In a horizontal pipe line pressure falls by 10 Pa between two points separated by a distance of 1 km. The change in kinetic energy per kg of oil when the oil flows from one point to the other is (Density of oil = 800 Kg/m³)

Given,

$$\begin{gathered} P_1-P_2=10 N/m ; \\ \rho =800kg{m}^{-3} \end{gathered}$$

according to Bernoulli's principle,

$\frac{P}{\rho }+gh+\frac{1}{2}{v}^2=cons\tan t$

h = 0 for a horizontal pipe.

$\frac{P}{\rho }+\frac{1}{2}{v}^2=cons\tan t$

so, at ends 1 and 2

$$\begin{gathered} \frac{P_1}{\rho }+\frac{{v_1}^2}{2}=\frac{P_2}{\rho }+\frac{{v_2}^2}{2} \\ \Rightarrow \frac{P_1-P_2}{\rho }=\frac{{v_2}^2-{v_1}^2}{2} \end{gathered}$$

Change in kinetic energy, $\frac{1}{2}m({v_1}^2-{v_2}^2)$

change in kinetic energy per kg $=\frac{1}{2}({v_2}^2-{v_1}^2)=\frac{P_1-P_2}{\rho }$

Change in kinetic energy per kg = $\frac{10}{800}=\frac{1}{80}J$