Question

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

• A. $\frac{1}{40}J$
• B. $\frac{1}{80}J$
• C. $\frac{3}{40}J$
• D. $\frac{9}{50}J$

Answer 1

The correct option is B $\frac{1}{80}J$

According to Bernoulli's theorem, total energy is conserved.

$\frac{P}{\rho }+gh+\frac{1}{2}{v}^2=$constant.

For a horizontal pipe,$h=0$

$\frac{P}{\rho }+\frac{1}{2}{v}^2=$constant.

At ends $1$ and $2$:

$\frac{P_1}{\rho }+\frac{v_1^2}{2}=\frac{P_2}{\rho }+\frac{v_2^2}{2}$

$\frac{P_1-P_2}{\rho }=\frac{v_1^2-v_2^2}{2}$ ......$\left(1\right)$

Change in K.E$=\frac{1}{2}m(v_1^2-v_2^2)$

Change in K.E per kg$=\frac{1}{2}(v_1^2-v_2^2)=\frac{P_1-P_2}{\rho }$

Given:$P_1-P_2=10$N/m, $\rho =800$kg$m^{-3}$

Change in K.E$=\frac{10}{800}=\frac{1}{80}$J