Question

Two identical right circular cones each of height $2cm$ are placed as shown in diagram (each is vertical apex downward). At the start, the upper cone is full of water and lower cone is empty. Then water drips down through a hole in the apex of upper come into the lower cone. The height of water in the lower cone at the moment when height of water in upper cone is $1cm$ is:

• A. $1cm$
• B. $\sqrt{\frac{1}{2}}cm$
• C. $\sqrt{\frac{1}{4}}cm$
• D. $\sqrt[3]{37}cm$

Answer 1

The correct option is D $\sqrt[3]{37}cm$

Let the radius of the two cone be $R$ cm and the height of the water in the lower cone be $h$ cm .

Now,

$△OAB\sim △OCD\therefore \frac{AB}{CD}=\frac{OA}{OC}=\frac{OB}{OD}\Rightarrow \frac{AB}{R}=\frac{1}{2}\Rightarrow AB=\frac{R}{2}\because △O^{\prime }{A}^{\prime }{B}^{\prime }\sim △O^{\prime }{C}^{\prime }{D}^{\prime }\because \frac{A^{\prime }{B}^{\prime }}{C^{\prime }{D}^{\prime }}=\frac{O^{\prime }{A}^{\prime }}{O^{\prime }{C}^{\prime }}=\frac{O^{\prime }{B}^{\prime }}{O^{\prime }{D}^{\prime }}\Rightarrow \frac{A^{\prime }{B}^{\prime }}{R^{\prime }}=\frac{h}{2}\Rightarrow A^{\prime }{B}^{\prime }=\frac{h{R}^{\prime }}{2}$

Volume of frustum of upper cone $=$ volume of water in lower cone

$\frac{1}{3}\pi \left(1\right)[R^2+{\frac{R}{2}}^2+R.\frac{R}{2}]=\frac{1}{3}\pi {\frac{Rh}{2}}^2\times h$
$R^2+\frac{R^2}{4}+\frac{R^2}{2}=\frac{R^2{h}^3}{4}$
$\frac{7R^2}{4}=\frac{R^2{h}^3}{4}$
$h^3=7$
$h=\sqrt[3]{37}cm$