The velocity of kerosene oil in a horizontal pipe is `5 m//s`. If `g = 10m//s^(2)` then the velocity head of oil wlill be

Options

A. 1.25 m
B. 12.5 m
C. 0.125 m
D. 125 m

Answer

To find the velocity head of kerosene oil in a horizontal pipe, we can use the formula for velocity head, which is given by:

\[ \text{Velocity Head} = \frac{v^2}{2g} \]

where:
- \( v \) is the velocity of the fluid (in m/s),
- \( g \) is the acceleration due to gravity (in m/s²).

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of kerosene oil, \( v = 5 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Substitute the values into the formula:**
\[ \text{Velocity Head} = \frac{v^2}{2g} = \frac{(5 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} \]

3. **Calculate \( v^2 \):**
\[ v^2 = 5^2 = 25 \, \text{m}^2/\text{s}^2 \]

4. **Calculate \( 2g \):**
\[ 2g = 2 \times 10 = 20 \, \text{m/s}^2 \]

5. **Now, substitute these values back into the velocity head formula:**
\[ \text{Velocity Head} = \frac{25 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} \]

6. **Perform the division:**
\[ \text{Velocity Head} = \frac{25}{20} = 1.25 \, \text{m} \]

### Final Answer:
The velocity head of the kerosene oil will be \( 1.25 \, \text{m} \).

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