Water is poured into a cylindrical tank of base radius 3 m at the rate of \(18\pi\) m3/min. The rate at which the water level in the tank is rising is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 m/min
Let h be the depth of the water at time t. The radius stays 3 m, so the volume of water is
\( V = \pi (3)^{2} h = 9\pi h \)
Differentiating with respect to t:
\( \displaystyle \frac{dV}{dt} = 9\pi \frac{dh}{dt} \)
With \(\frac{dV}{dt} = 18\pi\):
\( \displaystyle \frac{dh}{dt} = \frac{18\pi}{9\pi} = 2 \) m/min
Since the cross-section is the same at every height, the level rises at this constant rate at every depth.