Question

An inverted cone (vertex is down) with height 12 inches and base of radius 8 inches is being filled with water. What is the height of the water when the cone is half filled?

• A. $6$
• B. $6\sqrt[3]{34}$
• C. $8\sqrt[3]{36}$
• D. $9\sqrt[3]{34}$
• E. $9\sqrt[3]{36}$

Answer 1

The correct option is B $6\sqrt[3]{34}$

$△OAB\sim △OCB$

$\Rightarrow \frac{OA}{O^{1}C}=\frac{OB}{O^{1}B}\Rightarrow \frac{8}{O^{1}C}=\frac{12}{h}$

$\Rightarrow \frac{O^{1}C}{h}=\frac{2}{3}\Rightarrow O^{1}C=\frac{2h}{3}$ and depth (height) of the cone half full $h=\frac{{30}^{1}C}{2}$. So the radius of half volume, $O^{1}C=\frac{2h}{3}$.

$\therefore$ Volume of cone = $\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 8\times 8\times 12=256\pi cu.in$

$\Rightarrow \frac{256\pi }{2}=\frac{1}{3}\pi {\left(\frac{2h}{3}\right)}^2\times h$

So, the volume of the half full cone is : $128\pi =\frac{1}{3}{\left(\frac{2h}{3}\right)}^2\times h\times \pi$

$\Rightarrow 128=\frac{1}{3}\times \frac{4{\pi }^3}{9}\Rightarrow h^3=\frac{128\times 27}{4}=32\times 27$

$\Rightarrow h=\sqrt[3]{38\times 4\times 27}=2\times 3\sqrt[3]{34}=6\sqrt[3]{34}$