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

To solve the problem, we need to use the principle of conservation of mass, specifically the equation of continuity for incompressible fluids, which states that the product of the cross-sectional area of the tube and the velocity of the fluid flow through that area is a constant. Mathematically, it is represented as:

\( A_1 v_1 = A_2 v_2 \)

 

where:

• \( A_1 \) and \( A_2 \) are the cross-sectional areas of the first and second tubes, respectively.
• \( v_1 \) and \( v_2 \) are the flow velocities in the first and second tubes, respectively.

Given the radii:

• The radius of the first tube is \( r \), thus its cross-sectional area \( A_1 = \pi r^2 \).
• The radius of the second tube is \( r/2 \), thus its cross-sectional area \( A_2 = \pi (r/2)^2 = \pi r^2/4 \).

Substituting the areas into the continuity equation:

\( \pi r^2 \cdot v = \frac{\pi r^2}{4} \cdot v_2 \)

 

The \( \pi r^2 \) terms cancel out, leaving:

\( v = \frac{v_2}{4} \)

 

Rearranging to solve for \( v_2 \), we find:

\( v_2 = 4v \)

 

Therefore, the speed of water in the second tube is \( 4v \).

Answer 2

For an incompressible fluid moving through a tube whose cross-section changes, the volume flow rate must stay the same everywhere along the tube: \( A_1v_1 = A_2v_2 \). Since the cross-sectional area of a tube scales as the square of its radius, this can be rewritten purely as a ratio of radii:

\[ \frac{v_2}{v_1} = \frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 \]

Here \( r_1 = r \) and \( r_2 = r/2 \), so \( r_1/r_2 = 2 \), giving:

\[ \frac{v_2}{v_1} = 2^2 = 4 \quad\Rightarrow\quad v_2 = 4v \]

Checking each option against this ratio:

1. Option A, \( 2v \): would follow if the speed scaled linearly with the radius ratio instead of its square, so it is incorrect.
2. Option B, \( v/4 \): has the right factor of 4 but inverted, this is what you'd get if the tube widened instead of narrowed.
3. Option C, \( v/2 \): also has the speed decreasing, which is inconsistent with flowing into a narrower tube.
4. Option D, \( 4v \): matches the squared radius ratio computed above and is correct.

The correct answer is \( 4v \).