2

Note: I am trying to prepare, i already see the answer to this question but i dont understand it.

It says volume of earth used for making the embarkment = $\pi (R^2 - r^2)h$

But i dont understand, what R? What r?The whole part solves out to 28$\pi$ but i dont understand where the two Rr come from and what are their values

The question is: A well of diameter 3m is dug 14m deep. The earth taken out has been spread evenly all around it in the shape of a circular ring of width 4m. Find the height of the embarkment.

I think that width refers to the diameter but i am not sure about that either, would be really helpful if someone explained the answer to me properly, my guide book sure doesnt D:

Aayush Agrawal
  • 1,207
  • 3
    It's not the answer but the question that you don't seem to understand. After the well was dug, the earth was spread around the hole forming a ring. $R$ is the outer radius and equals $3/2+4$. $r$ is the inner radius (radius of the well) and equals $3/2$. – Karolis Juodelė Mar 21 '13 at 13:02
  • Thanks alot for that, the diagram was misleading -.- – Aayush Agrawal Mar 21 '13 at 13:06

2 Answers2

2

$d$ = the inner diameter of the ring, defined as $3$ meters

$r$ = the inner radius of the ring, defined as $\frac{d}{2} = 1.5$ meters

$R$ = the outer radius of the ring, defined as $r+4 = 5.5$ meters

$h_d$ = the height of the well, defined as $14$ meters

The volume of earth taken out of the well is defined by $\pi \times r^2 \times h_d = \pi \times 31.5$

Now, we need to solve the equation, which gives us: $$\pi \times r^2 \times h_d = \pi(R^2 - r^2)h\\ h = \frac{\pi \times r^2 \times h_d}{\pi(R^2 - r^2)}\\ h = \frac{\pi \times 31.5}{\pi \times 28} = \frac{9}{8} \pi $$

Therefore, the height of the embarkment is $\frac{9}{8} \pi \approx 3.5$ meters.

Roelof van den Berg
  • 126
1

The total volume of earth taken out is $r^2\pi\cdot h=3^2\pi\cdot14$. The volume of the new pile is $(R^2-r^2)\pi\cdot H=(4^2-3^2)\pi H$, where you need to find $H$. Compare the two, and express $H$.

Dennis Gulko
  • 15,640