When a tap is closed, the manometer attached to the pipe reads `3.5xx10^(5)Nm^(-2)`. When the tap is opened, the reading of manometer falls to `3.0xx10^(5)Nm^(-2)`. The velocity of water in the pipe is

Options

A. `0.1ms^(-1)`
B. `1ms^(-1)`
C. `5ms^(-1)`
D. `10ms^(-1)`

Explanation

To solve the problem, we will use Bernoulli's principle, which relates the pressure difference in a fluid to its velocity. Here are the steps to find the velocity of water in the pipe:

### Step 1: Identify the pressure readings
When the tap is closed, the pressure reading from the manometer is:
- \( P_1 = 3.5 \times 10^5 \, \text{N/m}^2 \)

When the tap is opened, the pressure reading is:
- \( P_2 = 3.0 \times 10^5 \, \text{N/m}^2 \)

### Step 2: Calculate the pressure difference
The pressure difference (\( \Delta P \)) when the tap is opened can be calculated as:
\[ \Delta P = P_1 - P_2 = (3.5 \times 10^5) - (3.0 \times 10^5) = 0.5 \times 10^5 \, \text{N/m}^2 \]

### Step 3: Use Bernoulli's equation
According to Bernoulli's principle, the pressure difference can be related to the velocity of the fluid:
\[ \Delta P = \frac{1}{2} \rho v^2 \]
Where:
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( v \) is the velocity of water in the pipe

### Step 4: Rearrange the equation to solve for velocity
We can rearrange the equation to solve for \( v \):
\[ v^2 = \frac{2 \Delta P}{\rho} \]
Substituting the values we have:
\[ v^2 = \frac{2 \times (0.5 \times 10^5)}{1000} \]

### Step 5: Simplify the equation
Calculating the right-hand side:
\[ v^2 = \frac{1 \times 10^5}{1000} = 100 \]

### Step 6: Calculate the velocity
Taking the square root of both sides gives:
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer
The velocity of water in the pipe is:
\[ \boxed{10 \, \text{m/s}} \]

---