Question

A conical vessel with vertex downward contains water to a depth of 5 cm. On adding 720 cu cm of water, the level rises to 10 cm. Find the quantity of water which must be added in order that the level may rise to 15 cm.

Answer 1

Hint: Here we need to find the quantity of water which must be added such that the level rises to the given height. We will first assume the radius of the cone to be any variable and then we will form the first equation using the first condition and then we will form the second equation using the second condition in the question. From there, we will get the required answer.

Complete step-by-step answer:
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.
\[ \Rightarrow V = 5 \times 144 = 720c{m^3}\]
Hence, the required quantity of water added is equal to \[720c{m^3}\].

Note: Since we have calculated the volume of the cone. Volume of any object is defined as the amount of space that an object occupies. Thus, we can say that the space available in the cone can be determined by its volume. Also the total amount of water added to a cone is equal to the total volume of the cone.