A water tank open at the top is hemispherical at the bottom and cylindrical above it. If radius of the hemisphere is 12 m and the total capacity of the tank is $3312\pi$ m$^3$, then the ratio of the surface areas of the hemispherical and the cylindrical portions is

### Concept & Composite Volumes The total volume of the tank is the sum of the volume of the hemisphere and the cylinder. Once the height of the cylinder is found, the surface areas can be compared. Volume of hemisphere: $$\frac{2}{3}\pi r^3$$ Volume of cylinder: $$\pi r^2 h$$ Curved Surface Area of hemisphere: $$2\pi r^2$$ Curved Surface Area of cylinder: $$2\pi r h$$ ### Step-by-Step Solution 1. The radius of both the hemisphere and cylinder is $r = 12$ m. 2. The total volume $V = \frac{2}{3}\pi r^3 + \pi r^2 h = 3312\pi$. 3. Substitute $r = 12$: $\frac{2}{3}\pi (12)^3 + \pi (12)^2 h = 3312\pi$ $\frac{2}{3}(1728) + 144h = 3312$ $1152 + 144h = 3312$ 4. Solve for $h$: $144h = 3312 - 1152 = 2160$ $h = \frac{2160}{144} = 15$ m. 5. Surface area of hemispherical portion (bottom) = $2\pi r^2 = 2\pi(12)^2 = 288\pi$. 6. Surface area of cylindrical portion (curved) = $2\pi r h = 2\pi(12)(15) = 360\pi$. 7. Find the ratio: $\frac{288\pi}{360\pi} = \frac{288}{360}$. Dividing by 72 gives $\frac{4}{5}$. ### Exam Strategy & Shortcut Instead of calculating full surface areas, use the ratio formula directly: $\frac{2\pi r^2}{2\pi rh} = \frac{r}{h}$. Once you find $h = 15$ and $r = 12$, the ratio is simply $\frac{12}{15} = \frac{4}{5}$. ### Common Pitfall A frequent error is including the flat circular top in the surface area of the cylinder, but the problem states the tank is "open at the top". ### Final Answer Therefore, the correct answer is **4 : 5**.