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

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.

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Chapter wise NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

CBSE Class 9 Math Chapter 13 Other Exercises