To solve this problem, we need to calculate the work required to pump all the water from the tank over its top. The work done in pumping water out of the tank depends on the height that each small volume of water needs to be lifted.

Let's start by defining the problem mathematically.

Step-by-Step Approach:

1. Volume of a thin slice of water: Consider a small slice of water at a height $x$ meters above the bottom of the tank. The thickness of this slice is $dx$, and its radius is 5 meters (since it's a cylinder). The volume of this thin slice of water is given by: $dV = \pi r^2 dx = \pi (5^2) dx = 25\pi dx$

2. Mass of the slice: The mass of this slice of water can be found using the formula: $dm = \rho dV$ where $\rho = 1000 \, \text{kg/m}^3$ is the density of water. Thus, the mass of the slice is: $dm = 1000 \times 25\pi dx = 25000\pi dx$

3. Force required to lift the slice: The force required to lift this small slice of water is the weight of the slice, given by: $dF = dm \cdot g = 25000\pi dx \cdot 9.8 = 245000\pi dx$

4. Work required to lift the slice: The work to lift this small slice of water a distance $x$ to the top of the tank is: $dW = dF \cdot (12 - x) = 245000\pi (12 - x) dx$ where $12 - x$ is the height the water must be lifted to reach the top.

5. Total work to pump all the water: To find the total work, integrate $dW$ from the bottom of the water level (at $x = 0$) to the top of the water (at $x = 9$): $W = \int_0^9 245000\pi (12 - x) dx$

6. Performing the integration: Let's compute the integral: $W = 245000\pi \int_0^9 (12 - x) dx$ The integral of $(12 - x)$ is: $\int (12 - x) dx = 12x - \frac{x^2}{2}$ So, $W = 245000\pi \left[ 12x - \frac{x^2}{2} \right]_0^9$

7. Substitute the limits: $W = 245000\pi \left[ \left(12(9) - \frac{9^2}{2}\right) - \left(12(0) - \frac{0^2}{2}\right) \right]$ Simplifying the expression:

   = 245000\pi \left(108 - 40.5\right) = 245000\pi \times 67.5$$

8. Final calculation: Now, multiply the values: $W = 245000 \times 67.5 \times \pi \approx 245000 \times 67.5 \times 3.1416 = 52,003,797.43 \, \text{J}$

Thus, the total work required to pump all the water over the top of the tank is approximately 52,003,797.43 Joules.

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Would you like more details or have any questions?

Related Questions:

1. What is the work required if the water level was 6 meters instead of 9 meters?
2. How would the work change if the tank's radius was doubled?
3. What would be the total work if the tank were 15 meters high instead of 12 meters?
4. How would the work change if the tank were conical instead of cylindrical?
5. How much work is required to pump half of the water out of the tank?

Tip:

Always remember that work calculations for pumping fluids involve integrating the force over the distance the fluid is moved, accounting for the varying heights in the case of large tanks.