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

$\frac{2}{y} = \frac{4}{6+y}$

$2y = 6 + y$

$y = 6 cm$

now volume of the coffee will be

$V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)$

here we know that

$\frac{r}{y+6} = \frac{2}{6}$

$r = \frac{y+6}{3}$

$V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)$

now we know that volume flow rate is given as

$Q = \frac{dV}{dt}$

$20 cm^3/s = \frac{1}{3}\pi (\frac{1}{9})(3(y+6)^2)\frac{dy}{dt}$

$20 \times 9 = \pi (y + 6)^2 v$

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