The pressure of water in a pipe when tap is closed is `5.5xx10^5 Nm^(-2)`. When tap gets open, pressure reduces to `5xx10^5Nm^(-2)`. The velocity with which water comes out on opening the tap is

Options

A. `10ms^(-1)`
B. `5ms^(-1)`
C. `20ms^(-1)`
D. `15ms^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of energy, specifically relating the change in pressure to the kinetic energy of the water as it flows out of the tap.

### Step-by-Step Solution:

1. **Identify the Initial and Final Pressures:**
- When the tap is closed, the pressure \( P_1 = 5.5 \times 10^5 \, \text{N/m}^2 \).
- When the tap is open, the pressure \( P_2 = 5.0 \times 10^5 \, \text{N/m}^2 \).

2. **Calculate the Change in Pressure:**
\[ \Delta P = P_1 - P_2 = (5.5 \times 10^5) - (5.0 \times 10^5) = 0.5 \times 10^5 \, \text{N/m}^2 \]

3. **Use the Relationship Between Pressure and Kinetic Energy:**
The change in pressure is related to the kinetic energy of the water:
\[ \Delta P = \frac{1}{2} \rho v^2 \]
where \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( v \) is the velocity of the water.

4. **Substituting the Values:**
\[ 0.5 \times 10^5 = \frac{1}{2} \times 1000 \times v^2 \]

5. **Rearranging the Equation to Solve for \( v^2 \):**
\[ v^2 = \frac{2 \times (0.5 \times 10^5)}{1000} \]
\[ v^2 = \frac{1 \times 10^5}{1000} \]
\[ v^2 = 100 \, \text{m}^2/\text{s}^2 \]

6. **Taking the Square Root to Find \( v \):**
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer:
The velocity with which water comes out when the tap is opened is \( 10 \, \text{m/s} \).

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