Question

Water flows through a horizontal pipe of variable cross-section at the rate of 12\(\pi\) litres per minute. The velocity of the water at the point where the diameter of the pipe becomes 2 cm is:

• A. \(6 \, \text{ms}^{-1}\)
• B. \(8 \, \text{ms}^{-1}\)
• C. \(4 \, \text{ms}^{-1}\)
• D. \(2 \, \text{ms}^{-1}\)

Hint:

The equation of continuity relates the velocity and cross-sectional area of a fluid in a pipe.
- When the area decreases, the velocity must increase, and vice versa.

Answer 1

The Correct Option is D

The equation of continuity for fluid flow is: \[ A_1 v_1 = A_2 v_2 \] where \( A_1 \) and \( A_2 \) are the cross-sectional areas at two points, and \( v_1 \) and \( v_2 \) are the velocities at those points. We are given the volume flow rate: \[ Q = 12\pi \, \text{litres per minute} = 12\pi \times 10^{-3} \, \text{m}^3/\text{minute} \] Convert it to cubic meters per second: \[ Q = \frac{12\pi \times 10^{-3}}{60} = 2\pi \times 10^{-4} \, \text{m}^3/\text{s} \] At the point where the diameter of the pipe is 2 cm, the radius is 1 cm = 0.01 m. Thus, the area is: \[ A_2 = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \] Using the equation of continuity, we find the velocity \( v_2 \): \[ v_2 = \frac{Q}{A_2} = \frac{2\pi \times 10^{-4}}{\pi \times 10^{-4}} = 2 \, \text{m/s} \] Thus, the velocity is \(\boxed{2 \, \text{ms}^{-1}}\).