Question

A vertical cone of volume v with vertex downwards is filled with water up to half of its height. The volume of the water is
(a) $ \dfrac{v}{16} $
(b) $ \dfrac{v}{8} $
(c) $ \dfrac{v}{4} $
(d) $ \dfrac{v}{2} $

Answer 1

Hint: First we need to draw the diagram with the conditions provided, then use the AA criteria to prove the similarity of triangles. Use the ratio of sides of the similar triangles and find the radius of the half-filled cone with water and then find the Volume of water to how many times the volume of the cone.

Complete step-by-step answer:
Here we have a vertical cone with its vertex pointed downwards and it is filled with water up to half of its height. Let us calculate the volume occupied by the water.
Let us first draw the figure according to the conditions given,

We know that volume of a cone, $ v=\dfrac{1}{3}\pi {{\text{r}}^{2}}\text{h} $ ................ (i)
where r = radius of the base, h = height of the cone.
Also, we have the water filled up to half of the cone, so its height will be $ \dfrac{1}{2}\times \text{h} $ .
Here, if observed properly, we have two triangles, $ \vartriangle \text{AOB} $ and $ \vartriangle \text{COD} $
We can see over here, $ \angle \text{AOB = }\angle \text{COD} $
                                          $ \angle \text{OAB = }\angle \text{OCD = 90}{}^\circ $
We also see that side AB is parallel to side CD
So, according to AA criterion, $ \vartriangle \text{AOB}\sim \vartriangle \text{COD} $
Therefore, we can say that both the triangles are similar triangles.
Now, let us take the ratio of the sides
 $ \dfrac{\text{AB}}{\text{CD}}=\dfrac{\text{AO}}{\text{CO}} $
We know, AB = r, AO = h, CD = $ \dfrac{\text{h}}{2} $
After substituting the values, $ \dfrac{\text{r}}{\text{CD}}=\dfrac{\text{h}}{\dfrac{\text{h}}{2}} $
                                                      $ \dfrac{\text{r}}{\text{CD}}=2 $
Therefore, the value of CD = $ \dfrac{\text{r}}{2} $
To find the volume occupied by water,
 $ \begin{align}
  & V=\dfrac{1}{3}\pi {{\left( \text{CD} \right)}^{2}}\left( \text{OC} \right) \\
 & =\dfrac{1}{3}\pi {{\left( \dfrac{\text{r}}{2} \right)}^{2}}\left( \dfrac{\text{h}}{2} \right) \\
 & =\dfrac{1}{3}\pi \left( \dfrac{{{\text{r}}^{2}}}{4} \right)\left( \dfrac{\text{h}}{2} \right)
\end{align} $
 $ V=\dfrac{1}{8}\cdot \dfrac{1}{3}\pi \cdot \left( {{\text{r}}^{2}} \right)\left( \text{h} \right) $ …………………… (ii)
Substitute the equation (ii) in equation (i), we get
 $ \begin{align}
  & V=\dfrac{1}{8}\cdot v \\
 & =\dfrac{v}{8}
\end{align} $
Hence, the volume of water is $ \dfrac{v}{8} $ .

Note: In this question, the description of the diagram is important and it solves most of the problems. There is also one more measurement which is not mentioned, which is slant height, it can be found out by using pythagoras theorem.