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

# 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.

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.

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

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

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

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

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

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 = $\frac{4}{3}\times\frac{22}{7}\times (0.63)^3$

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

$\\ = 1.048\ m^3$

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

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

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 $0.004851\ m^3$

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 = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

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 $1: 64$

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 = $= \frac{303.1875}{1000} \approx 0.303\ litres\ \ \ (approx.)$

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

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

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

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

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

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 = $\frac{4}{3}\pi r^3$

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

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

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 $249.48\ m^2$

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 = $\frac{2}{3}\pi r^3$

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

$= 523.908\ m^3$

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,

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

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

$\therefore r' = 3r$

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 $1:9$

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.

## More About NCERT Solutions for Class 9 Maths Exercise 13.8

In order to find the volume of sphere in NCERT solutions for Class 9 Maths exercise 13.8 , first, we need to check the radius of the given sphere. Also if the diameter of the sphere is given, then we need to divide it by 2, to get the radius of the sphere. Then we need to find the cube of the radius that is r3. Next, we need to multiply it (cube of the radius ) with 43. Thus the final answer will be the volume of the sphere. Similarly, the volume of a hemisphere is given by 23r3 cubic units.

Also Read| Surface Areas And Volumes Class 9 Notes

## 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

1. 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.

2. 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.

3. Real-World Applications: The exercise includes problems with practical applications, showing students how these mathematical concepts are relevant in everyday life.

4. CBSE Syllabus Alignment: The solutions adhere to the CBSE syllabus, making them ideal for homework, assignments, and exam preparation.

5. PDF Format: Solutions are available in PDF format, allowing students to download and access them offline for convenience and accessibility.

## NCERT Solutions of Class 10 Subject Wise

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.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

#### National Institute of Open Schooling 12th Examination

Exam Date:20 September,2024 - 07 October,2024

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9