Question

A pipe has a radius of 2 cm at one end and 1 cm at the other end. The velocity of the water at the wider end is 5 m/s. What is the velocity of the water at the narrower end, assuming incompressible flow?

• A. \( 10 \, \text{m/s} \)
• B. \( 20 \, \text{m/s} \)
• C. \( 15 \, \text{m/s} \)
• D. \( 25 \, \text{m/s} \)

Hint:

In incompressible flow, the product of cross-sectional area and velocity at any point in the pipe remains constant. Use the principle of continuity to relate velocities at different points in the pipe.

Answer 1

The Correct Option is A

According to the principle of continuity for incompressible fluids, the flow rate (volume per unit time) must remain constant. This is expressed as: \[ A_1 v_1 = A_2 v_2 \] Where: - \( A_1 = \pi r_1^2 \) is the cross-sectional area at the wider end, - \( A_2 = \pi r_2^2 \) is the cross-sectional area at the narrower end, - \( v_1 = 5 \, \text{m/s} \) is the velocity at the wider end, - \( v_2 \) is the velocity at the narrower end. Now, substitute the values: \[ A_1 v_1 = A_2 v_2 \] \[ \pi (2)^2 \times 5 = \pi (1)^2 \times v_2 \] \[ 4 \times 5 = 1 \times v_2 \] \[ v_2 = 20 \, \text{m/s} \] Thus, the velocity of the water at the narrower end is \( 20 \, \text{m/s} \).