Question

Water is flowing through a tube of radius \( r \) with a speed \( v \). If this tube is joined to another tube of radius \( r/2 \), the speed of water in the second tube is:

• A. \( 4v \)
• B. \( \frac{v}{4} \)
• C. \( \frac{v}{2} \)
• D. \( 2v \)

Hint:

When a fluid flows through a pipe, the speed increases when the cross-sectional area decreases, assuming steady flow.

Answer 1

The Correct Option is D

By the principle of conservation of mass, the volume flow rate must remain constant: \[ A_1 v_1 = A_2 v_2 \] Where: - \( A_1 = \pi r^2 \) is the cross-sectional area of the first tube, - \( A_2 = \pi \left( \frac{r}{2} \right)^2 = \frac{\pi r^2}{4} \) is the cross-sectional area of the second tube. Thus: \[ \pi r^2 v = \frac{\pi r^2}{4} v_2 \] Solving for \( v_2 \): \[ v_2 = 4v \] So the speed of water in the second tube is \( 4v \).