Question

At a cement plant, sand is stored in a container called a hopper. The shape of the container is an inverted cone. The height is 18 feet and the radius is 12 feet. If the hopper is full and sand is emptied at the rate of 4 ft3/min, how fast is the level of sand dropping when the sand level is 6 feet high? V = 1/3r2h​

Answer 1

`v=(0,25)/(\pi ) ft/min`

Explanation:

There is a relation with the initial and final dimensions of the hopper according to the sand level when its high is 6 feet as follows:

`(h_(1) )/(h_(2) )=(r_(1) )/(r_2)\\{r_2}=(12ft*6ft)/(18ft)\\ r_2=4ft`

Where the calculated
`r_(2)` is given with the 6 feet hgh.

Then we have the sand flow formula which is:

`Q=A*v`

Where A represents the area of the transversal section and v the velocity that we need to know, the area is:

`A=\pi r^(2)\\ A=\pi *4^(2)\\ A=16\pi ft^(2)`

And finally the sand is dropping when the level is 6 feet high with the velocity (v) :

`v=(Q)/(A) \\v=(4)/(16\pi )`

`v=(0.25)/(\pi ) ft/min`