A problem by Rohit Dhiman

Sorry about the diagrams. I do not have a great drawing tool for figures.

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Consider, the top cone, as shown above in the diagram. From the simple geometry of the figure, we have,

$\dfrac{H}{h} = \dfrac{R}{r} \implies \dfrac{2}{1} = \dfrac{R}{r} \implies r = \dfrac{R}{2}$

$\begin{aligned} \text{Initial volume of water in the top cone} & = & \dfrac{1}{3} \pi R^{2} H \\ & = & \dfrac{1}{3} \pi R^{2} \cdot 2 \\ & = & \dfrac{2}{3} \pi R^{2} \\ \text{Final volume of water in the top cone} & = & \dfrac{1}{3} \pi r^{2} h \\ & = & \dfrac{1}{3} \pi {\dfrac{R}{4}}^{2} \cdot 1 \\ & = & \dfrac{\pi R^{2}}{12} \\ \text{Change in volume of water} & = & \pi R^{2} \left( \dfrac{2}{3} - \dfrac{1}{12} \right) \\ & = & \dfrac{7 \pi R^{2}}{12} \end{aligned}$

This volume of water is now filled in the next cone.

Consider the bottom cone after the water has filled some portion of it (which has height $h$ to be calculated), as shown below in the diagram. again, we have,

$\dfrac{H}{h} = \dfrac{R}{r} \implies \dfrac{2}{h} = \dfrac{R}{r} \implies \color{#D61F06}{\dfrac{R^{2}}{r^{2}} = \dfrac{4}{h^{2}}}$

Now, volume of water filled in this cone can be represented as: $\dfrac{1}{3} \pi r^{2} h$

$\begin{aligned} \therefore \dfrac{1}{3} \pi r^{2} h & = & \dfrac{7 \pi R^{2}}{12} \\ \implies h & = & \dfrac{7}{4} \color{#D61F06}{\left( \dfrac{R^{2}}{r^{2}} \right)} \\ \implies h & = & \dfrac{7}{4} \color{#D61F06}{\left( \dfrac{4}{h^{2}} \right)} \\ \implies h & = & \dfrac{7}{h^{2}} \\ \implies h^{3} & = & 7 \\ \implies h & = & \boxed{\sqrt[3]{7}~ \text{cm}} \end{aligned}$

Please give solution