Q. An engine pump is used to pump a liquid of density $\rho$ continuously through a pipe of cross-sectional area $A$. If the speed of flow of the liquid in the pipe is $v$, then the rate at which kinetic energy is being imparted to the liquid is
A) $\frac{1}{2}\ A\rho v^{3}$
B) $\frac{1}{2}\ A\rho v^{2}$
C) $\frac{1}{2}\ A\rho v$
D) $A\rho v$
Explanation
(a) Energy supplied to liquid per second by the pump
$= \frac{1}{2}\frac{mv^{2}}{t} = \frac{1}{2}\frac{V\rho v^{2}}{t} = \frac{1}{2}A \times \left( \frac{l}{t} \right) \times \rho \times v^{2}\left\lbrack \frac{l}{t} = v \right\rbrack$
$= \frac{1}{2}A \times v \times \rho \times v^{2} = \frac{1}{2}\ A\rho v^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.