In a horizontal pipe line of uniform cross-section, pressure falls by 5 Pa between two points separated by 1 km. The change in the kinetic energy per kg of the oil flowing at these points is (density of oil `=800kgm^(-3))`

Options

A. `6.25xx10^(-3)Jkg^(-1)`
B. `5.25xx10^(-4)Jkg^(-1)`
C. `3.25xx10^(-5)Jkg^(-1)`
D. `4.25xx10^(-2)Jkg^(-1)`

Explanation

To solve the problem, we will use Bernoulli's theorem, which relates the pressure, kinetic energy, and potential energy in a fluid flow. Since the pipe is horizontal, we can ignore potential energy changes.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a horizontal pipe with a pressure drop of 5 Pa over a distance of 1 km. The density of the oil is given as 800 kg/m³. We need to find the change in kinetic energy per kilogram of the oil flowing through the pipe.

2. **Applying Bernoulli's Equation**:
According to Bernoulli's theorem for a horizontal flow:
\[ P_A + \frac{1}{2} \rho V_A^2 = P_B + \frac{1}{2} \rho V_B^2 \]
Rearranging this gives us:
\[ P_A - P_B = \frac{1}{2} \rho V_B^2 - \frac{1}{2} \rho V_A^2 \]
This can be rewritten as:
\[ P_A - P_B = \frac{1}{2} \rho (V_B^2 - V_A^2) \]

3. **Finding Change in Kinetic Energy**:
The left side of the equation represents the pressure difference, which is given as:
\[ P_A - P_B = 5 \text{ Pa} \]
Thus, we can express the change in kinetic energy per unit mass as:
\[ \frac{P_A - P_B}{\rho} = \frac{1}{2} (V_B^2 - V_A^2) \]
Therefore, the change in kinetic energy per kilogram is:
\[ \Delta KE = \frac{5 \text{ Pa}}{800 \text{ kg/m}^3} \]

4. **Calculating the Change in Kinetic Energy**:
Now, we perform the calculation:
\[ \Delta KE = \frac{5}{800} = 0.00625 \text{ J/kg} \]

5. **Final Result**:
We can express this in scientific notation:
\[ \Delta KE = 6.25 \times 10^{-3} \text{ J/kg} \]

### Conclusion:
The change in kinetic energy per kg of the oil flowing at these points is \( 6.25 \times 10^{-3} \text{ J/kg} \).