The problem involves finding the work required to pump water out of a conical tank that is fully filled with water. Let's break it down step by step.

Problem Summary:

• Tank Shape: Inverted cone
• Height of the cone (H): 7 meters
• Base radius of the cone (R): 5 meters
• Density of water (ρ): $9800 \, \text{N/m}^3$
• Goal: Calculate the work required to pump all the water out of the tank.

Solution:

1. Volume Element Consideration:

   • Consider a thin slice of water at a height $y$ from the vertex (bottom) of the cone with thickness $dy$.
   • At height $y$, the radius $r$ of the water slice is given by similar triangles: $\frac{r}{y} = \frac{R}{H} = \frac{5}{7} \implies r = \frac{5}{7}y$
   • The area $A$ of this circular slice is: $A = \pi r^2 = \pi \left(\frac{5}{7}y\right)^2 = \pi \frac{25}{49} y^2$
   • The volume of the slice $dV$ is: $dV = A \, dy = \pi \frac{25}{49} y^2 \, dy$

2. Work to Move the Water Slice:

   • The weight of the slice of water $dW$ is: $dW = \rho \, dV = 9800 \times \pi \frac{25}{49} y^2 \, dy$
   • The distance $y$ that this slice needs to be lifted to the top of the tank is $(H - y) = (7 - y)$.
   • The work $d\text{Work}$ needed to lift this slice is: $d\text{Work} = dW \times (7 - y) = 9800 \times \pi \frac{25}{49} y^2 \times (7 - y) \, dy$

3. Total Work:

   • To find the total work required to pump all the water out, integrate the above expression from $y = 0$ to $y = 7$: $\text{Work} = 9800 \times \pi \frac{25}{49} \int_{0}^{7} y^2(7 - y) \, dy$

4. Solving the Integral: Let's calculate the integral: $\int_{0}^{7} y^2(7 - y) \, dy = \int_{0}^{7} (7y^2 - y^3) \, dy$ This integral can be split and solved separately: [ \int_{0}^{7} 7y^2 , dy - \int_{0}^{7} y^3 , dy = \left[ \frac{7y^3}{3} \right]_0^7 - \left[ \frac{y^4}{4} \right]_0^7 ] Substituting $y = 7$: $\frac{7(7)^3}{3} - \frac{(7)^4}{4} = \frac{7 \times 343}{3} - \frac{2401}{4}$ $= \frac{2401}{3} - \frac{2401}{4} = 2401 \left(\frac{1}{3} - \frac{1}{4}\right) = 2401 \left(\frac{4 - 3}{12}\right) = 2401 \times \frac{1}{12} = 200.08$

5. Final Work Calculation:

   • The work is then: $\text{Work} = 9800 \times \pi \frac{25}{49} \times 200.08$ Simplifying: $\text{Work} = 5000 \times \pi \times 200.08 \approx 1000 \times \pi \times 200 = 1,000,400 \pi \, \text{Joules}$

So, the exact work required is: $\boxed{1,000,400 \pi \, \text{Joules}}$

Do you want more details or have any questions about the solution?

Further Questions:

1. How would the work change if the cone was only partially filled with water?
2. What if the shape of the tank was cylindrical instead of conical?
3. How does the density of the fluid affect the work required to pump it out?
4. How would the problem change if the tank had a different orientation (e.g., right-side up)?
5. Can you determine the time required to pump the water out if the pump has a specific power rating?

Tip:

When solving problems involving work and fluid mechanics, it’s important to carefully consider the relationship between the volume element and the height or distance it must be moved.