Height of a tank in the form of an inverted cone is 10 m and radius of its circular base is 2 m. The tank contains water and it is leaking through a hole at its vertex at the rate of `0.02m^(3)//s.` Find the rate at which the water level changes and the rate at which the radius of water surface changes when height of water level is 5 m.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Tank
The tank is in the shape of an inverted cone with:
- Height (H) = 10 m
- Radius (R) = 2 m

### Step 2: Establish the Relationship Between Radius and Height
From the geometry of the cone, we can establish a relationship between the radius (r) of the water surface and the height (h) of the water level:
\[ \frac{r}{h} = \frac{R}{H} = \frac{2}{10} = \frac{1}{5} \]
This implies:
\[ r = \frac{h}{5} \]

### Step 3: Write the Volume of the Cone
The volume (V) of the cone filled with water is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \(r\) from the previous step:
\[ V = \frac{1}{3} \pi \left(\frac{h}{5}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{25} h = \frac{\pi h^3}{75} \]

### Step 4: Differentiate the Volume with Respect to Time
To find the rate at which the height of the water level changes, we differentiate the volume with respect to time (t):
\[ \frac{dV}{dt} = \frac{\pi}{75} \cdot 3h^2 \frac{dh}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = \frac{\pi h^2}{25} \frac{dh}{dt} \]

### Step 5: Substitute Known Values
We know that the volume is leaking at a rate of:
\[ \frac{dV}{dt} = -0.02 \, \text{m}^3/\text{s} \]
We need to find \(\frac{dh}{dt}\) when the height \(h = 5 \, \text{m}\):
\[ -0.02 = \frac{\pi (5)^2}{25} \frac{dh}{dt} \]
This simplifies to:
\[ -0.02 = \frac{\pi \cdot 25}{25} \frac{dh}{dt} = \pi \frac{dh}{dt} \]
Thus:
\[ \frac{dh}{dt} = -\frac{0.02}{\pi} \, \text{m/s} \]

### Step 6: Find the Rate of Change of Radius
Using the relationship \(r = \frac{h}{5}\), we differentiate with respect to time:
\[ \frac{dr}{dt} = \frac{1}{5} \frac{dh}{dt} \]
Substituting \(\frac{dh}{dt} = -\frac{0.02}{\pi}\):
\[ \frac{dr}{dt} = \frac{1}{5} \left(-\frac{0.02}{\pi}\right) = -\frac{0.004}{\pi} \, \text{m/s} \]

### Final Answers
- The rate at which the water level changes when the height is 5 m is:
\[ \frac{dh}{dt} = -\frac{0.02}{\pi} \, \text{m/s} \]
- The rate at which the radius of the water surface changes is:
\[ \frac{dr}{dt} = -\frac{0.004}{\pi} \, \text{m/s} \]