# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.6) Exercise 13.6

For students, Class 9 is one of the most essential academic years. Some experts believe that Class 9 is more significant than Class 10 because in Class 9, foundational concepts are taught that are very valuable in Classes 10, 11, and 12. As a result, students frequently find Class 9 difficult. Class 9 is important for students to comprehend the subjects in the long run. For the first time, students in Class 10 sit for board exams. Students who want to do well in their Class 10 board exams should prepare thoroughly for their Class 9 subjects.

It also aids students in determining which subjects they are most interested in. Class 9 studies assist students in deciding which subjects to study in Class 11 and Class 12. As a result, students must efficiently prepare for Class 9. Students can focus on their interests by exploring different subject streams.

Mathematics is an essential topic for students to study. It improves their cognitive and thinking skills. Students improve their analytical thinking skills by using Mathematics. Mathematics also helps to create mental discipline. Mathematics has numerous applications. Mathematics is employed in everyday life in some form or another. As a result, children begin studying this subject at a young age.

Mathematical ability can be beneficial for pursuing employment in a range of fields. Its concepts can be used in domains such as Computer Science, Engineering, Astronomy, Statistics, Data Science, and others. Students interested in these vocations should concentrate on mathematics in Class 9. Class 9 Math is quite helpful in understanding the principles of other subjects as well. It is generally viewed as difficult by students. To overcome their concern, students must practice a considerable number of questions. Students will find it simple and easy after solving many problems. It is suggested that students answer a variety of questions. They can acquire confidence in their mathematical ability by solving practice problems. Students must thoroughly prepare for Class 9 Mathematics because it will benefit them in other subjects. Many scientific fields, including Physics and Chemistry, require mathematical talents. Business and Statistics can also benefit from Mathematics. Students who are struggling with this subject should commit more time and effort to their studies.

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.6) Exercise 13.6**

Chapter 13 of Class 9 Mathematics is “Surface Areas and Volumes.” To prepare this chapter thoroughly students must solve all the NCERT exercises in this chapter. Students can find the most accurate answers for Class 9 Maths Chapter 13 Exercise 13.6 in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are available on the website and mobile application of Extramarks. Students can download the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 in PDF format. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 contain the most accurate solutions for Maths Class 9 Chapter 13 Exercise 13.6.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are written in a stepwise manner for the clarity of students. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are written by expert subject teachers. Students must refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6.

**Access NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes**

Students can access the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 from the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are written by expert Mathematics teachers for the benefit of students. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 will help students earn good marks in their Class 9 Math exams. Students are advised to download the

NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 in PDF format.

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6**

Chapter 13 of Class 9 Mathematics has 9 exercises. All the exercises are important and must be solved by the students. Since the chapter is a bit longer than the other chapters, students must devote adequate time to it. Class 9 Chapter 13 Maths Exercise 13.6 contains 8 questions. The various concepts covered in Chapter 13 are Surface Area, Lateral Surface Area, Volume etc. A very important component of this chapter is the units. Students must carefully write the correct units to avoid mistakes. Students must regularly practice questions from this chapter to remember all the formulas in this chapter.

**NCERT Solutions for Class 9**

Because Class 9 is such an important academic year for students, they should organise their studies carefully. Here are some pointers to help students prepare for their Class 9 exams:

Consolidation of Study Material: There is a lot of study material available for Class 9 Mathematics both online and offline, but students should only choose the best study material for their preparation. They should select study resources that provide them with the most straightforward and accurate solutions. They should also think about whether the study material they select covers the complete curriculum. Because CBSE issues new guidelines each year, students should always obtain updated versions of study materials. The NCERT book for Class 9 Mathematics should be the starting point for students. NCERT books are recommended for all CBSE topics. The NCERT books cover the complete syllabus in detail and give students a plethora of practice questions of varying difficulty levels. Along with the NCERT books, students must also refer to the resources provided by Extramarks. For Class 9 Mathematics Chapter 13, students can use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. They can download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 from the website and mobile application of Extramarks. Students can completely rely on the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 for solving questions from this exercise. Moreover, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 can be downloaded in PDF format.

Following a Timetable: Students are more likely to complete topics on time when they have a strategy and deadlines. Higher topic weightage and difficulty level subjects may demand additional study time. As a result, students must set aside time for different topics and chapters. Many students find mathematics difficult; therefore, they should commit more time to practise it. Students in Class 9 Mathematics should focus on various chapters and set weekly and monthly targets for themselves. This will allow them to monitor their progress and ensure that they complete the full course. Chapter 13 of Class 9 Mathematics has 9 exercises. Since this chapter has considerably more exercises, students should devote more time to preparing this chapter. To solve questions from Exercise 13.6 of this chapter, students must take help from the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are solved by expert Mathematics teachers for the benefit of students. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6, have the most accurate answers for this exercise.

NCERT Exercises: The most important resource for Class 9 school examinations is NCERT questions. These solutions and exercises provide students with an idea of the types of questions they will face in the exams. NCERT exercises include questions of varying levels of complexity. Students must practice the NCERT questions because questions in the CBSE board exams usually come directly from the NCERT exercises. If students face any problems, they can use the Extramarks resources to get help. Extramarks materials are reliable and trustworthy. NCERT practice questions can also help learners prepare for competitive exams. Exercise 13.6 Class 9 Maths can be solved by students with the help of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. Students are advised to download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 from the website and mobile application of Extramarks in PDF format. Parents can also download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 to assist their children in exam preparation.

Mock Examinations: Mock examinations help students assess their preparation. Students are encouraged to take unit exams after finishing a chapter. They should try to answer questions to determine how well they comprehend the chapter. Unit tests must be analysed to discover weak areas that students can subsequently work on. Students should also prepare for the term-end exam by taking full-length mock exams. The full mock exams will help them manage their time on exam day. Mock tests will help demonstrate to students how much information they can retain at one time. Mock examinations should be taken in an exam-like setting so that students are not too nervous on exam day. Students can use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 to answer mock exam questions. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are available on the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 can be downloaded in PDF format for easy access.

Revision: Revision is the most important component of exam preparation. Revision helps students remember a large amount of information for a longer period of time. Students should be able to recall the majority of the formulas from their Class 9 Mathematics textbooks with proper revision. Revisions must be carried out on a regular basis. Students should review their brief notes as well as their mock tests and unit tests. After multiple changes, students will gain confidence. The most effective way to revise mathematics is to practice questions on a regular basis. Regular question practice will reduce the time it takes students to solve each question, allowing them to complete their papers on time. Students must also regularly revise the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. The difficult questions must be revisited again and again. With the help of NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6, students can also revise important formulas.

CBSE Sample Papers And Past Year Papers: CBSE sample papers are beneficial to students. CBSE publishes sample papers every year before the board exams. The sample papers are crucial because they are up-to-date with the most recent CBSE requirements. CBSE sample papers must be completed by students. Students must solve past years’ papers on the topics in addition to sample papers. Past years’ papers provide students with an idea of the types of questions the examiner will ask. If students have difficulty solving the sample problems or past year papers, they can refer to the Extramarks study resources. Moreover, if they face any challenges in answering the questions from past year papers, they can refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 offered by Extramarks. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are very beneficial for the students’ preparation.

Short Notes: Short notes aid in the revision of formulas and theorems prior to the exam. Students must summarise all of the formulas and take brief notes on them. This will assist them in reviewing all relevant formulas and their applications prior to the exam when students have limited time. Students should go over the formulas as many times as possible. Students can prepare short notes for important formulas, with the help of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are very useful resources for students. Students must refer to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6.

Extracurricular Activities: Extracurricular activities are an important part of academic life. Extracurricular activities are required of all students. They help students build practical skills and refresh their minds. As a result, students are required to devote time to extracurricular activities. Extracurricular activities help kids develop their complete personalities. Athletics, debates, theatrical plays, dancing, and other activities are some of the activities students can participate in.

**CBSE Study Materials for Class 9**

Mathematics, Social Science, Science, English, and Hindi/Sanskrit are the core subjects in Class 9. All of the subjects are required to pass the Class 9 exams. The Extramarks website and mobile app provide access to study tools for all Class 9 topics. Students can acquire revision notes, past years’ papers, and NCERT solutions based on their needs. All of the resources will assist them in better understanding the subjects and answering questions from the NCERT exercises. Students should save the resources in PDF format for future reference. Extramarks resources can help students improve their grades. Students must read the resources on a regular basis. The revision will help students retain what they have studied for a longer period of time.

**CBSE Study Materials**

NCERT books are recommended by all CBSE-affiliated schools. NCERT books should be used as the primary source of preparation for students. Students can use Extramarks’ study resources in addition to NCERT books. Extramarks offers study materials in both Hindi and English. Extramarks’ resources are extremely valuable to students because they cover the full curriculum in depth.

Subject experts curate Extramarks materials for the benefit of students. All of Extramarks’ resources are regularly proofread. These solutions are tailored to the specific needs of the questions, allowing students to study more successfully. The materials from Extramarks include all of the relevant illustrations, chemical reactions, and formulas. Extramarks’ study materials can be used for all of a student’s preparation needs. It gives revision notes and past years’ papers for revision and analysis. As part of their total preparation, students must study all of these resources. All of the resources are available on Extramarks’ website and mobile application. The resources are available in PDF format for students to download.

**Q.1 **The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm^{3}=1 *l *)

**Ans**

$\begin{array}{l}\mathrm{Circumference}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{a}\mathrm{cylindrical}\mathrm{vessel}\\ =132\mathrm{\hspace{0.17em}}\mathrm{cm}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{\pi r}=132\mathrm{\hspace{0.17em}}\mathrm{cm}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\times \frac{22}{7}\times \mathrm{r}=132\mathrm{\hspace{0.17em}}\mathrm{cm}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{r}=\frac{132\times 7}{44}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{r}=21\mathrm{\hspace{0.17em}}\mathrm{cm}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Height}\mathrm{of}\mathrm{cylinderical}\mathrm{vessel}=25\mathrm{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{vessel}={\mathrm{\pi r}}^{2}\mathrm{h}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{22}{7}\times {\left(21\right)}^{2}\times 25\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=34650\mathrm{\hspace{0.17em}}{\mathrm{cm}}^{3}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{34650}{1000}\mathrm{\hspace{0.17em}}\mathrm{litres}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\because 1{\mathrm{cm}}^{3}=\frac{1}{1000}\hspace{0.17em}\hspace{0.17em}\mathrm{litres}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=34.65\mathrm{\hspace{0.17em}}\mathrm{litres}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Thus},\mathrm{the}\mathrm{cyliderical}\mathrm{vessel}\mathrm{can}\mathrm{hold}34.65\mathrm{litres}\mathrm{water}\mathrm{.}\end{array}$

**Q.2** The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm^{3} of wood has a mass of 0.6 g.

**Ans**

$\begin{array}{l}\text{The inner diameter of a cylinderical wooden pipe}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}The inner radius of a cylinderical wooden pipe}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\text{\hspace{0.17em}}\mathrm{cm}\\ \text{The outer diameter of a cylinderical wooden pipe}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=28\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}The outer radius of a cylinderical wooden pipe}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}The length of pipe}=\text{35 cm}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mass of 1 cm}}^{\text{3}}\text{wood}=0.6\text{\hspace{0.17em}}\mathrm{gm}\\ \text{Volume of wood in cylinderical wooden pipe}\\ =\frac{22}{7}\{{\left(14\right)}^{2}-{\left(12\right)}^{2}\}\times 35\\ =\frac{22}{7}\times 2\times 26\times 35\\ =5720\text{\hspace{0.17em}}{\mathrm{cm}}^{3}\\ \text{Mass of wood in cylinderical wooden pipe}\\ =0.6\times 5720\text{\hspace{0.17em}}\mathrm{gm}\\ =3432\text{\hspace{0.17em}}\mathrm{gm}\\ =3.432\text{\hspace{0.17em}}\mathrm{kg}\\ \text{Thus},\text{the mass of wood in cylinderical wooden pipe is 3.432 kg.}\end{array}$

**Q.3 **A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

**Ans**

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.33em}}\mathrm{Capacity}\text{of cuboidal tin can}=5\times 4\times 15\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=300\text{\hspace{0.17em}}{\mathrm{cm}}^{3}\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}}\mathrm{Capacity}\text{of plastic cylinder}={\mathrm{\pi r}}^{2}\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^{2}\times 10\text{\hspace{0.17em}}{\mathrm{cm}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=385\text{\hspace{0.17em}}{\mathrm{cm}}^{3}\\ \mathrm{In}\text{this way we see that capacity of plastic container is more}\\ \text{than tin container.}\\ \text{More capacity of cylinderical container than tin container}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=385-300\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=85\text{\hspace{0.17em}}{\mathrm{cm}}^{3}\end{array}$

**Q.4 **If the lateral surface of a cylinder is 94.2 cm^{2} and its height is 5 cm, then find

(i) radius of its base

(ii) its volume.

(Use π = 3.14)

**Ans**

$\begin{array}{l}\left(i\right)\text{The height of cylinder}=\text{5 cm}\\ \text{The lateral surface area of cylinder}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=94.2\text{\hspace{0.17em}}c{m}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\pi rh=94.2\text{\hspace{0.17em}}{\text{cm}}^{\text{2}}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\times 3.14\times r\times 5=94.2{\text{cm}}^{\text{2}}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{94.2}{2\times 3.14\times 5}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\text{cm}\\ \text{Thus},\text{the radius of cylinder is 3 cm}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}Volume of cylinder}=\pi {r}^{2}h\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3.14\times 3\times 3\times 5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=141.3\text{\hspace{0.17em}}c{m}^{3}\\ \text{Thus},\text{the volume of cylinder is 141}{\text{.2 cm}}^{\text{3}}\text{.}\end{array}$

**Q.5 **It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m^{2}, find

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel.

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Depth of cylinder}=\text{10 m}\\ \text{Cost of painting the inner curved surface area of cylinder}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\u20b9 2200}\\ {\text{Rate of painting of 1 m}}^{\text{2}}=\text{\hspace{0.17em}\hspace{0.17em}\u20b9 20}\end{array}$

$\begin{array}{l}\text{(i)Inner curved surface area of cylinder}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{Cost}\text{of painting}}{\mathrm{Rate}\text{of painting}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2200}{20}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=110\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\mathrm{ii}\right)\because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi rh}=110\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\times \frac{22}{7}\times \mathrm{r}\times 10=110\\ \Rightarrow \mathrm{r}=\frac{110\times 7}{44\times 10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.75\text{\hspace{0.17em}}\mathrm{m}\\ \text{Thus},\text{the radius of the base of cylinder is 1.75 m.}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Capacity of vessel}={\mathrm{\pi r}}^{2}\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22}{7}\times 1.75\times 1.75\times 10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=22\times 0.25\times 17.5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=96.25\text{\hspace{0.17em}}{\mathrm{m}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=96.25\text{\hspace{0.17em}}\mathrm{kl}[\because 1{\mathrm{m}}^{3}=1\mathrm{kl}]\end{array}$

**Q.6** The volume capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Height of a closed cylinder}=1\text{\hspace{0.17em}}m\\ \text{The capacity of a closed cylinder}=15.4\text{\hspace{0.17em}}litres\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\pi {r}^{2}h\text{\hspace{0.17em}}=\frac{15.4}{1000}\text{\hspace{0.17em}}{m}^{3}\text{\hspace{0.17em}}[1\text{\hspace{0.17em}}litre=\frac{1}{1000}\text{\hspace{0.17em}}{m}^{3}]\\ \frac{22}{7}\times {r}^{2}\times 1=\frac{15.4}{1000}\text{\hspace{0.17em}}{m}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=\frac{15.4}{1000}\times \frac{7}{22}\\ =\frac{7\times 7}{10000}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\frac{7}{100}=0.07\text{\hspace{0.17em}}m\\ \text{Required metal sheet for closed cylinder}\\ =2\pi r(h+r)\\ =2\times \frac{22}{7}\times 0.07(1+0.07)\\ =44\times 0.01\times 1.07\\ =0.4708\text{\hspace{0.17em}}{m}^{2}\\ \text{Thus},\text{}0.4708\text{\hspace{0.17em}}{m}^{2}\text{\hspace{0.17em}}\text{of metal sheet would be needed to}\text{make}\\ \text{closed cylinder}.\end{array}$

**Q.7 **A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

**Ans**

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Diameter of the pencil}=\text{7 mm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Radius}\left({r}_{1}\right)\text{of the pencil}=\frac{\text{7}}{2}\text{mm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Diameter of the graphite}=\text{1 mm}\\ \text{Radius}\left({r}_{2}\right)\text{of the graphite}=\frac{\text{1}}{2}\text{mm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{The length of the pencil}=\text{14 cm}\\ =\text{140}\text{\hspace{0.17em}}\text{mm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Volume of the wood}=\pi ({r}_{1}{}^{2}-{r}_{2}{}^{2})h\\ =\frac{22}{7}\{{\left(\frac{7}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}\}\times 140\\ =22(\frac{7}{2}+\frac{1}{2})(\frac{7}{2}-\frac{1}{2})\times 20\\ =44\times 4\times 30\\ =5280\text{\hspace{0.17em}}m{m}^{3}\\ =\frac{5280}{1000}\text{\hspace{0.17em}}c{m}^{3}[\because 1\text{\hspace{0.17em}}cm=10\text{\hspace{0.17em}}mm]\\ =5.28\text{\hspace{0.17em}}c{m}^{3}\end{array}$

Thus, the volume of wood in the pencil is 5.28 cm^{3}.

\begin{array}{c}Volume\text{of graphite}=\pi {r}_{2}{}^{2}h\\ =\frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\times 140\\ =11\times 10\\ =110\text{\hspace{0.17em}}m{m}^{3}\\ =\frac{110}{1000}\text{\hspace{0.17em}}c{m}^{3}\\ =0.11\text{\hspace{0.17em}}c{m}^{3}\end{array}

Thus, the volume of graphite in the pencil is 0.11 cm^{3}.

**Q.8** A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare (in litres) daily to serve 250 patients?

**Ans**

$\begin{array}{c}\text{The diameter of bowl}=7\text{\hspace{0.17em}}cm\\ \text{The radius of bowl}=\frac{7}{2}\text{\hspace{0.17em}}cm\\ \text{Height of soup in the bowl}=4\text{\hspace{0.17em}}cm\\ \text{Volume of soup in bowl}=\pi {r}^{2}h\\ =\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 4\\ =154\text{\hspace{0.17em}}c{m}^{3}\\ \text{Volume of soup required for one patient}=154\text{\hspace{0.17em}}c{m}^{3}\\ \text{Volume of soup required for 250 patients}=250\times 154\text{\hspace{0.17em}}c{m}^{3}\\ =\frac{250\times 154}{1000}\text{\hspace{0.17em}}litres\\ [\because 1\text{\hspace{0.17em}}c{m}^{3}=\frac{1}{1000}\text{\hspace{0.17em}}litres]\\ =38.5\text{\hspace{0.17em}}litres\end{array}$

Thus, 38.5 litres soup the hospital has to prepare daily to serve 250 patients.

## FAQs (Frequently Asked Questions)

### 1. Is Class 9 Mathematics tough?

It is true that Class 9 has an increased difficulty level in all subjects but students can make it easier for themselves by studying regularly and practising NCERT exercises. Class 9 Mathematics must also be prepared in this manner. Students can solve NCERT exercises with the help of resources provided by Extramarks. To solve the questions of Chapter 13 Class 9, students can take the help of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6.

### 2. From where can students download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6?

Students can download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 from the website and mobile application of Extramarks.

### 3. Can the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 be downloaded in PDF format?

Yes, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 are available in PDF format. Students are advised to download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 in PDF format for easy access.

### 4. Is the Class 9 Mathematics syllabus important for Class 10 board examinations?

Yes, the Class 9 Mathematics syllabus is very important for Class 10 board examinations as students learn the foundations in Class 9. If they understand the basics well they will be able to learn the more complex topics easily. Moreover, Class 9 and Class 10 are further very important for Class 11 and Class 12 Mathematics.

### 5. Will the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 help students to solve mock papers?

Yes, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 will help students in solving the mock examination questions. They can also refer to Extramarks’ past years’ papers to find authentic solutions. Students must download the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 for preparing the chapter efficiently.

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### 7. Are solutions of other Chapter 13 Exercises also provided by Extramarks?

Yes, along with the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6, solutions for all the 9 exercises of Chapter 13 are available on the website and mobile application of Extramarks for the benefit of students. Students can refer to the solutions in case of any challenges.

### 8. Are the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 also provided in Hindi by Extramarks?

Extramarks does provide study materials in Hindi for students. They can visit the website and mobile application to download the study material in Hindi.

### 9. Are the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 easy to understand?

Yes, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 provided by Extramarks are written in an easy-to-understand language for the benefit of students. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 must be downloaded from the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.6 can be downloaded in PDF format.