NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2) - Future Study Point

NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2)

NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2) have been brought to you by India’s growing website Future Study Point.NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2) are created here to boost your preparation for Term-2 CBSE Board exam.NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2) will help you in clearing your doubts on solving unsolved questions of your NCERT Maths exercise 13.6 Surface areas and the volumes. Exercise 13.6 Surface areas and Volumes are based on the questions related to volumes of cylindrical objects which are used in our day-to-day lives. You can study here NCERT solutions of maths and science, solutions of last year’s question papers, and sample papers, important science notes, and articles related to your carrier.

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

NCERT Solutions Class 9 Maths-All Chapters

NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes (Term 2)

Q1.The circumference of the base of cylindrical vessel is 132cm and its height is 25cm.How many litres of water can it hold? (1000 cm3= 1L) (Assume ฯ = 22/7)

Ans. Circumference of the base of the cylindrical vessel is = 2ฯr

The given circumference of the base of the cylindrical vessel is 132 cm and height(h) =25 cm

2ฯr = 132

2(22/7)รr = 132

r = (132ร7)44 =21

Volume ofย cylindrical vessel =ฯrยฒh =(22/7)ร21ร21ร25 =22ร3ร21ร25=34650 cmยณ

34650 cmยณ =34650 cmยณ/1000 =34.65 l

Q2.The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm3ย of wood has a mass of 0.6g. (Assume ฯ = 22/7)

Ans. The inner diameter of a cylindrical wooden pipe is 24cm, therefore inner radius (r) of the wooden pipe is 12 cm

The outer diameter of a cylindrical wooden pipe is 28cm, therefore outer radius (r) of the wooden pipe is 14 cm

The mass of the pipe = mass of 1cmยณ of woodรvolume of cylindrical wooden pipe

The volume of cylindrical wooden pipe

=(22/7)[14ยฒ- 12ยฒ]ร35

=(22/7)[196- 144]ร35

=(22/7)ร52ร35

= 110 ร52

=5720

The mass of 1 cmยณ cylindrical wooden pipe is = 0.6 g

The mass of the pipe

=0.6ร5720

=3432.0 g

1kg =1000g

โด 3432 g =3432/1000 =3.432 kg

Hence the mass of the pipe is 3.432 kg

Q3.A soft drink is available in two packs โ (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume ฯ=22/7)

Ans.

(i) The length(l) of soft drink container is 5 cm,breadth(b) 4 cm and height(h) is 15 cm

The capacity of the soft drink container

=lรbรh

=5ร4ร15

=300

Hence the capacity of the soft drink container with rectangular base is 300 cmยณ

(ii) The diameter of the cylindrical soft drink container is 7 cm,therefore its radius(r) is 3.5 cm

Height(h) of the container is 10 cm

The capacity of the container

=ฯrยฒh

=(22/7)ร3.5ร3.5ร10

=22ร0.5ร3.5ร10

=385

Hence the capacity of the cylindrical soft drink container with circular base is 385 cmยณ

385 -300 =85

The capacity of cylindrical soft drink is 85 cmยณ more than the capacity of soft drink with rectangular base

Q4.If the lateral surface of a cylinder is 94.2cm2ย and its height is 5cm, then find

(i) radius of its base (ii) its volume.[Use ฯ= 3.14]

Ans. (i)Lateral surface area of the cylinder

=2ฯrh,where h =5cm

The given lateral surface of a cylinder is 94.2cmยฒ

2ฯrh =94.2

2ร3.14รrร5 =94.2

31.4r =94.2

r = 94.2/31.4 =3

(ii) Volume of the cylinder

=ฯrยฒh

=3.14ร3ร3ร5

=3.14ร45

=141.3

Hence the capacity of the cylinderย  is 141.3 cmยณ

Q5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find

(i) inner curved surface area of the vessel

(iii) capacity of the vessel

(Assume ฯ = 22/7)

Ans.The cost of painting a cylindrical vessel is Rs 2200

The depth(h) of the vessel is 10 m

The rate of painting is Rs 20 per mยฒ

(i) Inner curved surface area of the vesselย

= The cost of painting a cylindrical vessel/The rate of painting

=2200/20

=110

Hence the Inner curved surface area of the vessel is 110 mยฒ

(ii) Inner curved surface area of the cylindrical vessel

= 2ฯrh,where r is the radius of the vessel and h =10 m

The inner curved surface area of the cylindrical vessel

=110 mยฒ

2ฯrh =110

2(22/7)รrร10 =110

r =(110ร7)/(44ร10)

r =7/4 =1.75 cm

Hence inner curved surface area of the cylindrical vessel is 1.75 cm

(iii) Capacity of the cylindrical vessel

=ฯrยฒh

=(22/7)ร1.75ร1.75ร10

=22ร0.25ร1.75ร10

=96.25

Hence the capacity of the cylindrical vessel is 96.25 mยณ

Q6.The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume ฯ = 22/7)

Ans.The height(h) of closed cylindrical vessel is 1 m=100 cm

The capacity of a closed cylindrical vessel =15.4 liters=15.4ร1000=15400 cmยณ

ฯrยฒh = 15400

(22/7)รrยฒร100= 15400

rยฒ = (15400ร7)/(22ร100)

rยฒ =49

r =7

Radius of the vessel is 7 cm

The area of the metal sheet required to make the vessel

= Total surface area of the closed cylindrical vessel

=2ฯr(r+h)

=2(22/7)ร7(7+100)

=44ร107

=4708 cmยฒ

4708=4708/10000=0.4708 mยฒ

Hence 0.4708 mยฒ metal sheet would be needed to make the vessel

Q7.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. (Assume ฯ = 22/7)

Ans.The diameter of the pencil is 7 mm =0.7 cm therefore radius (R) of the pencil is 0.7/2 =0.35 cm

Diameter of the graphite is 1mm =0.1 cm ,therefore radius (r) of the graphite is 0.1/2 =0.05 cm

The length(h) of the pencil is 14 cm

Volume of the graphite

=ฯrยฒh

=(22/7)ร0.05ร0.05ร14

=22ร0.05ร0.05ร2

=0.11

The volume of the graphite is 0.11 cmยณ22ร0.05ร0.35ร14

Volume of the pencil is

= ฯRยฒh

=(22/7)ร0.35ร0.35ร14

=22ร0.05ร0.35ร14 =5.39 cmยณ

The volume of the wood in the pencil is

= Volume of the pencil – Volume of the graphite

=5.39 – 0.11

=5.28

Hence the volume of the wood in the pencil is 5.28 cmยณ

Q8.A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume ฯ = 22/7)

Ans.

The diameter of the cylindrical bowl is 7 cm, therefore radius(r) of the bowl is 7/2 =3.5 cm

The bowl filled with soup to a height(h) of 4 cm22ร0.05ร3.5ร4

= ฯrยฒh

=(22/7)ร3.5ร3.5ร4

=22ร0.5ร3.5ร4=154 cmยณ

The soup in a bowl is 154 cmยณ

The soup served to 250 patients

154ร 250 =38500 cmยณ=38500/1000 =38.5 litres

Hence the soup served to 250 patients is 38.5 litres

You can compensate us by donating any amount of money for our survival

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

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

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

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

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

CBSE Class 10 – Question paper of science 2020 with solutions

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

Scroll to Top