Step 1: Understanding the Concept
This problem involves the principle of conservation of mass for a fluid in motion, which is described by the equation of continuity. This equation relates the cross-sectional area of a pipe and the speed of the fluid flowing through it. The pressure information is extra and not needed for this specific question.
Step 2: Key Formula or Approach
The equation of continuity states that for an incompressible fluid flowing through a pipe, the product of the cross-sectional area (A) and the fluid speed (v) is constant.
\[ A_1 v_1 = A_2 v_2 \]
where subscripts 1 and 2 refer to two different points along the pipe.
The cross-sectional area of a pipe with diameter D (or radius r) is \(A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}\).
Step 3: Detailed Explanation
1. Define the initial and final states.
- Let the initial section be section 1 and the narrow section be section 2.
- Initial speed, \(v_1 = 3 \text{ ms}^{-1}\).
- Let the initial diameter be \(D_1\). The initial area is \(A_1 = \frac{\pi D_1^2}{4}\).
- The pipe narrows to half its original diameter, so the final diameter is \(D_2 = \frac{D_1}{2}\).
- The final area is \(A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (D_1/2)^2}{4} = \frac{\pi D_1^2/4}{4} = \frac{A_1}{4}\).
- We need to find the final speed, \(v_2\).
2. Apply the equation of continuity.
\[ A_1 v_1 = A_2 v_2 \]
3. Solve for the final speed \(v_2\).
\[ v_2 = v_1 \left( \frac{A_1}{A_2} \right) \]
Substitute \(A_2 = \frac{A_1}{4}\):
\[ v_2 = v_1 \left( \frac{A_1}{A_1/4} \right) = v_1 \times 4 \]
Now substitute the value of \(v_1\):
\[ v_2 = 3 \text{ ms}^{-1} \times 4 = 12 \text{ ms}^{-1} \]
Step 4: Final Answer
The speed of water in the narrow section is 12 ms\(^{-1}\).