NCERT Solutions for Class 9 Maths Exercise 13.2 -Chapter 13 Surface Areas and Volumes - Future Study Point

# 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

Class 9 Maths Exercise-13.1

Class 9 Maths Exercise-13.2

Class 9 Maths Exercise-13.3

Class 9 Maths Exercise -13.5

Class 9 Maths Exercise -13.7

Class 9 Maths Exercise -13.8

Class 9 Maths Exercise 13.9

NCERT Solutions Class 9 Maths-All Chapters

### 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 cmยฒ. Find the diameter of the base of the cylinder.(Assume ฯ = 22/7).

Ans. The curved surface area of a right circular cylinder = 2ฯrh

Height,h given =14 cm

CSA is given = 88 cmยฒ

2ฯrh = 88

2ร(22/7) รrร14 = 88

r = 88ร7/(2ร22ร14) = 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 =2ฯr(r +h)

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

= 2ร22ร0.10ร1.7

= 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  =2ฯrh =2ร(22/7)ร2ร77 = 2ร22ร2ร11 = 44ร22 =968 cmยฒ

(ii)Outer curve surface area of the pipeย  =2ฯrh =2ร(22/7)ร2.2ร77 = 2ร22ร2.2ร11 = 44ร24.2 =1064.8 cmยฒ

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

= 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 m2? (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 = 2ฯrh = 2ร(22/7)ร 0.42 ร1.20 =2 ร22ร0.06ร1.20 =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 m2.

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 =2ฯrh = 2ร(22/7)ร 0.25ร3.5 = 2ร22ร0.25ร0.5=5.5 mยฒ

Cost of painting the curved surface of the pillar=5.5 ร12.50=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 m2. 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 =2ฯrh =2ร(22/7)ร0.7 h

โด2ร(22/7)ร0.7 h = 4.4

2ร22 ร0.1h =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 m2.

(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 =2ฯrh

= 2ร(22/7)ร1.75 ร10

=2ร22ร0.25ร10=44ร2.5=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 = 40ร110 =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

= 2ฯrh

= 2ร(22/7)ร0.025 ร28

=2ร22ร0.025ร4

=44ร0.1

=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

=2ฯrh

= 2ร(22/7)ร2.1 ร4.5

= 2ร22ร0.3ร4.5

=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

=2ฯr(r +h)

=2ร(22/7)ร2.1(2.1+4.5)

=44ร0.3ร6.6

=87.12

11x/12 = 87.12

x = (87.12ร12)/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

=2ฯrH ,where H is length of the cover

= 2ร(22/7)ร10[ h + length of margin at the top and base)]

= 2ร(22/7)ร10[ 30+ 2ร2.5)]

=(440/7)[35]

= 440ร5

=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

=2ฯrh +ฯrยฒ

=ฯr(2h +r)

=(22/7) ร3(2ร10.5 +3)

=(66/7)(21 +3)

=1584/7

Cardboard required forย 35 competitors isย

=(35 ร1584)/7

=5 ร1584

=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

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

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### NCERT Solutions of class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

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### NCERT solutions of class 10 science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### Solutions of class 10 last years Science question papers

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### 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

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