Question

Coffee is poured at a uniform rate at 20 cm3 / s into a cup whose inside is shaped like a truncated cone. if the upper and lower radii of the cup are 4 cm and 2 cm and the height of the cup is 6 cm, how fast will the coffee be rising when the coffee is halfway up (hint: extend the cup downward to form a cone.)

Answer 1

Let the cup is filled to height h after some time

now the total volume of coffee filled in the cup is given as

`(2)/(y) = (4)/(6+y)`

`2y = 6 + y`

`y = 6 cm`

now volume of the coffee will be

`V = (1)/(3)\pi r^2(y + 6) - (1)/(3)\pi 2^2 (6)`

here we know that

`(r)/(y+6) = (2)/(6)`

`r = (y+6)/(3)`

`V = (1)/(3)\pi ((y+6)/(3))^2(y+6) - (1)/(3)\pi 2^2(6)`

now we know that volume flow rate is given as

`Q = (dV)/(dt)`

`20 cm^3/s = (1)/(3)\pi ((1)/(9))(3(y+6)^2)(dy)/(dt)`

`20 * 9 = \pi (y + 6)^2 v`

here y = 3 cm

`180 = \pi (9)^2 v`

`v = 0.71 cm/s`

so water will rise up with speed 0.71 cm/s