Question

Water flows through a horizontal pipe of varying cross-section at a rate of $0.314 m^3s^{-1}$. The velocity of water at a point where the radius of the pipe is 10 cm is

• A. 0.1$ms^{-1}$
• B. 1$ms^{-1}$
• C. 10$ms^{-1}$
• D. 100$ms^{-1}$

Answer 1

The Correct Option is C

Given:
The flow rate of water through the pipe is \( Q = 0.314 \, \text{m}^3/\text{s} \)
The radius of the pipe at the point of interest is \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) We are asked to find the velocity \( v \) of the water at this point. Step 1: Use the continuity equation The continuity equation for incompressible flow states that the flow rate \( Q \) is constant along the pipe, and is given by: \[ Q = A \cdot v \] where:
\( A \) is the cross-sectional area of the pipe
\( v \) is the velocity of the water The cross-sectional area \( A \) of the pipe at the point where the radius is \( r \) is: \[ A = \pi r^2 \] Step 2: Solve for the velocity Substitute the expression for \( A \) into the continuity equation: \[ Q = \pi r^2 \cdot v \] Rearrange to solve for \( v \): \[ v = \frac{Q}{\pi r^2} \] Substitute the given values: \[ v = \frac{0.314}{\pi (0.1)^2} \] \[ v = \frac{0.314}{\pi \times 0.01} = \frac{0.314}{0.0314} = 10 \, \text{m/s} \] Thus, the velocity of water at the point where the radius is 10 cm is \( 10 \, \text{m/s} \). Therefore, the correct answer is (C) 10 m/s.

Answer 2

We are given the volumetric flow rate \( Q = 0.314 \, \text{m}^3\text{s}^{-1} \), and we need to find the velocity \( V \) of water at a point where the radius of the pipe is \( r = 10 \, \text{cm} = 0.1 \, \text{m} \). Using the equation for the flow rate of a fluid in a pipe: \[ Q = A V \] where \( A \) is the cross-sectional area of the pipe and \( V \) is the velocity of the water. The cross-sectional area \( A \) of the pipe is given by the formula for the area of a circle: \[ A = \pi r^2 \] Substitute \( r = 0.1 \, \text{m} \): \[ A = \pi (0.1)^2 = 3.14 \times 10^{-2} \, \text{m}^2 \] Now, substitute the values of \( Q \) and \( A \) into the equation for flow rate: \[ 0.314 = 3.14 \times 10^{-2} \times V \] Solving for \( V \): \[ V = \frac{0.314}{3.14 \times 10^{-2}} = 10 \, \text{ms}^{-1} \] Thus, the velocity of the water at the point where the radius of the pipe is 10 cm is \( 10 \, \text{ms}^{-1} \).