
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?
$\begin{align}
& \text{A}\text{. }\dfrac{1}{2}{{m}^{2}}{{v}^{2}} \\
& \text{B}\text{. }\dfrac{1}{2}m{{v}^{2}} \\
& \text{C}\text{. }\dfrac{m{{v}^{3}}}{2} \\
& \text{D}\text{. }\dfrac{1}{6}{{m}^{2}}{{v}^{2}} \\
\end{align}$
Answer
565.8k+ views
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.
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

