Step 1: Understanding the Concept:
For an incompressible fluid (like water) flowing through a pipe, the mass flow rate remains constant.
This leads to the Equation of Continuity, which states that the product of the cross-sectional area and the fluid velocity is constant at any point in the pipe.
Basically, the same amount of fluid that enters one end per second must exit the other end per second.
Step 2: Key Formula or Approach:
\[ A_1 v_1 = A_2 v_2 \]
Where \(A\) is cross-sectional area and \(v\) is velocity.
Step 3: Detailed Explanation:
From the problem data:
- Area of wider section (\(A_1\)) = 4 \(\text{cm}^2\).
- Velocity in wider section (\(v_1\)) = 2 m/s.
- Area of narrower section (\(A_2\)) = 1 \(\text{cm}^2\).
Substitute these values into the continuity equation:
\[ (4 \text{ cm}^2) \times (2 \text{ m/s}) = (1 \text{ cm}^2) \times v_2 \]
\[ 8 \text{ cm}^2 \cdot \text{m/s} = 1 \text{ cm}^2 \cdot v_2 \]
Solve for \(v_2\):
\[ v_2 = \frac{8}{1} = 8 \text{ m/s} \]
The units of area (\(\text{cm}^2\)) cancel out, so conversion to SI units was not strictly necessary here, though it is good practice in complex problems.
Step 4: Final Answer:
The velocity in the narrower section is 8 m/s.