The cross-sectionss of a pipe at two points A and B are in the ratio 1.4. If the speed of water flowing through the pipe at point A is v, its speed at point B is

Options

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

Answer

To solve the problem, we will use the principle of continuity for fluid flow, which states that the product of the cross-sectional area and the velocity of the fluid at any two points along a streamline is constant. This can be expressed mathematically as:

\[ A_1 v_1 = A_2 v_2 \]

Where:
- \( A_1 \) is the cross-sectional area at point A,
- \( v_1 \) is the velocity at point A,
- \( A_2 \) is the cross-sectional area at point B,
- \( v_2 \) is the velocity at point B.

### Step-by-Step Solution:

1. **Identify the given information:**
- The ratio of cross-sectional areas at points A and B is given as \( A_1 : A_2 = 1 : 4 \).
- The velocity at point A is given as \( v_1 = v \).

2. **Express the areas in terms of a variable:**
- Let \( A_1 = A \) (some arbitrary area).
- Then, according to the ratio, \( A_2 = 4A \).

3. **Apply the continuity equation:**
- Using the continuity equation \( A_1 v_1 = A_2 v_2 \):
\[ A v = 4A v_2 \]

4. **Cancel out the common area (A):**
- Since \( A \) is common on both sides and is not zero, we can simplify:
\[ v = 4 v_2 \]

5. **Solve for \( v_2 \):**
- Rearranging the equation gives:
\[ v_2 = \frac{v}{4} \]

### Final Answer:
The speed of water flowing through the pipe at point B is \( v_2 = \frac{v}{4} \).

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