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

NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes are created here for students of class 9 for the preparation of term 2 CBSE Board exams. Chapter 13-Surface Areas and Volumes is the most important chapter in the Class 9 term 2 maths exam, this chapter is based on surface areas and volumes of three-dimensional figures like cubes, cuboids, cylinders, cones, and the sphere. Here in this chapter, you will study the problems on surface areas and volumes one by one from exercises 13.1 to 13.9.NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes are the most important inputs to correct the maths concepts. These questions are also useful for the preparation of competitive entrance exams like SSC, CPO, Railway, Bank clerk, PO, etc.

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**NCERT Solutions for Class 9 Mathsย Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam**

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

**Q1.The curved surface area of a right circular cylinder of height 14 cm is ****88 cm**ยฒ**. Find the diameter of the base of the cylinder.(Assume ฯ = 22/7).**

Ans. The curved surface area of a right circular cylinder = 2ฯrh

Height,h given =14 cm

CSA is given = 88 cmยฒ

2ฯrh = 88

2ร(22/7) รrร14 = 88

r = 88ร7/(2ร22ร14) = 14/14 =1

The radius of the right circular cylinder is 1 cm

Hence the diameter of the cylinder is 2(1) = 2cm

**Q2.It is required to make a closed cylindrical tank of height 1 mย and base diameter 140 cmย from a metal sheet. How many square meters of the sheet are required for the same?**

Ans. The height(h) of closed cylindrical tank is = 1m

Base diameter of the cylinder is =140 cm

Radius of the base is = 140/2 = 70 cm = 0.70 m

Total surface area of the cylinder =2ฯr(r +h)

TSA of cylinder = 2(22/7) 0.70(0.70 + 1)

= 2ร22ร0.10ร1.7

= 7.48 cmยฒ

**Therefore , required metallic sheet to make a closed cylindrical tank is 7.48 cmยฒ**

**Q3.A metal pipe is ****77 cm** **long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find the following:**

**(i) Inner curved surface areaย **

**(ii)Outer curved surface areaย **

**(iii)Total surface area**

Ans. The length of metal pipe = 77 cm

Inner diameter of the pipe = 4 cm,corresponding radius = 4/2 = 2 cm

Outer diameter of the pipe = 4.4 cm,corresponding radius =4.4/2 = 2.2 cm

(i)Inner curve surface area of the pipeย =2ฯrh =2ร(22/7)ร2ร77 = 2ร22ร2ร11 = 44ร22 =**968 cmยฒ**

(ii)Outer curve surface area of the pipeย =2ฯrh =2ร(22/7)ร2.2ร77 = 2ร22ร2.2ร11 = 44ร24.2 =**1064.8 cmยฒ**

(iii) Total surface area =Inner surface area + Outer surface area + Area of both cross-sections

Area of both cross-sections =2ฯ(outer radiusยฒ – Inner radiusยฒ)

= 2ร(22/7) [2.2ยฒ -2ยฒ] =(44/7) [4.84 – 4] =(44/7)ร.84 =44ร.12 =5.28 cmยฒ

Total surface area = 968 + 1064.8 +5.28** =2038.08 cmยฒ**

**Q4.The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m ^{2}? (Assume ฯ = 22/7)**

Ans. The diameter of the roller is given 84 cm,therefore radius r = 84/2=42 cm=0.42 m

ย Length,h is 120 cm =1.20 m

Area of the playground = No.of revolution ร Area covered in a single revolution

Area covered in a single revolution = curve surface area of the cylinder

Curve surface area of the cylinder = 2ฯrh = 2ร(22/7)ร 0.42 ร1.20 =2 ร22ร0.06ร1.20 =3.168 mยฒ

Area of the playground =500 ร 3.168 =1584 mยฒ

**Q5.A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m ^{2}.**

Ans. Diameter of cylindrical pillar =50 cm ,therefore radius r =50/2=0.25 m

Height (h) of cylindrical pillar is =3.5 m

Curve surface area of cylindrical pillar =2ฯrh = 2ร(22/7)ร 0.25ร3.5 = 2ร22ร0.25ร0.5=5.5 mยฒ

Cost of painting the curved surface of the pillar=5.5 ร12.50=68.75

Hence the cost of painting the curved surface of Piller is Rs 68.75

**Q6.Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the cylinder is 0.7 m, find its height. (Assume ฯ = 22/7)**

Ans.Curved surface area of a right circular cylinder = 4.4 mยฒ

Radius(r) of the base of the cylinder =0.7 m

Let the height of circular cylinder =h

Curve surface area of the cylinder =2ฯrh =2ร(22/7)ร0.7 h

โด2ร(22/7)ร0.7 h = 4.4

2ร22 ร0.1h =4.4

h = 44/44 =1

Hence height of the cylinder is 1 m

**Q7.The inner diameter of a circular well is 3.5m. It is 10m deep. Find**

**(i) its inner curved surface area,**

**(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m ^{2}.**

**(Assume ฯ = 22/7)**

Ans. The inner diameter of a circular well is 3.5m, therefore radius r =3.5/2 =1.75 m

The depth (h) of the well is = 10 m

(i) Inner curved surface area =2ฯrh

= 2ร(22/7)ร1.75 ร10

=2ร22ร0.25ร10=44ร2.5=110 mยฒ

(ii) The rate of plastering inner curve surface area of the well =Rs. 40 per mยฒ

Cost of plastering inner curved surface area of the well = 40ร110 =4400

Hence the cost of plastering inner curved surface area of the well is Rs 4400

**Q8.In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find**

**the total radiating surface in the system. (Assume ฯ = 22/7)**

Ans. The length, h of the cylindrical pipe = 28 m

Diameter of the pipe is 5 cm,therefore radius ,r of the pipe= 5/2 =2.5 cm =0.025 m

Total radiating surface of the pipe = Curve surface area of the pipe

Total radiating surface of the pipe

= 2ฯrh

= 2ร(22/7)ร0.025 ร28

=2ร22ร0.025ร4

=44ร0.1

=4.4

**Hence the total** **ย radiating surface of the pipe is 4.4 m**

**Q9.Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in**

**diameter and 4.5m high.**

**(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume ฯ = 22/7)**

Ans. The diameter of the closed cylindrical petrol storage tank is 4.2 m,therefore its radius(r) is 4.2/2 =2.1 m

Height(h) of the cylindrical petrol storage tank is =4.5 m

Curved surface area of a closed cylindricalย tank

=2ฯrh

= 2ร(22/7)ร2.1 ร4.5

= 2ร22ร0.3ร4.5

=59.4

**Hence the** **lateral or curved surface area of a closed cylindrical petrol storage tank is 59.4 mยฒ**

(ii) Let the total ย steel actually used in making the tank is x mยฒ

The steel wasted in making the tank is (1/12) of x =x/12

Total surface area of the tank=Total steel used – wasted steel =x -x/12 =11x/12

Total surface area of the tank

=2ฯr(r +h)

=2ร(22/7)ร2.1(2.1+4.5)

=44ร0.3ร6.6

=87.12

11x/12 = 87.12

x = (87.12ร12)/11 = 95.04

**Total steel used in making the tank is 95.04 mยฒ**

**Q10.In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.****The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 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. (Assume ฯ = 22/7)**

Ans. The diameter of the lampshade frame is 20 cm, therefore radius r of the lampshade frame is =20/2 =10 cm

Height h of the lampshade frame is 30 cm

The cloth required for covering the lampshade

=2ฯrH ,where H is length of the cover

= 2ร(22/7)ร10[ h + length of margin at the top and base)]

= 2ร(22/7)ร10[ 30+ 2ร2.5)]

=(440/7)[35]

= 440ร5

=2200

**Hence cloth is required for covering the lampshade is 2200 cmยฒ**

**Q11.The students of 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 cm and height 10.5 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? (Assume ฯ =22/7)**

Ans.ย Rdius r of the penholder is 3 cm and its height h is 10.5 cm

Cardboard required for a single penholder =Curved surface area of penholder +area of its circular base

Cardboard required for a single penholder

=2ฯrh +ฯrยฒ

=ฯr(2h +r)

=(22/7) ร3(2ร10.5 +3)

=(66/7)(21 +3)

=66ร24/7

=1584/7

Cardboard required forย 35 competitors isย

=(35 ร1584)/7

=5 ร1584

=7920

Hence cardboard was required to be bought for the competition is 7920 cmยฒ

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**NCERT Solutions ofย Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 scienceย **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

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**CBSE Class 10-Question paper of maths 2021 with solutions**

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**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |