Answer

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Hint: Study about the concepts of work done and power. Understand the definitions and try to obtain the mathematical equations that we can use to solve this question. Express the equation in terms of the given quantities in the question and put the values to find the required solution.

Complete Step-by-Step solution:

Velocity of the water when it leaves the pipe is given as, $v=2m/s$

The mass per unit length of the water in the pipe is, $100kg/m$

Now, to find the power of the engine we can find out the change in kinetic energy of the water per unit time.

Kinetic energy is given by, $K.E.=\dfrac{1}{2}m{{v}^{2}}$

Power can be defined as the amount of energy required per unit time to a work.

So, power is equal to the kinetic energy of water flowing out of the pipe per unit time.

$\begin{align}

& power=\dfrac{d\left( K.E. \right)}{dt} \\

& power=\dfrac{d}{dt}\left( \dfrac{1}{2}m{{v}^{2}} \right) \\

\end{align}$

Now, the velocity of the water will be constant. So, we can put the velocity-square out of the derivative.

$power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dt}$

We are given the mass of water per unit length of the pipe. In the above equation we will change the quantity from mass per unit time to mass per unit length.

$power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dl}\dfrac{dl}{dt}$

Where, $\dfrac{dl}{dt}$ is the velocity of the water through the pipe and $\dfrac{dm}{dl}$ is the mass per unit length of water in the pipe.

$\begin{align}

& power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dl}v \\

& power=\dfrac{1}{2}{{v}^{3}}\dfrac{dm}{dl} \\

\end{align}$

Putting the values of all the quantities in the above equation, we get

$\begin{align}

& power=\dfrac{1}{2}\times {{2}^{3}}\times 100 \\

& power=400W \\

\end{align}$

So, the power of the engine is 400 W.

The correct option is (C).

Note: Mass per unit length is like the mass gradient expressed in one-dimension.

With the help of the work energy principle we get that, the change in kinetic energy of an object is equal to the net work done on the object. That’s why we are taking the kinetic energy of the water to find the power of the machine.

Complete Step-by-Step solution:

Velocity of the water when it leaves the pipe is given as, $v=2m/s$

The mass per unit length of the water in the pipe is, $100kg/m$

Now, to find the power of the engine we can find out the change in kinetic energy of the water per unit time.

Kinetic energy is given by, $K.E.=\dfrac{1}{2}m{{v}^{2}}$

Power can be defined as the amount of energy required per unit time to a work.

So, power is equal to the kinetic energy of water flowing out of the pipe per unit time.

$\begin{align}

& power=\dfrac{d\left( K.E. \right)}{dt} \\

& power=\dfrac{d}{dt}\left( \dfrac{1}{2}m{{v}^{2}} \right) \\

\end{align}$

Now, the velocity of the water will be constant. So, we can put the velocity-square out of the derivative.

$power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dt}$

We are given the mass of water per unit length of the pipe. In the above equation we will change the quantity from mass per unit time to mass per unit length.

$power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dl}\dfrac{dl}{dt}$

Where, $\dfrac{dl}{dt}$ is the velocity of the water through the pipe and $\dfrac{dm}{dl}$ is the mass per unit length of water in the pipe.

$\begin{align}

& power=\dfrac{1}{2}{{v}^{2}}\dfrac{dm}{dl}v \\

& power=\dfrac{1}{2}{{v}^{3}}\dfrac{dm}{dl} \\

\end{align}$

Putting the values of all the quantities in the above equation, we get

$\begin{align}

& power=\dfrac{1}{2}\times {{2}^{3}}\times 100 \\

& power=400W \\

\end{align}$

So, the power of the engine is 400 W.

The correct option is (C).

Note: Mass per unit length is like the mass gradient expressed in one-dimension.

With the help of the work energy principle we get that, the change in kinetic energy of an object is equal to the net work done on the object. That’s why we are taking the kinetic energy of the water to find the power of the machine.

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