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# NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.2

Last updated date: 09th Aug 2024
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## NCERT Solutions for Class 9 Maths Chapter 13 (Ex 13.2)

Free PDF download of NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 (Ex 13.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all NCERT Book Solutions in your emails. Download free NCERT Solution Class 9 Maths to amp up your preparations and to score well in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website.

 Class: NCERT Solutions for Class 9 Subject: Class 9 Maths Chapter Name: Chapter 13 - Surface Areas and Volumes Exercise: Exercise - 13.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.2
Surface Area and Volume L-1 | Surface Area & Volume of Cuboid & Cube | CBSE 9 Maths Ch 13 | Term 2
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## Access NCERT Solutions for Class 9 Mathematics Chapter 13 – Surface Areas and Volume

### Exercise 13.2

1. The curved surface area of a right circular cylinder of height $14 \mathrm{~cm}$ is $88 \mathrm{~cm}^{2}$. Find the diameter of the base of the cylinder. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of cylinder $-14 \mathrm{~cm}$

Let the diameter of the cylinder be $\mathrm{d}$

Curved surface area of cylinder $=88~\text{c}{{\text{m}}^{2}}$

$\Rightarrow 2\pi r\text{h}=88~\text{c}{{\text{m}}^{2}}(r$ is the radius of the base of the cylinder)

$\Rightarrow \pi dh=88~\text{c}{{\text{m}}^{2}}\$

$\Rightarrow \left[ \dfrac{22}{7}\times d\times 14 \right]\text{cm}\ \text{=}\ 88~\text{c}{{\text{m}}^{2}}$

$\Rightarrow \ d\ =\ \dfrac{88\ \times \ 7}{22\ \times \ 14}\ =\ 2\ cm$

Therefore, the diameter of the base of the cylinder is $2 \mathrm{~cm}$.

2. It is required to make a closed cylindrical tank of height $1 \mathrm{~m}$ and base diameter $140 \mathrm{~cm}$ from a metal sheet. How many square meters of the sheet are required for the same? $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of cylindrical tank - $1 \mathrm{~m}$

Base radius (r) of cylindrical tank $=\ \dfrac{140}{2}~\text{cm}\ \text{=}\ 70~\text{cm}\ \text{=}\ 0.7~\text{m}$

Area of sheet required = Total surface area of tank $=2\pi r(r+h)$

$=\ 2\ \times \ \dfrac{22}{7}\ \times \ 0.7\ \left( 0.7\ +\ 1 \right)\ {{m}^{2}}$

$=\ 4.4\ \times 1.7\ =\ 7.48\ {{m}^{2}}$

Therefore, it will require $7.48 \mathrm{~m}^{2}$ area of sheet.

3.  A metal pipe is $77 \mathrm{~cm}$ long. The inner diameter of a cross-section is $4 \mathrm{~cm}$, the outer diameter being $4.4 \mathrm{~cm}$, Find:

i) Inner curved surface area,

ii) Outer curved surface area,

iii) Total surface area. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Inner radius $\left( {{\text{r}}_{1}} \right)$ of cylindrical pipe $=\dfrac{4}{2}~\text{cm}=2~\text{cm}$

Outer radius $\left(\mathrm{r}_{2}\right)$ of cylindrical pipe $=\dfrac{4.4}{2} \mathrm{~cm}=2.2 \mathrm{~cm}$

Height (h) of cylindrical pipe = Length of cylindrical pipe $=77~\text{cm}$

(i) $\mathrm{CS} \mathrm{A}$ of inner surface of pipe $=2\pi {{\text{r}}_{1}}\text{h}$

$=\left( 2\times \dfrac{22}{7}\times 2\times 77 \right)\text{c}{{\text{m}}^{2}}=968~\text{c}{{\text{m}}^{2}}$

(ii) $\mathrm{CS} \mathrm{A}$ of outer surface of pipe $=2\pi {{\text{r}}_{2}}\text{h}$

$=\left( 2\times \dfrac{22}{7}\times 2.2\times 77 \right)\text{c}{{\text{m}}^{2}}=(22\times 22\times 2.2)\text{c}{{\text{m}}^{2}}\ =\ 1064.8\ c{{m}^{2}}$

(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe.

$=\ 2\pi {{r}_{1}}h+2\pi {{r}_{2}}h+2\pi \left( r_{2}^{2}-r_{1}^{2} \right)$

$=\ \left[ 968+1064.8+2\pi \left\{ {{(2.2)}^{2}}-{{(2)}^{2}} \right\} \right]\text{c}{{\text{m}}^{2}}$

$=\left( 2032.8+2\times \dfrac{22}{7}\times 0.84 \right)\text{c}{{\text{m}}^{2}}$

$=\ (2032.8+5.28)\ \text{c}{{\text{m}}^{2}}=\ \ 2038.08~\text{c}{{\text{m}}^{2}}$

Therefore, the total surface area of the cylindrical pipe is $2038.08 \mathrm{~cm}^{2}$.

4. The diameter of a roller is $84 \mathrm{~cm}$ and its length is $120 \mathrm{~cm}$. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in $\mathrm{m}^{2}$? $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: It can be observed that a roller is cylindrical.

Height (h) of cylindrical roller = Length of roller $=120 \mathrm{~cm}$

Radius (r) of the circular end of roller $=\dfrac{84}{2}~\text{cm}=42~\text{cm}$

CSA of roller $=2 \pi t h$

$=\left( 2\times \dfrac{22}{7}\times 42\times 120 \right)\text{c}{{\text{m}}^{2}}\ =\ 31680~\text{c}{{\text{m}}^{2}}$

Area of field $=500\times$ CSA of roller

$=\ (500\times 31680)\text{c}{{\text{m}}^{2}}$

$=15840000~\text{c}{{\text{m}}^{2}}=1584~{{\text{m}}^{2}}$

5. A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12 .50 per $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) cylindrical pillar $=3.5 \mathrm{~m}$

Radius (r) of the circular end of pillar $=\dfrac{50}{2}~\text{cm}\ \text{=}\ 25~\text{cm}\ \text{=}\ 0.25~\text{m}$

CSA of pillar $=2\pi rh$

$=\left(2 \times \dfrac{22}{7} \times 0.25 \times 3.5\right) \mathrm{m}^{2}$

$=(44\times 0.125){{\text{m}}^{2}}=5.5~{{\text{m}}^{2}}$

Cost of painting $1 \mathrm{~m}^{2}$ area $\text{=}\ \text{Rs}.12.50$

Cost of painting $5.5 \mathrm{~m}^{2}$ area $=(5.5\times 12.50)$

$=\mathrm{Rs} .68 .75$

Therefore, the cost of painting the CSA of the pillar is Rs. 68.75.

6. Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$. If the radius of the base of the cylinder is $0.7 \mathrm{~m}$, find its height. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Let the height of the circular cylinder be $h$

Radius (r) of the base of cylinder $=0.7 \mathrm{~m}$

CSA of cylinder $=4.4~{{\text{m}}^{2}}$

$\Rightarrow \ \ 2\pi \text{rh}=4.4~{{\text{m}}^{2}}$

$\Rightarrow \ \left[ 2\times \dfrac{22}{7}\times 0.7\times h \right]=4.4~{{\text{m}}^{2}}$

$~\Rightarrow \ \text{h}\ =\ \dfrac{4.4\ \times \ 7}{2\ \times 22\ \times \ 0.7}\ =\ 1m$

Therefore, the height of the cylinder is$1 \mathrm{~m}$.

7. The inner diameter of a circular well is$3.5 \mathrm{~m}$. It is $10 \mathrm{~m}$ deep. Find

(i)  Inner curved surface area,

(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m2 . $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Inner radius (r) of circular well $=\left( \dfrac{3.5}{2} \right)\text{m}=1.75~\text{m}$

Depth of well = $=\ 10\ m$

(i) Inner curved surface area $=2\pi \text{rh}$

$=\left( 2\times \dfrac{22}{7}\times 1.75\times 10 \right){{\text{m}}^{2}}$

$=\ (44\times 0.25\times 10){{\text{m}}^{2}}=110~{{\text{m}}^{2}}$

Therefore, the inner curved surface area of the circular well is $110 \mathrm{~m}^{2}$.

(ii) Cost of plastering $1 \mathrm{~m}^{2}$ area = Rs. 40

Cost of plastering $100 \mathrm{~m}^{2}$ area $\text{=Rs}(110\times 40)=\text{Rs}.4400$

Therefore, the cost of plastering the CSA of this well is Rs.4400.

8. In a hot water heating system, there is a cylindrical pipe of length $28 \mathrm{~m}$ and diameter $5 \mathrm{~cm}$. Find the total radiating surface in the system. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of cylindrical pipe  = Length of cylindrical pipe $=28~\text{m}$

Radius $(\mathrm{r})$ of circular end of pipe $=\dfrac{5}{2}=2.5~\text{cm}\ \text{=}\ \,0.025~\text{m}$

CSA of cylindrical pipe  $=\ 2\pi rh$

$=\ \left( 2\times \dfrac{22}{7}\times 0.025\times 28 \right){{\text{m}}^{2}}=4.4~{{\text{m}}^{2}}$

The area of the radiating surface of the system is$4.4 \mathrm{~m}^{2}$.

9. Find:

(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2 \mathrm{~m}$ in diameter and $4.5 \mathrm{~m}$ high.

(ii) How much steel was actually used, if $\dfrac{1}{12}$ of the steel actually used was wasted in making the tank? $\left[ {} \right.$Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of cylindrical tank $=4.5 \mathrm{~m}$

Radius (r) of the circular end of cylindrical tank $=\left(\dfrac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m}$

(i) Lateral or curved surface area of tank $=2\pi rh$

$=\left( 2\times \dfrac{22}{7}\times 2.1\times 4.5 \right){{\text{m}}^{2}}\ =\ (44\times 0.3\times 4.5){{\text{m}}^{2}}\ =\ 59.4~{{\text{m}}^{2}}$

Therefore, the CSA of the tank is$59.4 \mathrm{~m}^{2}$.

(ii) Total surface area of tank $=2\pi r(r+h)$

$=\left( 2\times \dfrac{22}{7}\times 2.1\times (2.1+4.5) \right){{\text{m}}^{2}}$

$=\ (44\times 0.3\times 6.6){{\text{m}}^{2}}=87.12~{{\text{m}}^{2}}$

Let A $\mathrm{m}^{2}$ steel sheet be actually used in making the tank.

$\therefore A\left( 1-\dfrac{1}{12} \right)=87.12~{{\text{m}}^{2}}$

$\Rightarrow A=\left( \dfrac{12}{11}\times 87.12 \right){{\text{m}}^{2}}$

$\Rightarrow \text{A}\ \text{=}\ 95.04~{{\text{m}}^{2}}$

Therefore, $95.04 \mathrm{~m}^{2}$ steel was actually used while making such a tank.

10. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of $20 \mathrm{~cm}$ and height of $30 \mathrm{~cm}$. A margin of $2.5 \mathrm{~cm}$ is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of the frame of lampshade $=(2.5+30+2.5)\text{cm}\ \text{=}\ 35~\text{cm}$

Radius (r) of the circular end of the frame of lampshade $=\dfrac{20}{2}~\text{cm}\ \text{=}\ 10~\text{cm}$

Cloth required for covering the lampshade $=\ 2\pi rh$

$=\left( 2\times \dfrac{22}{7}\times 10\times 35 \right)\text{c}{{\text{m}}^{2}}=2200~\text{c}{{\text{m}}^{2}}$

Hence, for covering the lampshade, $2200 \mathrm{~cm}^{2}$ cloth will be required.

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius $3 \mathrm{~cm}$ and height $10.5 \mathrm{~cm}$. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? $\left[ Assume\ \pi \ =\ \dfrac{22}{7} \right]$

Ans: Radius (r) of the circular end of cylindrical penholder $=3~\text{cm}$

Height (h) of penholder $=10.5 \mathrm{~cm}$

Surface area of 1 penholder = CSA of penhold + Area of base of penholder $=2\pi rh\ +\ \pi {{r}^{2}}$

$=\left( 2\times \dfrac{22}{7}\times 3\times 10.5+\dfrac{22}{7}\times {{(3)}^{2}} \right)\text{c}{{\text{m}}^{2}}$

$=\left( 132\times 1.5+\dfrac{198}{7} \right)\text{c}{{\text{m}}^{2}}$

$=\left( 198+\dfrac{198}{7} \right)\text{c}{{\text{m}}^{2}}=\dfrac{1584}{7}~\text{c}{{\text{m}}^{2}}$

Area of card board sheet used by 1 competitor $=\dfrac{1584}{7} \mathrm{~cm}^{2}$

Area of cardboard sheet used by 35 competitors $=\left( \dfrac{1584}{7}\times 35 \right)\text{c}{{\text{m}}^{2}}$ $=7920~\text{c}{{\text{m}}^{2}}$

Therefore, $7920 \mathrm{~cm}^{2}$ cardboard sheet will be bought.

### Important Formulas to Remember to Solve Questions of Exercise 13.2 of NCERT Class 9 Math

Here is the list of most important formulas related to the right circular cylinder, the topic discussed in Exercise 13.2, that you must learn.

• The curved surface area of the right circular cylinder

= perimeter of the base of cylinder ✖ height of the cylinder

= 2πrh

Where r = base radius of the cylinder, h = height of the cylinder

• Total surface area of the right circular cylinder

= Curved surface area of the cylinder + area of base and top faces of cylinder

= 2πrh + πr2 + πr2

=  2πrh + 2πr2

= 2πr (r + h)

Where r = base radius of the cylinder, h = height of the cylinder

### Basic Formulas and Tips

While solving the questions, you also need to recall some basic formulas and tips. These are given below:

• Diameter = 2 ✖ Radius

• 1 meter = 100 cm

In question 3, while calculating the inner curved surface area, you must evaluate the radius of the inner cross-section whereas to calculate the outer curved surface area evaluate the radius of the outer cross-section.

### NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 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 13.2 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 13 Exercise 13.2 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 13 Exercise 13.2, 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 13 Exercise 13.2 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 that they can be accessed both online and offline as well.

### CBSE Class 9 Math Chapter 13 Other Exercises

 Chapter 13 Surface Areas and Volumes All Exercises in PDF Format Exercise 13.1 8 Questions & Solutions (4 Short Answers, 4 Long Answers). Exercise 13.3 8 Questions & Solutions. Exercise 13.4 9 Questions & Solutions (4 Short Answers, 5 Long Answers). Exercise 13.5 9 Questions & Solutions (4 Short Answers, 5 Long Answers). Exercise 13.6 8 Questions & Solutions (8 Short Answers). Exercise 13.7 9 Questions & Solutions (9 Long Answers). Exercise 13.8 10 Questions & Solutions (5 Short Answers, 5 Long Answers). Exercise 13.9 3 Questions & Solutions (3 Long Answers).

## FAQs on NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.2

1. What are the Concepts Students Learn in Chapter 13 Exercise 13.2?

There are a total of 8 questions in chapter 13 exercise 13.2. All these questions are pertaining to finding the surface area and volume of various geometrical objects in a clear and easy way. The exercise 13.2 in particle helps the student understand how to find the surface area of the cylinder in detail. These NCERT solutions contain an explicit answer to each and every question in the exercise. All the answers are given elaborately without missing a single step to make students understand them easily. All the related formulas and respective substitutions are clearly stated in these solutions ensuring that the students will not get confused in any of the steps.

2. What are Some of the Chief Features of NCERT Solutions of Class 9 Maths Chapter 13, Exercise 13.2?

The NCERT Solutions of Class 9 Maths Chapter 13, exercise 13.2 are made available here by Vedantu as a free PDF download to help students boost their study process. The features that make these solutions are as follows:

• Helps you to understand and solve the problems in multiple levels

• All the answers are solved in the exact same manner as expected in the exam

• Aids you in securing good marks in the exam

• Has answers to all the important exams related to questions as well as questions given in the NCERT books.

• Steps are detailed and simple allowing students to remember them for a longer period.

3. What makes the NCERT Solutions Provided by Vedantu the Best for Class 9 Maths Chapter 13, Exercise 13.2?

The NCERT Solutions provided by Vedantu the best for  Class 9 Maths Chapter 13, exercise 13.2 has the most accurate answers. It gives a comprehensive and thorough solution for all the problems asked in the NCERT textbook exercise. These problems are solved by maths subject matter experts with a very wide experience in the field of teaching. These solutions are intended at making studying an easy process for the students. Considering that Maths is a subject that is found difficult by most students, these solutions are designed in a simple manner with step by step explanation for each and every problem given in the book.

4. What are Some of the Best Study Materials to Score Well in Maths?

Maths is a scary subject for most of the students. This is mainly because of students' ignorance of not knowing how to study rather than the subject itself. The key to excelling in Maths is practice and hard work. Along with it, is the need for the right mindset in order to tackle the subject successfully in the exam. Following are the few exam study materials which when is a part of your preparation makes it easy for students to  score well in the exams :

• Previous years question papers with solutions.

• Mock papers with solutions

• NCERT Solutions for Class 8 Maths by Vedantu

• Sample papers for Class 8 Maths

• Live Online Classes by Vedantu

5. What do I do to score well in Class 9 Maths, Chapter 13- Surface Areas and Volumes, Exercise 13.2?

To score well in Class 9 Maths, Chapter 13- Surface Areas and Volumes, Exercise 13.2, first of all, understand the formulae of the cylinder and its application. Solve all the questions preceding the exercise and then move on to the NCERT Exercise 13.2 questions. If you get stuck on any question, refer to Vedantu's NCERT Solutions available free of cost for this exercise. Practice and revise the concepts of the cylinder regularly to master this exercise. If you follow this strategy, you will score high marks in the Maths exam.

6. Is Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2 difficult?

Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2 can be simplified by making a well-planned strategy and its implementation. The basic requirement of this exercise is the formulae of the cylinder. So, prepare a list of all the formulae from this exercise and revise them regularly. Next, practice each type of question from this exercise and focus more on the difficult questions. Practice the questions that are tough for you plenty of times to make this exercise easier for you. Students can check the answers from the NCERT Solutions available on the vedantu website and the app.

7. What are the practical applications of Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2?

Several practical applications of Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2 can be seen in our day-to-day lives. The concepts of the cylinder and its surface area are used to calculate the amount of paint required to paint a fuel cylinder. All these concepts are used extensively in the case of cylindrical figures such as beakers, cans, candles, water tanks, pipes etc. in our daily lives. These concepts are also used during the engineering and construction of cylindrical structures and objects.

8. What is a cylinder Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2?

As discussed in Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2, a cylinder is a solid figure which is three dimensional in nature. A cylinder has two identical circular faces at each end, it has a curved surface with no edges or vertices. Some real-life examples of a cylinder are a rolling pin, a fuel cylinder, a cylindrical water bottle, toilet paper rolls, pencil, bucket, dustbin etc.

9. What are the important questions from Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2?

The Class 9 Maths, Chapter 13-Surface Areas and Volumes, Exercise 13.2 has several important questions based on the concepts of the cylinder, its surface area- total surface area, curved/lateral surface area, volume and related concepts. Each type of question in this exercise is important from an examination point of view. Focus more on the questions that can't be solved directly with the help of formulae such as the area covered by a rolling cylinder when not all dimensions are given. Each question will teach you something new, so solve all NCERT questions from this exercise.