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.)

Respuesta :

Let the cup is filled to height h after some time

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

[tex]\frac{2}{y} = \frac{4}{6+y}[/tex]

[tex]2y = 6 + y[/tex]

[tex]y = 6 cm[/tex]

now volume of the coffee will be

[tex]V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)[/tex]

here we know that

[tex]\frac{r}{y+6} = \frac{2}{6}[/tex]

[tex]r = \frac{y+6}{3}[/tex]

[tex]V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)[/tex]

now we know that volume flow rate is given as

[tex]Q = \frac{dV}{dt}[/tex]

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

[tex]20 \times 9 = \pi (y + 6)^2 v[/tex]

here y = 3 cm

[tex]180 = \pi (9)^2 v[/tex]

[tex]v = 0.71 cm/s[/tex]

so water will rise up with speed 0.71 cm/s