NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas And Volumes Ex 11.3

Exercise - 11.3

1. Find the volume of the right circular cone with

i. Radius $\text{6 cm}$, height $\text{7 cm}$

Ans:

It is given the radius of cone $\text{r = 6 cm}$

The height of the cone $\text{h = 7 cm}$

The volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{6}\right)}^{\text{2}}\times \text{ 7}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{12 }\times \text{ 22})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 264 c}{\text{m}}^{\text{3}}$

The volume of the right circular cone is $\text{264 c}{\text{m}}^{\text{3}}$.

ii. Radius $\text{3}\text{.5 cm}$, height $\text{12 cm}$ $\left[\text{Assume }\pi \text{ =}\frac{\text{22}}{\text{7}}\right]$

Ans:

It is given the radius of cone $\text{r = 3}\text{.5 cm}$

The height of the cone $\text{h = 12 cm}$

The volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{3}\text{.5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{1}\text{.75 }\times \text{ 88})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 154 c}{\text{m}}^{\text{3}}$

The volume of the right circular cone is $\text{154 c}{\text{m}}^{\text{3}}$.

2. Find the capacity in litres of a conical vessel with

i. Radius $\text{7 cm}$, slant height $\text{25 cm}$

Ans:

It is given the radius of cone $\text{r = 7 cm}$

The slant height of the cone $\text{l = 25 cm}$

So, the height of the cone $\text{h = }\sqrt{{\text{l}}^{\text{2}}\text{ - }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{h = }\sqrt{\text{2}{\text{5}}^{\text{2}}\text{ - }{\text{7}}^{\text{2}}}\text{ cm}$

$\Rightarrow \text{h = 24 cm}$

The volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{7}\right)}^{\text{2}}\times \text{ 24}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{154 }\times \text{ 8})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 1232 c}{\text{m}}^{\text{3}}$

We know that $\text{1000 c}{\text{m}}^{\text{3}}\text{ = 1 litre}$

So, the capacity of the conical vessel $\text{= }\frac{\text{1232}}{\text{1000}}\text{ = 1}\text{.232 litres}$

Therefore, the capacity of the conical vessel is $\text{1}\text{.232 litres}$.

ii. Height $\text{12 cm}$, slant height $\text{13 cm}$ $\left[\text{Assume }\pi \text{ =}\frac{\text{22}}{\text{7}}\right]$

Ans:

It is given the height of cone $\text{h = 12 cm}$

The slant height of the cone $\text{l = 13 cm}$

So, the radius of the cone $\text{r = }\sqrt{{\text{l}}^{\text{2}}\text{ - }{\text{h}}^{\text{2}}}$

$\Rightarrow \text{r = }\sqrt{\text{1}{\text{3}}^{\text{2}}\text{ - 1}{\text{2}}^{\text{2}}}\text{ cm}$

$\Rightarrow \text{r = 5 cm}$

The volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }(\text{4 }\times \frac{\text{22}}{\text{7}}\times \text{ 25})\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = }\frac{2200}{7}\text{ c}{\text{m}}^{\text{3}}$

We know that $\text{1000 c}{\text{m}}^{\text{3}}\text{ = 1 litre}$

So, the capacity of the conical vessel $\text{= }\frac{\text{2200}}{\text{7}}\times \frac{\text{1}}{\text{1000}}\text{ = 0}\text{.314 litres}$

Therefore, the capacity of the conical vessel is $\text{0}\text{.314 litres}$.

3. The height of a cone is $\text{15 cm}$. It its volume is $\text{1570 c}{\text{m}}^{\text{3}}$, find the diameter of its base. $\left[\text{Use }\pi \text{ = 3}\text{.14}\right]$

Ans:

It is given the height of cone $\text{h = 12 cm}$

Let us assume the radius of the cone be $\text{r}$.

The volume of the cone is $\text{V = 1570 c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore \frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h = 1570 c}{\text{m}}^{\text{3}}$

$\Rightarrow [\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{r}\right)}^{\text{2}}\times \text{ 12}]\text{ cm = 1570 c}{\text{m}}^{\text{3}}$

$\Rightarrow {\text{r}}^{\text{2}}\text{ = 100 c}{\text{m}}^{\text{2}}$

$$\Rightarrow \text{r = 10 cm}$$

Diameter of base $$\text{= 2r = 20 cm}$$

Therefore, the diameter of the cone is  $$\text{20 cm}$$.

4. If the volume of right circular cone of height $\text{9 cm}$ is $\text{48 }\pi \text{ c}{\text{m}}^{\text{3}}$, find the diameter of its base.

Ans:

It is given the height of cone $\text{h = 9 cm}$

Let us assume the radius of the cone is $\text{r}$.

The volume of the cone is $\text{V = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore \frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow [\frac{\text{1}}{\text{3}}\times \pi \times {\left(\text{r}\right)}^{\text{2}}\times \text{ 9}]\text{ cm = 48 }\pi \text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow {\text{r}}^{\text{2}}\text{ = 16 c}{\text{m}}^{\text{2}}$

$$\Rightarrow \text{r = 4 cm}$$

Diameter of base $$\text{= 2r = 8 cm}$$

Therefore, the diameter of the base of the cone is $$\text{8 cm}$$.

5. A conical pit of top diameter $\text{3}\text{.5 m}$ is $\text{12 m}$ deep. What is the capacity in kilolitres? $\left[\text{Assume }\pi \text{ =}\frac{\text{22}}{\text{7}}\right]$

Ans:

It is given the height of conical pit $\text{h = 12 m}$

The radius of conical pit $\text{r = }\frac{\text{3}\text{.5}}{\text{2}}\text{ m = 1}\text{.75 m}$

We know the volume of the conical pit $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{\text{22}}{\text{7}}\times {\left(\text{1}\text{.75}\right)}^{\text{2}}\times \text{ 12}]{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 38}\text{.5 }{\text{m}}^{\text{3}}$

We know that $\text{1 kilolitre = 1 }{\text{m}}^{\text{3}}$

So, the capacity of the pit $\text{= }(\text{38}\text{.5 }\times \text{ 1})\text{ kilolitres = 38}\text{.5 kilolitres}$

Therefore, the capacity of the conical pit is $\text{38}\text{.5 kilolitres}$.

6. The volume of a right circular cone is $\text{9856 c}{\text{m}}^{\text{3}}$. If the diameter of the base is $\text{28 cm}$, find

i. Height of the cone

Ans:

It is given the diameter of base of cone $\text{= 28 cm}$

So, the radius $\text{r = }\frac{\text{28}}{\text{2}}\text{ = 14 cm}$

Let us assume the height of the cone is $\text{h}$.

The volume of the cone is $\text{V = 9856 c}{\text{m}}^{\text{3}}$

We know the formula for the volume of the cone $\text{= }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\therefore \frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h = 9856 c}{\text{m}}^{\text{3}}$

$\Rightarrow [\frac{\text{1}}{\text{3}}\times \frac{22}{7}\times {\left(\text{14}\right)}^{\text{2}}\times \text{ h}]\text{ c}{\text{m}}^{\text{2}}\text{ = 9856 c}{\text{m}}^{\text{3}}$

$$\Rightarrow \text{h = }\left(\frac{\text{9856 }\times \text{ 21}}{\text{22 }\times \text{ 196}}\right)\text{ cm}$$

$$\Rightarrow \text{h = 48 cm}$$

Therefore, the height of the cone is $$\text{48 cm}$$.

ii. Slant height of the cone

Ans:

The slant height of the cone $\text{l = }\sqrt{{\text{h}}^{\text{2}}\text{ + }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{l = }\sqrt{\text{4}{\text{8}}^{\text{2}}\text{ + 1}{\text{4}}^{\text{2}}}\text{ cm}$

$\Rightarrow \text{l = }\sqrt{\text{2304 + 196}}\text{ cm}$

$\Rightarrow \text{l = 50 cm}$

Therefore, the slant height of the cone is $\text{50 cm}$.

iii. Curved surface area of the cone. $\left[\text{Assume }\pi \text{ =}\frac{\text{22}}{\text{7}}\right]$

Ans:

The curved surface area of the cone $\text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }(\frac{\text{22}}{\text{7}}\times \text{ 14 }\times \text{ 50})\text{ c}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 2200 c}{\text{m}}^{\text{2}}$

Therefore, the curved surface area of the cone is $\text{2200 c}{\text{m}}^{\text{2}}$.

7. A right triangle $\mathrm{\Delta }\text{ ABC}$ with sides $\text{5 cm}$,$\text{12 cm}$ and $$\text{13 cm}$$ is revolved about the side $\text{12 cm}$. Find the volume of the solid so obtained.

Ans:

(Image will be uploaded soon)

If the triangle is revolved about the side $\text{12 cm}$, we will get a cone with:

Radius $\text{r = 5 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 12 cm}$

We know the volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \pi \times {\left(\text{5}\right)}^{\text{2}}\times \text{ 12}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 100 }\pi \text{ c}{\text{m}}^{\text{3}}$

Therefore, the volume of the cone will be $\text{100 }\pi \text{ c}{\text{m}}^{\text{3}}$.

8. If the triangle $\mathrm{\Delta }\text{ ABC}$ in the Question $\text{7}$ above is revolved about the side $\text{5 cm}$, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions $\text{7}$ and $\text{8}$.

Ans:

(Image will be uploaded soon)

If the triangle is revolved about the side $\text{5 cm}$, we will get a cone with:

Radius $\text{r = 12 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 5 cm}$

We know the volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \pi \times {\left(\text{12}\right)}^{\text{2}}\times \text{ 5}]\text{ c}{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 240 }\pi \text{ c}{\text{m}}^{\text{3}}$

Therefore, the volume of the cone will be $\text{240 }\pi \text{ c}{\text{m}}^{\text{3}}$.

The ratio of volume of cone from previous question an the one we obtained above $\text{= }\frac{\text{100 }\pi }{\text{240 }\pi }\text{ = }\frac{\text{5}}{\text{12}}\text{ = 5 : 12}$

Therefore, the required ratio is $\text{5 : 12}$.

9. A heap of wheat is in the form of a cone whose diameter is $\text{10}\text{.5 m}$and height is $\text{3 m}$. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

$\left[\text{Assume }\pi \text{ =}\frac{\text{22}}{\text{7}}\right]$

Ans:

It is given that diameter of the heap $\text{= 10}\text{.5 m}$

So, the radius of heap $\text{r = }\frac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 m}$

Height of heap $\text{h = 3 m}$

We know the volume of the cone $\text{V = }\frac{\text{1}}{\text{3}}\pi {\text{r}}^{\text{2}}\text{h}$

$\Rightarrow \text{V = }[\frac{\text{1}}{\text{3}}\times \frac{22}{7}\times {\left(\text{5}\text{.25}\right)}^{\text{2}}\times \text{ 3}]{\text{m}}^{\text{3}}$

$\Rightarrow \text{V = 86}\text{.625 }{\text{m}}^{\text{3}}$

Hence, the volume of the heap is $\text{86}\text{.625 }{\text{m}}^{\text{3}}$.

The area of canvas required is the same as the curved surface area of the cone.

$\therefore \text{A = }\pi \text{ rl}$

$\Rightarrow \text{A = }\pi \text{ r}\sqrt{{\text{h}}^{\text{2}}\text{ + }{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\frac{\text{22}}{\text{7}}\times \text{ 5}\text{.25 }\times \sqrt{{\left(\text{3}\right)}^{\text{2}}\text{ + }{\left(\text{5}\text{.25}\right)}^{\text{2}}}{\text{m}}^{\text{2}}$

$\Rightarrow \text{A = }(\frac{\text{22}}{\text{7}}\times \text{ 5}\text{.25 }\times \text{ 6}\text{.05}){\text{m}}^{\text{2}}$

$\Rightarrow \text{A = 99}\text{.825 }{\text{m}}^{\text{2}}$

Therefore, to protect the heap from the rain, the amount of canvas required is $\text{99}\text{.825 }{\text{m}}^{\text{2}}$.

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Exercise 11.3

Important Topics Covered in Exercise 11.3 of Class 9 Maths NCERT Solutions 

Surface areas and volumes are very important concepts in the study of geometry. Exercise 11.3 of Class 9 Maths NCERT Solutions is mainly based on the concept of deriving the formula of the volume of a right circular cone. Volume of a right circular cone is one third of the volume of the cylinder. One most popular example of a right circular cone is the caps that the Kids wear on their birthdays.

Below is the formula for the volume of a cone which is discussed in this exercise.

• Volume of a Cone = (1/3 )πr2h

This exercise consists of questions based on finding the volume of a cone. By practising the solutions provided in this exercise, students will have a better understanding of the concepts and it will also help them to build a solid base for the advanced Maths topics.

Opting for the NCERT solutions for Ex 11.3 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.3 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Surface Areas and Volumes textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 11 Exercise 11.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 11 Exercise 11.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 11 Exercise 11.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

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