NCERT Solutions for Exercise 13.6 Class 9 Maths Chapter 13

NCERT Solutions for Exercise 13.6 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 17, 2023 09:16 AM IST

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

NCERT Solutions for Class 9 Maths Exercise 13.6 deals with the concept of the volume of the right circular cylinder. A right circular cylinder is a cylinder with a closed circular surface, two parallel bases on both ends and elements that are perpendicular to the base. The volume of the right circular cylinder is the capacity of the cylinder in exercise 13.6 Class 9 Maths, which estimates the amount of material that it can carry. The volume of the right circular cylinder can be calculated by using the dimensions such as the radius and height of the right circular cylinder.

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NCERT Solutions for Class 9 Maths Exercise 13.6 Chapter 13 Surface Areas and Volumes- Download Free PDF
Download the PDF of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6
Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.6
Q1 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? ( )
Q8 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? Answer: Given, Height = The radius of the cylindrical bowl = The volume of soup in a bowl for a single person = The volume of soup given for 250 patients = Therefore, the amount of soup the hospital has to prepare daily to serve 250 patients is
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Key Features of 9th Class Maths Exercise 13.6 Answers
NCERT Solutions of Class 10 Subject Wise

NCERT Solutions for Exercise 13.6 Class 9 Maths Chapter 13 - Surface Area and Volumes

The product of the area of the base and the height of the right circular cylinder gives the volume of the right circular cylinder. That is V=πr²h where r is the radius of the cylinder's round base. The NCERT solutions for Class 9 Maths exercise 13.6 also focused on the volume of the hollow cylinder. A cylinder that is empty from the inside and has some difference between the internal and external radius is known as a hollow cylinder. NCERT Solutions for Class 9 Maths chapter 13 exercise 13.6 consists of 8 questions regarding the volume of the right circular cylinder. The concepts related to the volume of the right circular cylinder and hollow cylinder are well explained in this NCERT syllabus Class 9 Maths chapter 13 exercise 13.6. Along with NCERT book exercise 13.6 class 9 maths solution, the following exercises are also present.

**This chapter has been renumbered as Chapter 11 according to the CBSE Syllabus 2023-24.

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

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Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.6

Q1 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? ( $\small 1000\hspace{1mm}cm^3=1l$ )

Answer:

Given,

Circumference of the base of a cylindrical vessel = $132\ cm$

Height = $h = 25\ cm$

Let the radius of the base of the cylinder be $r\ cm$

Now, Circumference of the circular base of the cylinder = $2\pi r = 132\ cm$

$\\ \Rightarrow 2\times\frac{22}{7}\times r = 12\times11 \\ \\ \Rightarrow r = \frac{12\times7}{4} = 21\ cm$

$\therefore$ The volume of the cylinder = $\pi r^2 h$

$\\ = \frac{22}{7}\times (21)^2 \times25 \\ = 22\times3\times21\times25 \\ = 34650\ cm^3$

Also, $\small 1000\hspace{1mm}cm^3=1l$

$\therefore$ $34650\ cm^3 = \frac{34650}{1000} = 34.65\ litres$

Therefore, the cylindrical vessel can hold $34.65\ litres$ of water.

Q2 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 $\small 1\hspace{1mm}cm^3$ of wood has a mass of $\small 0.6\hspace{1mm}g$ .

Answer:

Given, A hollow cylinder made of wood.

The inner diameter of the cylindrical wooden pipe = $r_1 = 24\ cm$
Outer diameter = $r_2 = 28\ cm$

Length of the pipe = $h= 35\ cm$

$\therefore$ The volume of the wooden cylinder = $\pi r^2 h = \pi (r_2^2 - r_1^2)h$

$\\ = \frac{22}{7}\times (14^2-12^2) \times35 \\ = 22\times (14-12)(14+12) \times5 \\ = 22\times (2)(26) \times5 \\ = 5720\ cm^3$

Mass of $\small 1\hspace{1mm}cm^3$ wood = $\small 0.6\hspace{1mm}g$
Mass of $5720\ cm^3$ wood = $5720\times0.6\ g$
$\\ = 3432\ g \\ = 3.432\ kg$

Therefore, the mass of the wooden hollow cylindrical pipe is $3.432\ kg$

Q3 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 and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer:

Given,

(i) The dimension of the rectangular base of the tin can = $5\ cm\times4\ cm$

Height of the can = $h = 15\ cm$

$\therefore$ The volume of the tin can = $Rectangular\ area\times height$

$(5\times4\times15)\ cm^3 = 300\ cm^3$

(ii) The radius of the circular base of the plastic cylinder = $r = \frac{7}{2}= 3.5\ cm$

Height of the cylinder = $h = 10\ cm$

$\therefore$ The volume of the plastic cylinder = $\pi r^2 h$

$\\ = \frac{22}{7}\times (3.5)^2 \times10 \\ = 22\times0.5\times3.5\times10 \\ = 11\times35 \\ = 385\ cm^3$

Clearly, the plastic cylinder has more capacity than the rectangular tin can.

The difference in capacity = $(385-300)\ cm^3 = 85\ cm^3$

Q4 (i) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find radius of its base (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

(i) Let the radius of the base be $r\ cm$

We know,

The lateral surface area of a cylinder = $2\pi r h$

$\\ \therefore 2\pi r h = 94.2 \\ \Rightarrow 2\times3.14\times r\times5 = 94.2 \\ \\ \Rightarrow r = \frac{94.2}{31.4} = 3\ cm$

Therefore, the radius of the base is $3\ cm$

Q4 (ii) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find it's volume. (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

The radius of the base is $3\ cm$

(ii) We know,

The volume of a cylinder = $\pi r^2 h$

$\\ = 3.14\times3^2\times5 \\ = 3.14 \times 9 \times 5= 3.14 \times 45 \\ = 141.3\ cm^3$

Therefore, the volume of the cylinder is $141.3\ cm^3$ .

Q5 (i) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find inner curved surface area of the vessel.

Answer:

(i) Given,

Rs 20 is the cost of painting $1\small m^2$ area of the inner curved surface of the cylinder.

$\therefore$ Rs 2200 is the cost of painting $\frac{1}{20}\times2200 = 110\ m^2$ area of the inner curved surface of the cylinder.

$\therefore$ The inner curved surface area of the cylindrical vessel = $110\ m^2$

Q5 (ii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find radius of the base.

Answer:

The inner curved surface area of the cylindrical vessel = $110\ m^2$

Height of the cylinder = $h = 10\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The inner curved surface area of the cylindrical vessel = $2\pi r h$

$\\ \Rightarrow 2\times\frac{22}{7}\times r \times10 = 110 \\ \Rightarrow 44\times r = 11\times7 \\ \Rightarrow 4\times r = 7 \\ \Rightarrow r = \frac{7}{4} = 1.75\ m$

Therefore, the radius of the base of the vessel is $1.75 \ m$

Q5 (iii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find capacity of the vessel.

Answer:

(iii) Height of the cylinder = $h = 10\ m$

Radius of the base of the vessel = $r =1.75 \ m$

$\therefore$ Volume of the cylindrical vessel = $\pi r^2 h$

$\\ = \frac{22}{7}\times (1.75)^2 \times10 \\ = 22\times2.5\times1.75\times10 \\ = 55\times17.5 \\ = 96.25 \ m^3$

Therefore, the capacity of the cylindrical vessel is $96.25 \ m^3$

Q6 The capacity of a closed cylindrical vessel of height 1 m is $\small 15.4$ litres. How many square metres of metal sheet would be needed to make it?

Answer:

(Using capacity(volume), we will find the radius and then find the surface area)

The capacity of the vessel = Volume of the vessel = $\small 15.4$ litres

Height of the cylindrical vessel = $h = 1\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The volume of the cylindrical vessel = $\pi r^2 h$

$\\ \Rightarrow \frac{22}{7}\times r^2 \times1 = 15.4\ litres= 0.0154\ m^3 \\ \Rightarrow r^2 = \frac{0.0014\times11\times7}{22} \\ \Rightarrow r^2 = \frac{7\times7}{10000} \\ \Rightarrow r = \frac{7}{100}= 0.07\ m$

Therefore, the total surface area of the vessel = $2\pi r h+ 2\pi r^2 = 2\pi r(r+h)$

$\\ = 2\times\frac{22}{7}\times0.07\times(0.07+1) \\ = 0.44 \times 1.07 \\ = 0.4708\ m^2$

Therefore, square metres of metal sheet needed to make it is $0.4708\ m^2$

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.

Answer:

Given,

Length of the cylindrical pencil = $h = 14\ cm$

The radius of the graphite (Inner solid cylinder) = $r_1 = \frac{1}{2}\ mm = 0.05\ cm$

Radius of the pencil (Inner solid graphite cylinder + Hollow wooden cylinder) =

= $r_2 = \frac{7}{2}\ mm = 0.35\ cm$

We know, Volume of a cylinder= $\pi r^2 h$

$\therefore$ The volume of graphite = $\pi r_1^2 h$

$\\ = \frac{22}{7}\times 0.05^2 \times14 \\ = 44\times0.0025 \\ = 0.11\ cm^3$

And, Volume of wood = $\pi (r_2^2- r_1^2) h$

$\\ = \frac{22}{7}\times (0.35^2-0.05^2) \times14 \\ = 44\times(0.35-0.05)(0.35+0.05) \\ = 44\times(0.30)(0.40) \\ = 5.28 \ cm^3$

Therefore, the volume of wood is $5.28 \ cm^3$ and the volume of graphite is $0.11 \ cm^3$

Q8 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?
Answer:
Given,
Height = $h = 4\ cm$
The radius of the cylindrical bowl = $r_1 = \frac{7}{2}\ cm = 3.5\ cm$
$\therefore$ The volume of soup in a bowl for a single person = $\pi r^2 h$
$\\ = \frac{22}{7}\times (3.5)^2 \times4 \\ = 88\times0.5\times3.5 \\ = 154\ cm^3$
$\therefore$ The volume of soup given for 250 patients = $(250 \times 154)\ cm^3$
$\\ = 38500\ cm^3 \\ = 38.5\ litres$
Therefore, the amount of soup the hospital has to prepare daily to serve 250 patients is $38.5\ litres$

Benefits of NCERT Solutions for Class 9 Maths Exercise 13.6

• NCERT solutions for Class 9 Maths exercise 13.6 give us a general notion of several figures that are always comparable to one another, regardless of their size. Example: The shapes of a sphere, a cuboid, and a cone.

• Solving the NCERT solution for Class 9 Maths chapter 13 exercise 13.6 teaches us how to solve problems quickly and easily, as well as how to analyse situations and correctly apply formulas. It also improves our efficiency and speed in solving problems.

• Class 9 Maths Exercise 13.6 assists us in determining the volume of the cylinder as well as the radius and height of the cylinder using the volume formula.

Key Features of 9th Class Maths Exercise 13.6 Answers

Extensive Coverage: This exercise comprehensively covers topics related to surface areas and volumes of various geometrical solids, including spheres, hemispheres, and more.
Practical Applications: Class 9 ex 13.6 solutions provide practical insights, helping students understand the real-life relevance of these geometric concepts.
Expert-Crafted Solutions: The ex 13.6 class 9 solutions are expertly crafted to offer clear explanations and step-by-step problem-solving techniques.
CBSE 2023-24 Syllabus Alignment: The class 9 maths ex 13.6 aligns with the CBSE 2023-24 syllabus, ensuring that students are well-prepared for their examinations.
Homework and Assignment Support: These solutions are invaluable for completing homework and assignments with confidence.
Quick Reference Guide: Students can use these 9th class maths exercise 13.6 answers as a quick reference guide to reinforce their understanding of geometry, 3D shapes, and surface area calculations.

Also See:

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Frequently Asked Questions (FAQs)

1. Volume of the right circular cylinder is _________

Volume of the right circular cylinder is r2h.

2. Volume of the hollow cylinder is __________

Volume of the hollow cylinder is πhr12–r22

3. Define hollow cylinder.

A cylinder that is empty from the inside and has some difference between the internal and external radius is known as a hollow cylinder.

4. A pencil is ______ a)Cone b)Cylinder c)Sphere

A pencil is cylinder

5. The volume of the right circular cylinder can be calculated by using ________

The volume of the right circular cylinder can be calculated by using the dimensions.

6. What are the dimensions of the right circular cylinder ?

The dimensions of the right circular cylinder are radius and height.

7. Define a right circular cylinder, according to NCERT solutions for Class 9 maths chapter 13 exercise 13.6 .

A cylinder with a closed circular surface, two parallel bases on both ends, and elements that are perpendicular to its base is known as a right circular cylinder, according to NCERT solutions for Class 9 maths chapter 13 exercise 13.6.

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