Volume of cone questions

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.

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.

Slant height (l) = 5 cm
Height = 4 cm

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

volume of cone solved examples

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.

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.

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.

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.

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.

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