NCERT Solutions for Exercise 13.8 Class 9 Maths Chapter 13

NCERT Solutions for Exercise 13.8 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 18, 2023 09:02 AM IST

NCERT Solutions for Class 9 Maths Exercise 13.8 Chapter 13 Surface Areas And Volumes- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8- This 9th class maths exercise 13.8 answers delves into the world of surface areas and volumes, offering students a comprehensive set of problems and expert-crafted solutions. As part of the CBSE syllabus, these class 9 maths chapter 13 exercise 13.8 solutions are designed to assist students in homework, assignments, and exam preparation. They are presented in a clear and accessible format, allowing for easy comprehension and application of key mathematical concepts. In addition, students can download the exercise 13.8 class 9 maths solutions in PDF format, enabling offline access and convenient learning.

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NCERT Solutions for Class 9 Maths Exercise 13.8 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 13.8
Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.8
Q1 (i) Find the volume of a sphere whose radius is 7 cm
More About NCERT Solutions for Class 9 Maths Exercise 13.8
Benefits of NCERT Solutions for Class 9 Maths Exercise 13.8
Key Features of 9th Class Maths Exercise 13.8 Answers
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NCERT Solutions of Class 10 Subject Wise

NCERT Solutions for Exercise 13.8 Class 9 Maths Chapter 13 - Surface Area and Volumes

The concept of the volume of the sphere is discussed in NCERT Solutions for Class 9 Maths exercise 13.8. A sphere is a three-dimensional object that is circular in shape. The radius is the distance between the sphere’s surface and its centre, while the diameter is the distance from one point on the sphere’s surface to another, passing through the centre. 2r, where r is the radius of the sphere, gives the diameter of the sphere.

In exercise 13.8 Class 9 Maths, the amount of space occupied within the sphere is known as the volume of the sphere. Hemisphere refers to the precise half of a sphere. When a sphere is cut exactly in the middle along its diameter, two equal hemispheres result. NCERT solutions for Class 9 Maths chapter 13 exercise 13.8 consists of 10 questions regarding the volume of sphere and hemisphere. Along with the Class 9 Maths chapter, 13 exercise 13.8 the following exercises are also present.

**As per the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

Download the PDF of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

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Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.8

Q1 (i) Find the volume of a sphere whose radius is 7 cm

Answer:

Given,

The radius of the sphere = $r = 7\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (7)^3$

$\\ = \frac{4}{3}\times22\times 7\times 7$

$\\ = \frac{4312}{3}$

$\\ = 1437\frac{1}{3}\ cm^3$

Q1 (ii) Find the volume of a sphere whose radius is $\small 0.63\hspace{1mm}m$

Answer:

Given,

The radius of the sphere = $r = 0.63\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (0.63)^3$

$\\ = 4\times22\times 0.03\times 0.63\times 0.63$

$\\ = 1.048\ m^3$

$\\ = 1.05\ m^3\ \ \ (approx.)$

Q2 (i) Find the amount of water displaced by a solid spherical ball of diameter 28 cm

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{28}{2}\ cm = 14\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times (14)^3$

$\\ = \frac{4}{3}\times22\times 2\times 14\times 14$

$\\ = \frac{34469}{3} \\ = 11489\frac{2}{3}\ cm^3$

Therefore, the amount of water displaced will be $11489\frac{2}{3}\ cm^3$

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter $\small 0.21\hspace{1mm}m$

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{0.21}{2}\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times \left(\frac{0.21}{2} \right )^3$

$\\ = 4\times22\times \frac{0.01\times 0.21\times 0.21}{8}$

$\\ = 11\times 0.01\times 0.21\times 0.21$

$\\ =0.004851\ m^3$

Therefore, amount of water displaced will be $0.004851\ m^3$

Q3 The diameter of a metallic ball is $\small 4.2\hspace{1mm}cm$ . What is the mass of the ball, if the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ?

Answer:

Given,

The radius of the metallic sphere = $r = \frac{4.2}{2}\ cm = 2.1\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere = $\frac{4}{3}\times\frac{22}{7}\times 2.1^3$

$\\ = 4\times22\times 0.1\times 2.1\times 2.1$

$\\ =38.808\ cm^3$

Now, the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$ ,which means,

Mass of $\small 1\ cm^3$ of the metallic sphere = $\small 8.9\hspace{1mm}g$

Mass of $38.808\ cm^3$ of the metallic sphere = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

Q4 The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Answer:

Given,

Let $d_e$ be the diameters of Earth

$\therefore$ The diameter of the Moon = $d_m = \frac{1}{4}d_e$

We know, Volume of a sphere =

$\frac{4}{3}\pi r^3 =\frac{4}{3}\pi \left (\frac{d}{2} \right )^3 = \frac{1}{6}\pi d^3$

$\therefore$ The ratio of the volumes = $\frac{Volume\ of\ the\ Earth}{Volume\ of\ the\ Moon}$

$\\ = \frac{\frac{1}{6}\pi d_e^3}{\frac{1}{6}\pi d_m^3} \\ = \frac{ d_e^3}{(\frac{d_e}{4})^3} \\ = 64: 1$

Therefore, the required ratio of the volume of the moon to the volume of the earth is $1: 64$

Q5 How many litres of milk can a hemispherical bowl of diameter $\small 10.5\hspace{1mm}cm$ hold?

Answer:

The radius of the hemispherical bowl = $r = \frac{10.5}{2}\ cm$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

The volume of the given hemispherical bowl = $\frac{2}{3}\times\frac{22}{7}\times \left (\frac{10.5}{2} \right )^3$

$= \frac{2}{3\times8}\times22\times1.5\times10.5\times10.5$

$= 303.1875\ cm^3$

The capacity of the hemispherical bowl = $= \frac{303.1875}{1000} \approx 0.303\ litres\ \ \ (approx.)$

Q6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer:

Given,

Inner radius of the hemispherical tank = $r_1 = 1\ m$

Thickness of the tank = $1\ cm = 0.01\ m$

$\therefore$ Outer radius = Internal radius + thickness = $r_2 = (1+0.01)\ m = 1.01\ m$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

$\therefore$ Volume of the iron used = Outer volume - Inner volume

$= \frac{2}{3}\pi r_2^3 - \frac{2}{3}\pi r_1^3$

$= \frac{2}{3}\times\frac{22}{7}\times (1.01^3 - 1^3)$

$= \frac{44}{21}\times0.030301$

$= 0.06348\ m^3\ \ (approx)$

Q7 Find the volume of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$ .

Answer:

Given,

The surface area of the sphere = $\small 154\hspace{1mm}cm^2$

We know, Surface area of a sphere = $4\pi r^2$

$\therefore 4\pi r^2 = 154$

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 14\times11 \\ \Rightarrow r^2 = \frac{7\times7}{4} \\ \Rightarrow r = \frac{7}{2} \\ \Rightarrow r = 3.5\ cm$

$\therefore$ The volume of the sphere = $\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times (3.5)^3$

$= 179\frac{2}{3}\ cm^3$

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

$\small Rs\hspace{1mm}20$ is the cost of white-washing $1\ m^2$ of the inside area

$\small Rs\hspace{1mm}4989.60$ is the cost of white-washing $\frac{1}{20}\times4989.60\ m^2 = 249.48\ m^2$ of inside area

(i) Therefore, the surface area of the inside of the dome is $249.48\ m^2$

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$ . If the cost of white-washing is Rs 20 per square metre, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be $r\ m$

Inside the surface area of the dome = $249.48\ m^2$

We know, Surface area of a hemisphere = $2\pi r^2$

$\\ \therefore 2\pi r^2 = 249.48 \\ \Rightarrow r^2 = \frac{249.48\times7}{2\times22} \\ \Rightarrow r = 6.3\ m$

$\therefore$ The volume of the hemisphere = $\frac{2}{3}\pi r^3$

$= \frac{2}{3}\times\frac{22}{7}\times (6.3)^3$

$= 523.908\ m^3$

Q9 (i) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $\small S'$ . Find the radius $\small r'$ of the new sphere

Answer:

Given,

The radius of a small sphere = $r$

The radius of the bigger sphere = $r'$

$\therefore$ The volume of each small sphere= $\frac{4}{3}\pi r^3$

And, Volume of the big sphere of radius $r'$ = $\frac{4}{3}\pi r'^3$

According to question,

$27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi r'^3$

$\\ \Rightarrow r'^3 = 27\times r^3 \\ \Rightarrow r' = 3\times r$

$\therefore r' = 3r$

Q9 (ii) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area $\small S'$ . Find the ratio of S and $\small S'$ .

Answer:

Given,

The radius of a small sphere = $r$

The surface area of a small sphere = $S$

The radius of the bigger sphere = $r'$

The surface area of the bigger sphere = $S'$

And, $r' = 3r$

We know, the surface area of a sphere = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{4\pi r'^2}{4\pi r^2}$

$\\ = \frac{ (3r)^2}{ r^2} \\ = 9$

Therefore, the required ratio is $1:9$

Q10 A capsule of medicine is in the shape of a sphere of diameter $\small 3.5\hspace{1mm}mm$ . How much medicine (in $\small mm^3$ ) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule = $r =\frac{3.5}{2}$

$\therefore$ The volume of the capsule = $\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times(\frac{3.5}{2})^3$

$= \frac{4}{3}\times22\times\frac{0.5\times3.5\times3.5}{8}$

$= 22.458\ mm^3 \approx 22.46\ mm^3\ \ (approx)$

Therefore, $22.46\ mm^3\ \ (approx)$ of medicine is needed to fill the capsule.

Benefits of NCERT Solutions for Class 9 Maths Exercise 13.8

• NCERT solutions for Class 9 Maths exercise 13.8, helps in improving the speed in answering the problems and we can assess our problem-solving ability and prepare accordingly.

• Exercise 13.8 Class 9 Maths, helps us to find the volume of the sphere and hemisphere with given specifications.

• By solving the NCERT solution for Class 9 Maths chapter 13 exercise 13.8 exercises, the method of finding the radius of the sphere can be known.

Key Features of 9th Class Maths Exercise 13.8 Answers

Comprehensive Coverage: Ex 13.8 class 9 covers a variety of problems related to surface areas and volumes, providing a thorough understanding of these mathematical concepts.
Expertly Crafted Solutions: The class 9 ex 13.8 solutions are expertly designed by subject matter experts, offering step-by-step explanations that make complex topics easy to understand.
Real-World Applications: The exercise includes problems with practical applications, showing students how these mathematical concepts are relevant in everyday life.
CBSE Syllabus Alignment: The solutions adhere to the CBSE syllabus, making them ideal for homework, assignments, and exam preparation.
PDF Format: Solutions are available in PDF format, allowing students to download and access them offline for convenience and accessibility.

Also See:

NCERT Solutions of Class 10 Subject Wise

Frequently Asked Questions (FAQs)

1. The hemisphere's volume is ___________

The hemisphere's volume is is equal to 2pi r^3/3 .

2. The cube's volume is calculated using _______ units.

Cubic units are used to measure the volume of the cube .

3. How many faces are there in a sphere?

There is one face in a sphere.

4. Earth is __________ a)Cone b)Cylinder c)Sphere

Earth is a sphere.

5. Find the volume of the sphere with radius 1 m.

The volume of the sphere is equal to 43r3

Given, r=1

V=4/3×(3.14×1^3)

=4.2 m^3

6. Find the volume of the hemisphere with radius 1 m.

The volume of the hemisphere is 23r3

V=2/3×(3.14×1^3 )

=2.09 m^3

7. Define hemisphere, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.8 .

Hemisphere refers to the exact half of a sphere, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.8.

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