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, what is the speed of water in the second tube?

Options

A. ` v//4 `
B. `v //2 `
C. `2v `
D. `4 v `

Explanation

To solve the problem, we will use the principle of conservation of mass, which is expressed through the equation of continuity for fluid flow. The equation states that the product of the cross-sectional area of a tube and the velocity of the fluid flowing through it remains constant.

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Initial radius of the first tube, \( r \)
- Speed of water in the first tube, \( v \)
- Radius of the second tube, \( r' = \frac{r}{2} \)

2. **Calculate the cross-sectional area of the first tube:**
\[ A = \pi r^2 \]
Here, \( A \) is the area of the first tube.

3. **Calculate the cross-sectional area of the second tube:**
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]
Here, \( A' \) is the area of the second tube.

4. **Apply the equation of continuity:**
According to the equation of continuity:
\[ A \cdot v = A' \cdot v' \]
Where \( v' \) is the speed of water in the second tube.

5. **Substitute the areas into the equation:**
\[ \pi r^2 \cdot v = \pi \frac{r^2}{4} \cdot v' \]

6. **Cancel out the common terms:**
We can cancel \( \pi r^2 \) from both sides:
\[ v = \frac{r^2}{4} \cdot v' \]

7. **Rearrange to solve for \( v' \):**
\[ v' = 4v \]

8. **Conclusion:**
The speed of water in the second tube is \( 4v \).

### Final Answer:
The speed of water in the second tube is \( 4v \).

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