Question:
An engine pumps water continuously through a hose pipe. If the water leaves the pipe with velocity $v$ and $m$ is the mass per unit length of the water in the pipe, then the rate at which kinetic energy is imparted to the water is:
An engine pumps water continuously through a hose pipe. If the water leaves the pipe with velocity $v$ and $m$ is the mass per unit length of the water in the pipe, then the rate at which kinetic energy is imparted to the water is:
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Power can be represented as $F \cdot v$. The thrust force on a pipe discharging fluid is $F = v \frac{dM}{dt} = m v^2$. Thus, Power $= F \cdot v = (m v^2) v = m v^3$. However, only half of this work goes into the water's kinetic energy, yielding $\frac{1}{2} m v^3$.
Updated On: Jun 3, 2026
- $\frac{1}{2} m v^3$
- $\frac{1}{2} m v^2$
- $m v^3$
- $\frac{3}{2} m v^2$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The rate at which kinetic energy is imparted is equivalent to power, defined as $P = \frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2} M v^2\right) = \frac{1}{2} \left(\frac{dM}{dt}\right) v^2$.
Step 2: Meaning
Since $m$ is the mass per unit length of water in the pipe ($m = \frac{dM}{dx}$), we can rewrite the mass rate of flow $\frac{dM}{dt}$ in terms of $m$ and the velocity $v$.
Step 3: Analysis
Expressing $\frac{dM}{dt}$: \[ \frac{dM}{dt} = \frac{d(m \cdot x)}{dt} = m \frac{dx}{dt} = m v \] Substituting this into the power equation: \[ P = \frac{1}{2} \left(\frac{dM}{dt}\right) v^2 = \frac{1}{2} (mv) v^2 = \frac{1}{2} m v^3 \]
Step 4: Conclusion
The rate at which kinetic energy is imparted to the water is $\frac{1}{2} m v^3$.
Final Answer: (A)
The rate at which kinetic energy is imparted is equivalent to power, defined as $P = \frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2} M v^2\right) = \frac{1}{2} \left(\frac{dM}{dt}\right) v^2$.
Step 2: Meaning
Since $m$ is the mass per unit length of water in the pipe ($m = \frac{dM}{dx}$), we can rewrite the mass rate of flow $\frac{dM}{dt}$ in terms of $m$ and the velocity $v$.
Step 3: Analysis
Expressing $\frac{dM}{dt}$: \[ \frac{dM}{dt} = \frac{d(m \cdot x)}{dt} = m \frac{dx}{dt} = m v \] Substituting this into the power equation: \[ P = \frac{1}{2} \left(\frac{dM}{dt}\right) v^2 = \frac{1}{2} (mv) v^2 = \frac{1}{2} m v^3 \]
Step 4: Conclusion
The rate at which kinetic energy is imparted to the water is $\frac{1}{2} m v^3$.
Final Answer: (A)
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