's principle requires assumptions of steady, incompressible, and non-viscous flow. Always check if these assumptions hold in practical situations.