Question

Water flows from pipe diameter $200\text{ mm}$ to $100\text{ mm}$. If velocity in larger pipe is $2\text{ m/s}$, find velocity in smaller pipe.

• A. $2\text{ m/s}$
• B. $4\text{ m/s}$
• C. $6\text{ m/s}$
• D. $8\text{ m/s}$

Hint:

Velocity is inversely proportional to the square of the diameter ($V \propto 1/D^2$).
If the diameter is halved ($200\text{ mm} \rightarrow 100\text{ mm}$), the velocity must increase by a factor of $2^2 = 4$.
$2\text{ m/s} \times 4 = 8\text{ m/s}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the fluid velocity in a smaller pipe section, given the diameters of both the larger and smaller sections, and the velocity in the larger section.

Step 2: Key Formula or Approach:
For an incompressible fluid, we apply the Continuity Equation:
\[ A_1 V_1 = A_2 V_2 \]
where:
$A_1, A_2$ are the cross-sectional areas of the larger and smaller pipe sections, respectively.
$V_1, V_2$ are the corresponding fluid velocities.

Step 3: Detailed Explanation:

• The cross-sectional area of a circular pipe is:
\[ A = \frac{\pi}{4} D^2 \]

• Substituting this into the continuity equation:
\[ \frac{\pi}{4} D_1^2 V_1 = \frac{\pi}{4} D_2^2 V_2 \]
\[ D_1^2 V_1 = D_2^2 V_2 \]

• Given values:
$D_1 = 200\text{ mm}$ (diameter of the larger pipe)
$D_2 = 100\text{ mm}$ (diameter of the smaller pipe)
$V_1 = 2\text{ m/s}$ (velocity in the larger pipe)

• Solve for $V_2$:
\[ V_2 = V_1 \left(\frac{D_1}{D_2}\right)^2 \]
\[ V_2 = 2 \times \left(\frac{200}{100}\right)^2 \]
\[ V_2 = 2 \times (2)^2 \]
\[ V_2 = 2 \times 4 = 8\text{ m/s} \]

Step 4: Final Answer:
The velocity of water in the smaller pipe is $8\text{ m/s}$.