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It is given that the depth of the conical vessel with vertex downward is 5 cm.

Let the radius of the conical vessel be \[r\].

We know that volume of the cone is

\[V = \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot h\]

Therefore, the volume of the given conical vessel \[ = \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot 5\]

It is given that when we add 720 cu cm of water, the level rises to 10 cm.

Therefore, mathematically, we can write it as

\[ \Rightarrow \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot 5 + 720 = \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot 10\]

On further simplification, we get

\[ \Rightarrow 720 = \dfrac{{10}}{3}\pi {r^2} - \dfrac{5}{3}\pi {r^2}\]

On subtracting the like terms, we get

\[ \Rightarrow 720 = \dfrac{5}{3}\pi {r^2}\]

Dividing both sides by 5, we get

\[ \Rightarrow 144 = \dfrac{1}{3}\pi {r^2}\]

On rewriting the equation, we get

\[ \Rightarrow \dfrac{1}{3}\pi {r^2} = 144\] ……….. \[\left( 1 \right)\]

Here we need to find the quantity of water which must be added such that the level rises to 15 cm.

Let the required volume of added water be \[V\].

Therefore, mathematically, we can write the given condition as

\[ \Rightarrow \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot 10 + V = \dfrac{1}{3} \cdot \pi \cdot {r^2} \cdot 15\]

On further simplification, we get

\[ \Rightarrow V = \dfrac{{15}}{3}\pi {r^2} - \dfrac{{10}}{3}\pi {r^2}\]

On subtracting the like terms, we get

\[ \Rightarrow V = \dfrac{5}{3}\pi {r^2}\]

From equation 1, we have

\[ \Rightarrow \dfrac{1}{3}\pi {r^2} = 144\]

Now, we will substitute this value here.

\[ \Rightarrow V = 5 \times 144 = 720c{m^3}\]