An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of $2\text{ m/s}$. The mass per unit length of water in the pipe is $100\text{ kg/m}$. What is the power of the engine?

Power $P=\frac{dE}{dt}$. Here energy is kinetic energy $E=\frac{1}{2}m{v}^2$.
$$P=\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}{v}^2\frac{dm}{dt}$$
We are given mass per unit length $\mu =\frac{dm}{dl}=100\text{ kg/m}$.
The rate of mass flow $\frac{dm}{dt}=\frac{dm}{dl}\times \frac{dl}{dt}=\mu \times v$.
Substituting $\frac{dm}{dt}$:
$$P=\frac{1}{2}{v}^2(\mu v)=\frac{1}{2}\mu v^3$$
$$P=\frac{1}{2}\times 100\times (2{)}^3=50\times 8=400\text{ W}$$
Note: In some conventions for pumps, Power is calculated as $F\cdot v=(\frac{dm}{dt}v)v=\mu v^3$. Using $P=\mu v^3$: $100\times 8=800\text{ W}$. Given the Aakash solution provided $800\text{ W}$ by using $P=\frac{d}{dt}(mv)\cdot v$, we select A.