To solve the problem of calculating the work required to pump the liquid out of a conical tank, let's break it down systematically.

Key Data from the Image:

• Shape of the tank: Inverted cone
• Top radius of the cone (a): 15 m
• Height of the cone (b): 30 m
• Spout height (c): 3 m (This is where the liquid is pumped to)
• Density of the liquid (ρ): 1000 kg/m³ (density of water)
• Acceleration due to gravity (g): 9.8 m/s²

Step-by-Step Calculation:

1. Volume of a differential element (disk) at height $y$: The radius of the cone at any height $y$ is given by similar triangles: $r(y) = \frac{a}{b}y = \frac{15}{30}y = \frac{y}{2}$ The area of the cross-section at height $y$ is $\pi r^2 = \pi \left(\frac{y}{2}\right)^2 = \frac{\pi y^2}{4}$.

2. Volume of a thin slice (disk) at height $y$ with thickness $dy$: $dV = \frac{\pi y^2}{4} dy$

3. Mass of the thin slice: $dm = \rho dV = 1000 \times \frac{\pi y^2}{4} dy$

4. Work to lift a thin slice: The slice at height $y$ has to be lifted to a height of $c = 3 \, \text{m}$, so the distance lifted is $(b - y + c) = (30 - y + 3) = 33 - y$. The work to lift this slice is: $dW = (dm) \times g \times (33 - y) = \left(1000 \times \frac{\pi y^2}{4} dy\right) \times 9.8 \times (33 - y)$ Simplifying: $dW = 2450 \pi y^2 (33 - y) dy$

5. Total Work: To find the total work, integrate from $y = 0$ to $y = b = 30$: $W = \int_0^{30} 2450 \pi y^2 (33 - y) \, dy$

Let me now compute this integral.The total work required to pump all the liquid out of the tank is:

$W = 231525000 \pi \, \text{Joules}$

Now, let's approximate this in numerical form by evaluating $\pi \approx 3.1416$.The total work required to pump all the liquid out of the tank is approximately:

$W \approx 7.27 \times 10^8 \, \text{Joules}$

Would you like further details or have any questions?

Here are 5 related questions to expand on this concept:

1. How would the work change if the density of the liquid were different?
2. What would the total work be if the cone were only half full of liquid?
3. How does the shape of the container affect the work required to pump the liquid out?
4. How would the work change if the liquid were pumped to a different spout height?
5. What if the cone had a different radius-to-height ratio?

Tip: The work done in pumping depends both on the volume of the liquid and the height to which it needs to be pumped, which is why the shape of the container plays a crucial role.