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

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

Step 1: Write the continuity equation for incompressible flow: \( A_1v_1 = A_2v_2 \), with \( A_1=\pi r^2 \) for the first tube and \( A_2=\pi(r/2)^2=\pi r^2/4 \) for the second.

Step 2: Substitute the areas: \( \pi r^2 \times v = \dfrac{\pi r^2}{4} \times v_2 \).

Step 3: Cancel \( \pi r^2 \) from both sides: \( v = \dfrac{v_2}{4} \).

Step 4: Solve for \( v_2 \) by multiplying both sides by 4.\[ \boxed{v_2 = 4v} \]

Answer 2

Picture the water moving through each tube as a series of short cylindrical slugs. In one second, the first tube passes a cylinder of water of length \( v \) and cross-section \( \pi r^2 \), so the volume delivered per second is \( \pi r^2 v \). For the same volume to pass through the narrower tube in the same second, a slug of length \( v_2 \) with cross-section \( \pi(r/2)^2 \) must carry the identical volume:

\[ \pi r^2 v = \pi\left(\frac{r}{2}\right)^2 v_2 \]

Since \( (r/2)^2 = r^2/4 \), the equation simplifies to \( v = v_2/4 \), so the slug in the narrow tube must be four times longer to hold the same volume in the same second, meaning it moves four times faster.

1. Option A, \( 2v \): not enough of a speed increase to keep the volume per second constant when the area drops to a quarter.
2. Option B, \( v/4 \): a slower speed would mean less volume passes per second, violating conservation of mass.
3. Option C, \( v/2 \): same problem as B, just a smaller shortfall.
4. Option D, \( 4v \): exactly compensates for the area shrinking to one quarter, keeping the flow rate constant, so this is correct.

The correct answer is \( 4v \).