Question:
If a cylindrical tank of radius 3 m is filled with water at the rate of \(\frac{3}{2} \, m^3/sec \), then the rate of change of its water level in (m/sec) is:
If a cylindrical tank of radius 3 m is filled with water at the rate of \(\frac{3}{2} \, m^3/sec \), then the rate of change of its water level in (m/sec) is:
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When working with related rates in geometry, always identify the constant dimensions before differentiating the volume formula to simplify your calculations.
Updated On: Jun 9, 2026
- \( \frac{1}{3\pi} \)
- \( \frac{1}{2\pi} \)
- \( \frac{1}{\pi} \)
- \( \frac{1}{6\pi} \)
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The Correct Option is D
Solution and Explanation
Concept:
The volume \(V\) of a cylinder with radius \(r\) and height \(h\) is defined as:
\[
V = \pi r^2 h
\]
To find the rate of change of the water level (\(\frac{dh}{dt}\)), we differentiate this volume formula with respect to time \(t\), keeping in mind that the radius \(r\) is constant for a cylindrical tank.
Step 1: Identify the given physical parameters and the rate of change of volume.
We are given:
• Radius of the tank, \(r = 3 \, \text{m}\) (which is constant)
• Rate of volume filling, \(\frac{dV}{dt} = \frac{3}{2} \, \text{m}^3/\text{sec}\)
Step 2: Differentiate the volume formula with respect to time \(t\).
Using the chain rule for the volume equation \(V = \pi r^2 h\): \[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]
Step 3: Substitute the values and solve for the rate of change of the water level \(\frac{dh}{dt}\).
Substituting \(r = 3\) and \(\frac{dV}{dt} = \frac{3}{2}\) into the differentiated equation: \[ \frac{3}{2} = \pi (3)^2 \frac{dh}{dt} \] Simplifying the squared term: \[ \frac{3}{2} = 9\pi \frac{dh}{dt} \] Solving for \(\frac{dh}{dt}\) by dividing both sides by \(9\pi\): \[ \frac{dh}{dt} = \frac{3}{2 \times 9\pi} \] \[ \frac{dh}{dt} = \frac{3}{18\pi} \] \[ \frac{dh}{dt} = \frac{1}{6\pi} \]
Conclusion: The rate of change of the water level is \(\frac{1}{6\pi} \, \text{m/sec}\). center Final Answer: (D) center
Step 1: Identify the given physical parameters and the rate of change of volume.
We are given:
• Radius of the tank, \(r = 3 \, \text{m}\) (which is constant)
• Rate of volume filling, \(\frac{dV}{dt} = \frac{3}{2} \, \text{m}^3/\text{sec}\)
Step 2: Differentiate the volume formula with respect to time \(t\).
Using the chain rule for the volume equation \(V = \pi r^2 h\): \[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]
Step 3: Substitute the values and solve for the rate of change of the water level \(\frac{dh}{dt}\).
Substituting \(r = 3\) and \(\frac{dV}{dt} = \frac{3}{2}\) into the differentiated equation: \[ \frac{3}{2} = \pi (3)^2 \frac{dh}{dt} \] Simplifying the squared term: \[ \frac{3}{2} = 9\pi \frac{dh}{dt} \] Solving for \(\frac{dh}{dt}\) by dividing both sides by \(9\pi\): \[ \frac{dh}{dt} = \frac{3}{2 \times 9\pi} \] \[ \frac{dh}{dt} = \frac{3}{18\pi} \] \[ \frac{dh}{dt} = \frac{1}{6\pi} \]
Conclusion: The rate of change of the water level is \(\frac{1}{6\pi} \, \text{m/sec}\). center Final Answer: (D) center
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