Step 1: Understanding the Concept:
We use the volume of a cone, \( V = \frac{1}{3}\pi r^2 h \), and related rates. By similar triangles, \( r/h = 10/20 = 1/2 \), so \( r = h/2 \).
Step 2: Key Formula or Approach:
Substitute \( r = h/2 \) into the volume: \( V = \frac{1}{3}\pi (h/2)^2 h = \frac{\pi h^3}{12} \).
Differentiate with respect to time \( t \): \( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt} \).
Step 3: Detailed Explanation:
When water is 5 cm from the top, depth \( h = 20 - 5 = 15 \) cm.
Given \( dV/dt = -0.15\pi \) (leaking).
\[ -0.15\pi = \frac{\pi (15)^2}{4} \cdot \frac{dh}{dt} \]
\( -0.15 = \frac{225}{4} \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-0.15 \times 4}{225} = \frac{-0.6}{225} = -\frac{6}{2250} = -\frac{1}{375} \text{ cm/s} \).
The water level is dropping at \( 1/375 \) cm/s.
Step 4: Final Answer:
The rate is \( 1/375 \) cm/s.