Cups A and B are cone shaped and have heights of 16 cm and 15 cm and openings with radii of 6 cm and 4 cm, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Jan 28, 2018

$11.95$ cm high of cup A will be filled.

Explanation:

Height and radius of cup A is ${h}_{a} = 16$ cm , ${r}_{a} = 6$ cm

Volume of cup is $\frac{1}{3} \cdot \pi \cdot {r}^{2} \cdot h \therefore {V}_{a} = \frac{1}{3} \cdot \pi \cdot {6}^{2} \cdot 16$ or

${V}_{a} \approx 603.19$ cubic cm.

Height and radius of cup B is ${h}_{b} = 15$ cm , ${r}_{b} = 4$ cm

$\therefore {V}_{b} = \frac{1}{3} \cdot \pi \cdot {4}^{2} \cdot 15 \mathmr{and} {V}_{b} \approx 251.33$ cubic cm.

Since ${V}_{a} > {V}_{b}$ , the content will not overflow.

The ratio of radius and hight of cup A is

$\frac{r}{h} = \frac{12}{32} = \frac{3}{8}$. The ratio of radius and hight of cup A filled

with water is ${r}_{w} / {h}_{w} = \frac{6}{16} = \frac{3}{8} \mathmr{and} {r}_{w} = \frac{3 \cdot {h}_{w}}{8}$

The volume of water cone is ${V}_{w} = 251.33$ cubic cm.

$\therefore \frac{1}{3} \cdot \pi \cdot {r}_{w}^{2} \cdot {h}_{w} = 251.33$ or

$\frac{1}{3} \cdot \pi \cdot {\left(\frac{3 \cdot {h}_{w}}{8}\right)}^{2} \cdot {h}_{w} = 251.33$ or

${h}_{w}^{3} = \frac{251.33 \cdot 64}{3 \cdot \pi} \approx 1706.684$ or

${h}_{w} = \sqrt[3]{1706.684} \approx 11.95 \left(2 \mathrm{dp}\right)$ cm

$11.95$ cm high of cup A will be filled [Ans]