NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

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Textbook	NCERT
Board	CBSE
Category	NCERT Solutions
Class	Class 9
Subject	Maths
Chapter	Chapter 13
Exercise	Class 9 Chapter 13 Surface Areas and Volumes Exercise 13.6
Number of Questions Solved	8

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

NCERT TEXTBOOK EXERCISES

Question 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000cm3 =1 L.)

Solution:

We have, circumference of the base = 132 cm
∴ 2πr = 132

Question 2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Solution:

Given, inner diameter = 24 cm
Inner radius (r₁) = 12cm
Outer diameter = 28 cm
Outer radius (r₂) = 14cm
h = 35cm

Question 3. A soft drink is available in two packs

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm.
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:

(i) We have, h = 15cm
l = 5cm, b = 4cm, h = 15 cm
Volume of cuboidical body =l x b x h= 5 x 4 x 15 = 300cm3 …(i)
(ii) We have, Diameter = 7cm
Radius, r = 7/2 cm
Height, h = 10cm

∴ From Eqs. (i) and (ii), we see that volume of a plastic cylinder has greater capacity and its capacity is 385 – 300 = 85 cm3 is more than the tin can.

Question 4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find

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

Solution:

We have, lateral surface of a cylinder = 94.2 cm² and h = 5 cm
∴ 2πrh = 94.2
⇒ 2 x 3.14 x r x 5 = 94.2
⇒ r = 94.2/31.4
⇒ r = 3 cm
(i) Hence, radius of base, r = 3cm
(ii) Volume of a cylinder = πr2h= 3.14(3)²x 5
= 3.14 x 9 x 5
= 141.3 cm³

Question 5. It costs ₹2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹20 per m², find

(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.

Solution:

We have, cost to paint the inner curved surface = ₹2200
Cost to paint per m² = ₹20

(i) Inner curved surface area of the vessel = 110m²
(ii) Hence, radius of the base is 1.75 m.
(iii) Capacity of the vessel = Volume of the vessel

Question 6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Solution:

Capacity of a closed cylindrical vessel = 15.4 L
= 15.4 x 1000cm3 (∵ 1L = 1000 cm²)
∴ πr²h = 15400 cm³
πr² x 100 = 15400
r² = 154 x 7/22 (∵ h= 1m= 100cm)
⇒ r² = 7 x 7
⇒ r = 7 cm
Total surface area of closed cylindrical vessel = 2πr(r + h)

Question 7. 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.

Solution:

Question 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Solution:

We have, diameter = 7cm
Radius, r = 7/22 cm
h= 4 cm
Capacity of the cylindrical bowl = Volume of the cylinder = πr²h

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 PDF

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Other Chapter of Class 9 Maths Chapter 13 Surface Areas and Volumes

• Surface Areas and Volumes Class 9 Ex 13.1
• Surface Areas and Volumes Class 9 Ex 13.2
• Surface Areas and Volumes Class 9 Ex 13.3
• Surface Areas and Volumes Class 9 Ex 13.4
• Surface Areas and Volumes Class 9 Ex 13.5
• Surface Areas and Volumes Class 9 Ex 13.6
• Surface Areas and Volumes Class 9 Ex 13.7
• Surface Areas and Volumes Class 9 Ex 13.8
• Surface Areas and Volumes Class 9 Ex 13.9