In this chapter we will discuss questions related to volume of cone.

The **formula for volume of cone** is given as;

Volume = \mathtt{\frac{1}{3} \pi r^{2} h}

r = radius of base

h = height of cone

Memorize this formula as it would help to solve the below questions.

## Solved problems on volume of cone

**Question 01**

Find the volume of cone of radius of base is 5 cm and height is 14 cm.

**Solution**

Radius = 5 cm

Height = 14 cm

Volume = \mathtt{\frac{1}{3} \pi r^{2} h}

Putting the values in formula;

\mathtt{Volume\ =\ \frac{1}{3} \pi r^{2} h}\\\ \\ \mathtt{Volume\ =\ \frac{1}{3} \times \frac{22}{7} \times ( 5)^{2} \times 14}\\\ \\ \mathtt{Volume\ =\ 366.67\ cm^{3}}

Hence, **366.67 cu cm** is the volume of cone.

**Question 02**

Find the volume of right circular cone if slant height is 28 cm and height 21 cm.

**Solution**

Slant height (l) = 5 cm

Height = 4 cm

To calculate the volume, we need to find radius first.

Consider the right triangle AOB.

Applying Pythagoras theorem;

\mathtt{AB^{2} =AO^{2} +OB^{2}}\\\ \\ \mathtt{5^{2} =\ 4^{2} +r^{2}}\\\ \\ \mathtt{r^{2} =25-16}\\\ \\ \mathtt{r^{2} \ =\ 9\ }

Now use the volume formula for cone;

\mathtt{Volume\ =\frac{1}{3} \pi r^{2} h}\\\ \\ \mathtt{Volume\ =\ \frac{1}{3} \times \frac{22}{7} \times 9\times 4=37.71\ cm^{3}}

Hence, 37.71 cu cm is the volume of given cube.

**Question 03**

Given is the conical vessel of radius 9 cm and height 21 cm. Find the volume of water that can be filled in the vessel in liters.

**Solution**

Radius (r) = 9 cm

Height (h) = 21 cm

To find the capacity of conical vessel, we need to calculate the volume.

Volume = \mathtt{\frac{1}{3} \pi r^{2} h}

Putting the values;

\mathtt{Volume\ =\ \frac{1}{3} \times \frac{22}{7} \times 9^{2} \times 21=1782\ cm^{3}}

So the volume of vessel is 1782 cubic cm.

Now we need to get the capacity in liters.

1000 cu cm = 1 liter

1782 cu cm = 1782/ 1000 = 1.782 liters

Hence, the **conic vessel can fill 1.782 liters of water**.

**Question 04**

Two cones are given whose height are in ratio 1 : 4 and radius are in ratio 4 : 1. Calculate the ratio of their volumes.

**Solution**

Let r1 & h1 are radius and height of 1st cone.

Let r2 & h2 are radius and height of 2nd cone.

Its given that;

\mathtt{\frac{h1}{h2} =\frac{1}{4}}\\\ \\ \mathtt{\frac{r1}{r2} =\frac{4}{1}}

Now calculating the ratio of volume of two cubes;

\mathtt{\frac{v1}{v2} =\frac{\frac{1}{3} .\mathtt{\pi .r_{1}^{2} .\ h}{1}}{\frac{1}{3} .\mathtt{\pi .r{2}^{2} .\ h}{2}}}\\\ \\ \mathtt{\frac{v1}{v2} =\frac{\mathtt{r{1}^{2} .\ h}{1}}{\mathtt{r{2}^{2} .\ h}_{2}} \ =\ \frac{4^{2}}{4} =4}

Hence, the ratio of their volume is 4 : 1

**Question 05**

Consider a cone in which radius and height are in ratio 5 : 12. Find the value of radius of the base, if the volume of cone is 2512 cu cm.

**Solution**

Radius = 5x

Height = 12x

Volume = 2512 cu cm

To find the value of radius of base of the cone.

Volume = \mathtt{\frac{1}{3} \pi r^{2} h}

Putting the values in formula;

\mathtt{2512=\ \frac{1}{3} \times \frac{22}{7} \times ( 5x)^{2} \times 12x}\\\ \\ \mathtt{x^{3} =\frac{2512\times 21}{22\times 25\times 12}}\\\ \\ \mathtt{x^{3} =\ 8}\\\ \\ \mathtt{x\ =\ 2}

We know that;

Radius = 5x

Putting the value of x;

Radius = 5 (2) = 10 cm

Hence, **the radius of given cone is 10 cm.**

**Question 06**

Imagine that for a given cone, the radius get tripled and height is same. Find the ratio of volume of new cone to old cone.

**Solution**

The dimension of original cone is given as;

Radius = r

Height = h

The dimension of new cone is given as;

Radius = 3r

Height = h

Now let’s calculate the ratio of volume of new to old cube.

\mathtt{\frac{v( new)}{v( old)} =\frac{\frac{1}{3} .\mathtt{\pi .( 3r)^{2} .\ h}}{\frac{1}{3} .\mathtt{\pi .r^{2} .h}}}\\\ \\ \mathtt{\frac{v( new)}{v( old)} =\frac{9}{1}}

Hence, **9: 1 is the required ratio**.