Question

Water flows through a horizontal pipe of varying cross-section at the rate of \( \pi \times 10^{-1} \, \text{m}^3/\text{s} \). The velocity of water at a point where the radius of the pipe is 10 cm is \( (\pi = 3.14) \)

• A. 0.1 m/s
• B. 1 m/s
• C. 10 m/s
• D. 100 m/s

Hint:

The continuity equation relates the flow rate, cross-sectional area, and velocity of a fluid. For incompressible fluids, the flow rate must remain constant at any point in the pipe.

Answer 1

The Correct Option is B

Step 1: Use the principle of continuity.
According to the principle of continuity, the volume flow rate \( Q \) through a pipe is given by:
\[ Q = A v, \]
where:
- \( A \) is the cross-sectional area of the pipe,
- \( v \) is the velocity of water at that point.
For a pipe with circular cross-section, the area \( A \) at any point is given by:
\[ A = \pi r^2, \] where \( r \) is the radius of the pipe.

Step 2: Applying the given data.
We are given that the flow rate \( Q = \pi \times 10^{-1} \, \text{m}^3/\text{s} \), and the radius \( r = 10 \, \text{cm} = 0.1 \, \text{m} \). Substituting these values into the continuity equation:
\[ Q = A v = \pi r^2 v. \]

Step 3: Solving for the velocity.
Now, substitute the given values of \( Q \) and \( r \) into the equation:
\[ \pi \times 10^{-1} = \pi (0.1)^2 v. \]
Simplifying:
\[ 10^{-1} = 0.01 v. \]
Solving for \( v \):
\[ v = \frac{10^{-1}}{0.01} = 1 \, \text{m/s}. \]
Final Answer:
Thus, the velocity of water at the point where the radius of the pipe is 10 cm is:
\[ \boxed{1 \, \text{m/s}}. \]