Question

In a horizontal pipeline of uniform cross-section the pressure falls by $8{\text{N/m}}^2$ between two points separated by $1\text{km}$. If oil of density $800{\text{kg/m}}^3$ flows through the pipe, find the change in $K.E$ per $\text{kg}$ of oil at these points.

• A. ${10}^{-2}\text{J/kg}$
• B. ${10}^{-3}\text{J/kg}$
• C. ${10}^{-4}\text{J/kg}$
• D. ${10}^{-1}\text{J/kg}$

Answer 1

The correct option is A ${10}^{-2}\text{J/kg}$

Let $P_1$ and $P_2$ be the pressures at points $1$ and $2$ respectively on the horizontal pipe.$P_1-P_2=8N/m^2$
$h_1=h_2$ ($\because$ pipe is horizontal)
By using bernoulli's equation,
$P_1{V}_1+\frac{1}{2}m{v}_1^2+mg{h}_1=P_2{V}_2+\frac{1}{2}m{v}_2^2+mg{h}_2$
For per unit volume, divide it by $m$ and then we get
$\frac{P_1}{\rho }+\frac{v_1^2}{2}=\frac{P_2}{\rho }+\frac{v_2^2}{2}$
$\frac{P_1-P_2}{\rho }=\frac{v_2^2-v_1^2}{2}$
Right side of above equation is the change in $K.E$ unit mass
$\Delta K.E.=\frac{8}{800}=\frac{1}{100}={10}^{-2}J/kg$