A pressure meter attached to a closed water tap reads `1.5 xx 10^(5) Pa`. When the tap is opened, the velocity of flow of water is `10 ms^(-1)` and the reading of the pressure meter is

Options

A. `1.5xx10^(5)Pa`
B. `3xx10^(5)Pa`
C. `0.5xx10^(5)Pa`
D. `10^(5)Pa`

Explanation

To solve the problem, we will use Bernoulli's equation, which relates the pressure and velocity of a fluid in a streamline flow. The equation is given as:

\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]

Where:
- \( P_1 \) is the initial pressure,
- \( P_2 \) is the pressure after the tap is opened,
- \( \rho \) is the density of the fluid (water in this case),
- \( V_1 \) is the initial velocity of the fluid,
- \( V_2 \) is the velocity of the fluid after the tap is opened.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Initial pressure, \( P_1 = 1.5 \times 10^5 \, \text{Pa} \)
- Initial velocity, \( V_1 = 0 \, \text{m/s} \) (since the tap is closed)
- Final velocity, \( V_2 = 10 \, \text{m/s} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

2. **Substitute the Values into Bernoulli's Equation:**
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Substituting the known values:
\[ 1.5 \times 10^5 + \frac{1}{2} \times 1000 \times 0^2 = P_2 + \frac{1}{2} \times 1000 \times (10)^2 \]
Simplifying:
\[ 1.5 \times 10^5 + 0 = P_2 + \frac{1}{2} \times 1000 \times 100 \]
\[ 1.5 \times 10^5 = P_2 + 50000 \]

3. **Rearranging to Solve for \( P_2 \):**
\[ P_2 = 1.5 \times 10^5 - 50000 \]
\[ P_2 = 1.5 \times 10^5 - 5 \times 10^4 \]
\[ P_2 = 1.0 \times 10^5 \, \text{Pa} \]

4. **Final Answer:**
The reading of the pressure meter when the tap is opened is:
\[ P_2 = 1.0 \times 10^5 \, \text{Pa} \]