An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
$\begin{align}
  & \text{A}\text{. }\dfrac{1}{2}{{m}^{2}}{{v}^{2}} \\
 & \text{B}\text{. }\dfrac{1}{2}m{{v}^{2}} \\
 & \text{C}\text{. }\dfrac{m{{v}^{3}}}{2} \\
 & \text{D}\text{. }\dfrac{1}{6}{{m}^{2}}{{v}^{2}} \\
\end{align}$

Hint: To solve this question obtain the expression for kinetic energy of a moving object in terms of the mass and the velocity of the object. Then find the rate of kinetic energy by taking the time derivative. Put the required values given in the question to find the required answer.

Complete answer:
Given, that an engine pumps water continuously through a hose where the water leaves the hose with velocity v.
The mass per unit length of the water jet is m.
Now, the kinetic energy of a particle of mass M moving with velocity v is given as,
$KE=\dfrac{1}{2}M{{v}^{2}}$
So, the rate of kinetic energy can be defined as,
$K{{E}_{R}}=\dfrac{d}{dt}\left( \dfrac{1}{2}M{{v}^{2}} \right)$
Now, for the water jet, let the length of the water jet is L.
So, the mass of the water will be, $M=mL$
So, the rate of kinetic energy imparted on the water can be given as,
$K{{E}_{R}}=\dfrac{d}{dt}\left( \dfrac{1}{2}mL{{v}^{2}} \right)$
Now, the mass per unit length of water and the velocity of the water through the hose will be constant. So, we can simplify the above equation as,
\[\begin{align}
  & K{{E}_{R}}=\dfrac{1}{2}m{{v}^{2}}\dfrac{dL}{dt} \\
 & K{{E}_{R}}=\dfrac{1}{2}m{{v}^{3}} \\
\end{align}\]
So, the rate at which kinetic energy is imparted to the water will be \[\dfrac{1}{2}m{{v}^{3}}\].

So, the correct answer is “Option C”.

Note:
The rate of kinetic energy imparted can also be defined as the change in kinetic energy per unit time. Using the work energy theorem, it can be explained as the work done by the net force acting on it or by the machine to impart the kinetic energy on the water which passes through the hose.