Question

Fermentation tanks are designed in the form of a cylinder mounted on a cone as shown below:

The total height of the tank is 3.3 m and the height of the conical part is 1.2 m. The diameter of the cylindrical as well as the conical part is 1 m. Find the capacity of the tank. If the level of liquid in the tank is 0.7 m from the top, find the surface area of the tank in contact with liquid.

Hint:

Use the formulas for volume and lateral surface area of cylinders and cones: \( V_{\text{cyl}} = \pi r^2 h \), \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \), \( A_{\text{lateral cone}} = \pi r l \), \( A_{\text{lateral cyl}} = 2\pi r h \)

Answer 1

Given:

• Total height = 3.3 m
• Height of cone = 1.2 m
• Height of cylinder = \( 3.3 - 1.2 = 2.1 \, \text{m} \)
• Diameter = 1 m \( \Rightarrow \) Radius \( r = \frac{1}{2} = 0.5 \, \text{m} \)

Capacity of the tank: Volume of cylinder: \[ V_{\text{cyl}} = \pi r^2 h = \pi (0.5)^2 \cdot 2.1 = \pi \cdot 0.25 \cdot 2.1 = 0.525\pi \, \text{m}^3 \] Volume of cone: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (0.5)^2 \cdot 1.2 = \frac{1}{3} \cdot \pi \cdot 0.25 \cdot 1.2 = 0.1\pi \, \text{m}^3 \] Total volume: \[ V_{\text{total}} = 0.525\pi + 0.1\pi = 0.625\pi \approx \boxed{1.9635 \, \text{m}^3} \]
Surface area in contact with liquid: Liquid height = \( 3.3 - 0.7 = 2.6 \, \text{m} \) Since cone height = 1.2 m, and 2.6 > 1.2, liquid fills entire cone and \( 2.6 - 1.2 = 1.4 \, \text{m} \) of cylinder.

• Lateral surface area of cone: \[ l = \sqrt{r^2 + h^2} = \sqrt{0.5^2 + 1.2^2} = \sqrt{0.25 + 1.44} = \sqrt{1.69} = 1.3 \] \[ \text{LSA}_{\text{cone}} = \pi r l = \pi \cdot 0.5 \cdot 1.3 = 0.65\pi \]
• Lateral surface area of cylinder part filled = \( 2\pi r h = 2\pi \cdot 0.5 \cdot 1.4 = 1.4\pi \)

Total surface area in contact with liquid: \[ A = 0.65\pi + 1.4\pi = 2.05\pi \approx \boxed{6.443 \, \text{m}^2} \]