NCERT Solutions for Exercise 13.4 Class 9 Maths Chapter 13

NCERT Solutions for Exercise 13.4 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 16, 2023 09:01 AM IST

NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Areas And Volumes- Download Free PDF

9th class maths exercise 13.4 answers is a significant component of Chapter 13. It consists of nine questions, each with multiple parts. These solutions have been meticulously crafted by subject matter experts from Careers360, ensuring they are presented in simple and detailed language for enhanced understanding. Furthermore, students can conveniently access the PDF version of these class 9 maths chapter 13 exercise 13.4 solutions, allowing them to study offline at their convenience without the need for an internet connection.

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NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Areas And Volumes- Download Free PDF
Access Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.4
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Key Features of Class 9 Maths Ex 13.4
NCERT Solutions of Class 10 Subject Wise

NCERT Solutions for Exercise 13.4 Class 9 Maths Chapter 13 - Surface Area and Volumes

exercise 13.4 class 9 maths deals with the concept of the sphere, hemisphere, and it's surface area. A three-dimensional object which is round in shape is known as a sphere. Radius is the distance between the surface and centre of the sphere and diameter is the distance from one point to another point on the surface of the sphere, passing through the centre. The diameter of the sphere is given by 2r where r is the radius of the sphere.

In exercise 13.4 Class 9 Maths, The total area covered by the surface of a sphere in a three-dimensional space is known as the surface area of the sphere. The amount of space occupied by the sphere is known as the volume of the sphere . The nine questions in NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.4 are based on the notion of surface areas and volumes of spheres. Class 9 Maths chapter 13 exercise 13.4 thoroughly explains the concepts of surface area and volume. The following activities are included along with Class 9 Maths chapter 13 exercise 13.4 .

**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 13.4

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Access Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.4

Q1 (i) Find the surface area of a sphere of radius: $\small 10.5\hspace{1mm}cm$ .

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(10.5)^2$

$\\ =88\times1.5\times10.5 \\ = 1386\ cm^2$

Q1 (ii) Find the surface area of a sphere of radius: $\small 5.6\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(5.6)^2$

$\\ =88\times0.8\times5.6 \\ = 394.24\ cm^2$

Q1 (iii) Find the surface area of a sphere of radius: $\small 14\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(14)^2$

$\\ = 88\times2\times14 \\ = 2464 \ cm^2$

Q2 (i) Find the surface area of a sphere of diameter: 14 cm

Answer:

Given,

The diameter of the sphere = $14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{14}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{14}{2}\times\frac{14}{2}$

$\\ = 22\times2\times14 \\ = 616\ cm^2$

Q2 (ii) Find the surface area of a sphere of diameter: 21 cm

Answer:

Given,

The diameter of the sphere = $21\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{21}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{21}{2}\times\frac{21}{2}$

$\\ = 22\times3\times21 \\ = 1386\ cm^2$

Q2 (iii) Find the surface area of a sphere of diameter: $\small 3.5\hspace{1mm}m$

Answer:

Given,

The diameter of the sphere = $3.5\ m$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{3.5}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{3.5}{2}\times\frac{3.5}{2}$

$\\ = 22\times0.5\times3.5 \\ = 38.5\ m^2$

Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use $\small \pi =3.14$ )

Answer:

We know,

The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end

$= 2\pi r^2 + \pi r^2 = 3\pi r^2$

$\therefore$ The required total surface area of the hemisphere = $3\times3.14\times(10)^2$

$\\ = 942\ cm^2$

Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer:

Given,

$r_1 = 7\ cm$

$r_2 = 14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of surface areas of the ball in the two cases = $\frac{Initial}{Final} = \frac{4\pi r_1^2}{4\pi r_2^2}$

$= \frac{r_1^2}{r_2^2}$

$\\ = \left (\frac{7}{14} \right )^2 \\ \\ = \left (\frac{1}{2} \right )^2 \\ \\ = \frac{1}{4}$

Therefore, the required ratio is $1:4$

Q5 A hemispherical bowl made of brass has inner diameter $\small 10.5\hspace{1mm}cm$ . Find the cost of tin-plating it on the inside at the rate of Rs 16 per $\small 100\hspace{1mm}cm^2$ .

Answer:

Given,

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

We know,

The curved surface area of a hemisphere = $2\pi r^2$

$\therefore$ The surface area of the hemispherical bowl = $2\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2$

$=11\times1.5\times10.5$

$= 173.25 \ cm^2$

Now,

Cost of tin-plating $\small 100\hspace{1mm}cm^2$ = Rs 16

$\therefore$ Cost of tin-plating $\small 33\hspace{1mm}cm^2$ = $\small \\ Rs. \left (\frac{16}{100}\times173.25 \right )$

$\small = Rs. 27.72$

Therefore, the cost of tin-plating it on the inside is $\small Rs. 27.72$

Q6 Find the radius 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,

The surface area of a sphere of radius $r$ = $4\pi r^2$

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

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 154$

$\\ \Rightarrow r^2 = \frac{154\times7}{4\times22} = \frac{7\times7}{4}$

$\\ \Rightarrow r = \frac{7}{2}$

$\\ \Rightarrow r = 3.5\ cm$

Therefore, the radius of the sphere is $3.5\ cm$

Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Let diameter of Moon be $d_m$ and diameter of Earth be $d_e$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{Surface\ area\ of\ moon}{Surface\ area\ of\ Earth}$

$= \frac{4\pi \left (\frac{d_m}{2} \right )^2}{4\pi \left (\frac{d_e}{2} \right )^2}$

$= \frac{d_m^2}{d_e^2}$

$=\left ( \frac{\frac{1}{4}d_e}{d_e} \right )^2$

$= \frac{1}{16}$

Therefore, the ratio of the surface areas of the moon and earth is $= 1:16$

Q8 A hemispherical bowl is made of steel, $\small 0.25\hspace{1mm}cm$ thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer:

Given,

The inner radius of the bowl = $r_1 = 5\ cm$

The thickness of the bowl = $\small 0.25\hspace{1mm}cm$

$\therefore$ Outer radius of the bowl = (Inner radius + thickness) =

$r_2 = 5+0.25 = 5.25\ cm$

We know, Curved surface area of a hemisphere of radius $r$ = $2\pi r^2$

$\therefore$ The outer curved surface area of the bowl = $2\pi r_2^2$

$= 2\times\frac{22}{7}\times (5.25)^2$

$= 2\times\frac{22}{7}\times5.25\times5.25 = 173.25\ cm^2$

Therefore, the outer curved surface area of the bowl is $173.25\ cm^2$

Q9 (i) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find surface area of the sphere,

1640847480239

Answer:

Given,

The radius of the sphere = $r$

$\therefore$ Surface area of the sphere = $4\pi r^2$

Q9 (ii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find curved surface area of the cylinder,

1640847500491

Answer:

Given,

The radius of the sphere =

$\therefore$ The surface area of the sphere = $4\pi r^2$

According to the question, the cylinder encloses the sphere.

Hence, the diameter of the sphere is the diameter of the cylinder.

Also, the height of the cylinder is equal to the diameter of the sphere.

We know, the curved surface area of a cylinder = $2\pi rh$

$= 2\pi r(2r) = 4\pi r^2$

Therefore, the curved surface area of the cylinder is $4\pi r^2$

Q9 (iii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find ratio of the areas obtained in (i) and (ii).

1640847519663

Answer:

The surface area of the sphere = $4\pi r^2$

And, Surface area of the cylinder = $4\pi r^2$

So, the ratio of the areas = $\frac{4\pi r^2}{4\pi r^2} = 1$

Benefits of NCERT Solutions for Class 9 Maths Exercise 13.4 :

• NCERT solutions for Class 9 Maths exercise 13.4, helps in finding the radius and diameter of a sphere and hemisphere by using the formula of total surface area of sphere and hemisphere respectively.

• NCERT book Exercise 13.4 Class 9 Maths, the questions are explained clearly with proper geometric figures and explanations in a step-by-step procedure for our good understanding that will help us to secure more marks.

• By answering the NCERT syllabus Class 9 Maths chapter 13 exercise 13.4 exercises, we may improve our grades in the first and second terms, and the formulae for calculating surface area and volume can aid us in solving competitive problems, therefore we should remember them.

Key Features of Class 9 Maths Ex 13.4

Comprehensive Solutions: These 9th class maths exercise 13.4 answers cover and clarify all questions in Exercise 13.4.
Sphere and Hemisphere Calculations: They assist in determining the radius of a sphere and the total surface area of a hemisphere.
Alignment with NCERT Guidelines: The ex 13.4 class 9 solutions adhere to NCERT guidelines, aiding students in their exam preparation.
Stepwise Expert Solutions: Subject matter experts have provided step-by-step solutions, helping students understand and score better.

Also See:

NCERT Solutions of Class 10 Subject Wise

Frequently Asked Questions (FAQs)

1. Define sphere, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 .

A three-dimensional object with a round shape is called a sphere, according to NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 .

2. Define radius and diameter.

Radius is the distance between surface and centre of the sphere whereas the diameter is the distance from one point to another point on the surface of the sphere, passing through the centre.

3. The surface area of the sphere is _________

The sphere has a surface area of 4πr^2 .

4. What are the different kinds of surface areas?

Surface area can be divided into two categories. They are.

Area of Lateral Surface
Area of Total Surface

5. The total surface area of hemisphere is _________

The total surface area of hemisphere is 3πr^2

6. The Lateral surface area of the hemisphere is __________ .

The Lateral surface area of hemisphere is 2πr^2 .

7. According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 , How hemispheres are formed from a sphere ?

According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 , When a sphere is cut at the exact centre along its diameter which leaves two equal hemispheres.

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