Water flows through a hose (pipe) whose internal diameter is 2 cm at a speed of 1 m/s. What should be the diameter of the nozzle if the water is to emerge at a speed of 4 m/s

Explanation

To solve the problem, we will use the principle of continuity for fluid flow, which states that the product of the cross-sectional area of a pipe and the velocity of the fluid at that section is constant.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the hose pipe (d1) = 2 cm = 0.02 m (convert to meters)
- Velocity of water in the hose pipe (v1) = 1 m/s
- Velocity of water at the nozzle (v2) = 4 m/s
- We need to find the diameter of the nozzle (d2).

2. **Calculate the Cross-Sectional Area of the Hose Pipe (A1):**
The area (A) of a circle is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
For the hose pipe:
\[ A_1 = \frac{\pi (0.02)^2}{4} \]
\[ A_1 = \frac{\pi (0.0004)}{4} = \frac{0.0004\pi}{4} = 0.0001\pi \, \text{m}^2 \]

3. **Use the Continuity Equation:**
According to the continuity equation:
\[ A_1 v_1 = A_2 v_2 \]
We can express the area of the nozzle (A2) in terms of its diameter (d2):
\[ A_2 = \frac{\pi d_2^2}{4} \]
Substituting into the continuity equation:
\[ 0.0001\pi \cdot 1 = \frac{\pi d_2^2}{4} \cdot 4 \]

4. **Simplify the Equation:**
Cancel \(\pi\) from both sides:
\[ 0.0001 \cdot 1 = d_2^2 \]
\[ 0.0001 = d_2^2 \]

5. **Solve for d2:**
Taking the square root of both sides:
\[ d_2 = \sqrt{0.0001} = 0.01 \, \text{m} \]
Convert back to centimeters:
\[ d_2 = 0.01 \times 100 = 1 \, \text{cm} \]

### Final Answer:
The diameter of the nozzle should be **1 cm**.