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)

ex.13.5 class 9 maths

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.

surface areas and volume class 9 ex.13.6

NCERT Solutions for Class 9 Maths  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 π = 22/7)

Ans. Circumference of the base of the cylindrical vessel is = 2πr

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

2πr = 132

2(22/7)×r = 132

r = (132×7)44 =21

Volume of cylindrical vessel =πr²h =(22/7)×21×21×25 =22×3×21×25=34650 cm³

34650 cm³ =34650 cm³/1000 =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 1cm3 of wood has a mass of 0.6g. (Assume π = 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 1cm³ of wood×volume of cylindrical wooden pipe

The volume of cylindrical wooden pipe

= π(outer radius²-inner radius²)h

=(22/7)[14²- 12²]×35

=(22/7)[196- 144]×35

=(22/7)×52×35

= 110 ×52

=5720

The mass of 1 cm³ cylindrical wooden pipe is = 0.6 g

The mass of the pipe

=0.6×5720

=3432.0 g

1kg =1000g

∴ 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 – (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 π=22/7)

Ans.

Q3 class 9 maths ex.13.5

(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

=l×b×h

=5×4×15

=300

Hence the capacity of the soft drink container with rectangular base is 300 cm³

(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

=πr²h

=(22/7)×3.5×3.5×10

=22×0.5×3.5×10

=385

Hence the capacity of the cylindrical soft drink container with circular base is 385 cm³

385 -300 =85

The capacity of cylindrical soft drink is 85 cm³ more than the capacity of soft drink with rectangular base

Q4.If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find

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

Ans. (i)Lateral surface area of the cylinder

=2πrh,where h =5cm

The given lateral surface of a cylinder is 94.2cm²

2πrh =94.2

2×3.14×r×5 =94.2

31.4r =94.2

r = 94.2/31.4 =3

(ii) Volume of the cylinder

=πr²h

=3.14×3×3×5

=3.14×45

=141.3

Hence the capacity of the cylinder  is 141.3 cm³

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

(ii) radius of the base

(iii) capacity of the vessel

(Assume π = 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 m²

(i) Inner curved surface area of the vessel 

= 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 m²

(ii) Inner curved surface area of the cylindrical vessel

= 2πrh,where r is the radius of the vessel and h =10 m

The inner curved surface area of the cylindrical vessel

=110 m²

2πrh =110

2(22/7)×r×10 =110

r =(110×7)/(44×10)

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

=πr²h

=(22/7)×1.75×1.75×10

=22×0.25×1.75×10

=96.25

Hence the capacity of the cylindrical vessel is 96.25 m³

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 π = 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=15.4×1000=15400 cm³

πr²h = 15400

(22/7)×r²×100= 15400

r² = (15400×7)/(22×100)

r² =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

=2πr(r+h)

=2(22/7)×7(7+100)

=44×107

=4708 cm²

4708=4708/10000=0.4708 m²

Hence 0.4708 m² 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 π = 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

=πr²h

=(22/7)×0.05×0.05×14

=22×0.05×0.05×2

=0.11

The volume of the graphite is 0.11 cm³22×0.05×0.35×14

Volume of the pencil is

= πR²h

=(22/7)×0.35×0.35×14

=22×0.05×0.35×14 =5.39 cm³

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 cm³

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 π = 22/7)

Ans.

Q8 ex.13.5 class 9 maths

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 cm22×0.05×3.5×4

= πr²h

=(22/7)×3.5×3.5×4

=22×0.5×3.5×4=154 cm³

The soup in a bowl is 154 cm³

The soup served to 250 patients

154× 250 =38500 cm³=38500/1000 =38.5 litres

Hence the soup served to 250 patients is 38.5 litres

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