At what speed, the velocity head of water is equal to pressure head of 40 cm of mercury?

Options

A. 10.32 m/s
B. 10.23 m/s
C. 10.52 m/s
D. 10.54 m/s

Answer

To find the speed at which the velocity head of water is equal to the pressure head of 40 cm of mercury, we can follow these steps:

### Step 1: Understand the relationship between velocity head and pressure head
The velocity head (h_v) of a fluid is given by the formula:
\[ h_v = \frac{V^2}{2g} \]
where \( V \) is the velocity of the fluid and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

The pressure head (h_p) can be expressed as:
\[ h_p = \frac{P}{\rho g} \]
where \( P \) is the pressure, \( \rho \) is the density of the fluid, and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the pressure head of 40 cm of mercury
To find the pressure head corresponding to 40 cm of mercury, we use the density of mercury, which is approximately \( 13.6 \times 10^3 \, \text{kg/m}^3 \). The height in meters is:
\[ h = 40 \, \text{cm} = 0.4 \, \text{m} \]

Now, substituting into the pressure head formula:
\[ h_p = \frac{\rho_{Hg} \cdot g \cdot h}{\rho_{water} \cdot g} \]
Since \( g \) cancels out, we have:
\[ h_p = \frac{\rho_{Hg} \cdot h}{\rho_{water}} \]

### Step 3: Substitute the values
Using the values:
- \( \rho_{Hg} = 13.6 \times 10^3 \, \text{kg/m}^3 \)
- \( \rho_{water} = 1000 \, \text{kg/m}^3 \)

We can calculate:
\[ h_p = \frac{(13.6 \times 10^3) \cdot 0.4}{1000} \]
\[ h_p = \frac{5440}{1000} = 5.44 \, \text{m} \]

### Step 4: Set the velocity head equal to the pressure head
Now, we set the velocity head equal to the pressure head:
\[ \frac{V^2}{2g} = h_p \]
Substituting for \( h_p \):
\[ \frac{V^2}{2 \cdot 9.81} = 5.44 \]

### Step 5: Solve for V
Rearranging gives:
\[ V^2 = 5.44 \cdot 2 \cdot 9.81 \]
\[ V^2 = 106.4648 \]
Taking the square root:
\[ V = \sqrt{106.4648} \]
\[ V \approx 10.32 \, \text{m/s} \]

### Final Answer
The speed at which the velocity head of water is equal to the pressure head of 40 cm of mercury is approximately:
\[ V \approx 10.32 \, \text{m/s} \]

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