**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**

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**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 cm ^{3}= 1L) (Assume ฯ = 22/7)**

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

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

2ฯr = 132

2(22/7)รr = 132

r = (132ร7)44 =21

Volume ofย cylindrical vessel =ฯrยฒh =(22/7)ร21ร21ร25 =22ร3ร21ร25=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 1cm ^{3}ย 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 woodรvolume of cylindrical wooden pipe

The volume of cylindrical wooden pipe

= ฯ(outer radiusยฒ-inner radiusยฒ)h

=(22/7)[14ยฒ- 12ยฒ]ร35

=(22/7)[196- 144]ร35

=(22/7)ร52ร35

= 110 ร52

=5720

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

The mass of the pipe

=0.6ร5720

=3432.0 g

1kg =1000g

โด 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

=lรbรh

=5ร4ร15

=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

=ฯrยฒh

=(22/7)ร3.5ร3.5ร10

=22ร0.5ร3.5ร10

=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 94.2cm ^{2}ย 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

=2ฯrh,where h =5cm

The given lateral surface of a cylinder is 94.2cmยฒ

2ฯrh =94.2

2ร3.14รrร5 =94.2

31.4r =94.2

r = 94.2/31.4 =3

(ii) Volume of the cylinder

=ฯrยฒh

=3.14ร3ร3ร5

=3.14ร45

=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 m ^{2}, find**

**(i) inner curved surface area of the vessel**

**(ii) radius of the base**

**(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

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

The inner curved surface area of the cylindrical vessel

=110 mยฒ

2ฯrh =110

2(22/7)รrร10 =110

r =(110ร7)/(44ร10)

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

=ฯrยฒh

=(22/7)ร1.75ร1.75ร10

=22ร0.25ร1.75ร10

=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=15.4ร1000=15400 cmยณ

ฯrยฒh = 15400

(22/7)รrยฒร100= 15400

rยฒ = (15400ร7)/(22ร100)

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

=2ฯr(r+h)

=2(22/7)ร7(7+100)

=44ร107

=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

=ฯrยฒh

=(22/7)ร0.05ร0.05ร14

=22ร0.05ร0.05ร2

=0.11

**The volume of the graphite is 0.11 cmยณ22ร0.05ร0.35ร14**

Volume of the pencil is

= ฯRยฒh

=(22/7)ร0.35ร0.35ร14

=22ร0.05ร0.35ร14 =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 cm22ร0.05ร3.5ร4

= ฯrยฒh

=(22/7)ร3.5ร3.5ร4

=22ร0.5ร3.5ร4=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**

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