ncert solutions for class 9 maths chapter 13 exercise 13.8

Examinations are a way for educators to assess how well a student has learned the material that has been taught to them in school. They can also be employed to determine a student’s eligibility to continue in a course or to graduate. Exams can be written or oral, and they can be given individually or in groups. Students in high school and college spend a lot of time studying and preparing for examinations. These tests are a very important part of their education, as they can determine a student’s future and how successful they will be. Examinations can be demanding for students, but they are also a great way to learn and improve their skills. Similarly, students try to seek out help regarding enhancing their Mathematical abilities and getting help for their exams. For such scenarios, students can take a look at the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 provided by the Extramarks’ website. Students should make sure they study hard and prepare well for their Examinations, to get into the college of their dreams.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

The National Council of Educational Research and Training, often known as NCERT, is an independent institution in India that was established in 1961. The primary goal of this association is to create educational resources for Indian schools. NCERT supports students’ academic success by offering study materials and holding tests. Additionally, it administers examinations at the school level. It is a department of the Indian government that offers textbooks and other study aids to students. NCERT is renowned for offering excellent means of education that support students in their academic endeavours.

In India, a national-level organization called the National Council of Educational Research and Training is in charge of creating the academic curriculum for all of the nation’s CBSE schools. One of the main benefits of using NCERT resources is that they are closely aligned with the curriculum set by the Central Board of Secondary Education (CBSE). This means that the textbooks and other resources provided by NCERT are designed to cover all the topics and concepts that are taught in CBSE schools. As a result, using NCERT resources can help students prepare for their CBSE examinations. Besides the facilities of practice questions and study materials that Extramarks gives, it also provides online resources such as revision notes, sample papers, and past years’ exam papers. These resources can be accessed on the Extramarks website. By utilizing these resources, students can prepare for their exams and improve their intellectual performance.

On the other hand, it is no secret that Extramarks NCERT Solutions may be a tremendous help to students as they study for their examinations. Students may learn more easily about the ideas covered in their textbooks with the aid of the step-by-step instructions offered in these answers. Moreover, students might benefit from the practise questions that can be easily found in the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 to gain a better understanding of the kinds of questions that might be on the test.

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

Students must check the Extramarks website or mobile app to explore the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 and other apt answers. Students can download the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 as PDF files for later usage, as they are widely available in that format. The solutions include in-depth reasons for each problem and are fairly extensive. Students may also utilize the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 as a significant help while completing their test preparation. Every question given in the NCERT Textbook and their comprehensive solutions are included in the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8. Considering that it includes a substantial number of questions covering a wide range of topics, practise exams, and past years’ papers on the smartphone application of Extramarks.

Chapter wise NCERT Solutions for Class 9 Maths

The National Council of Educational Research and Training (NCERT) of India is the author of a number of textbooks known as NCERT Solutions. In addition, solutions such as NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 can help students develop their problem-solving capabilities and become more familiar with the types of questions that are likely to be asked in the examinations. Finally, the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 can assist students in becoming more confident in their own skills.

Class 9 Maths Chapter 13 Includes:

According to Class 9 Chapter 13 of Mathematics NCERT Textbook, students will study about the notions of Surface Areas and Volumes. The following points mentioned below are the topics that are covered in the textbook:

1) Surface Area of a Cuboid and a Cube – Lateral and Total Surface Area

2) Surface Area of a Right Circular Cylinder – Curved and Total Surface Area of a Cylinder

3) Surface Area of a Right Circular Cone – Curved and Total Surface Area of a Cone

4) Surface Area of a Sphere – Radius, Centre, and Sphere

5) Hemisphere – Curved Surface Area and Total Surface Area

6) Volume of a Cuboid – What is Volume and Volume of a Cube

7) Volume of a Cylinder

8) Volume of a Right Circular Cone – Volume of a Cone

9) Volume of a Sphere and Hemisphere

The above specified list of topics can be conveniently understood with the assistance of the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 advised by Extramarks website and mobile application.

NCERT Solutions for Class 9

Students may ensure a smooth preparation for their exams from using the NCERT Solutions, for example, Exercise 13.8 Class 9 Maths solutions to speed up their studying. By providing compact and straightforward solutions for complex concepts, these solutions might help students grasp concepts they may be struggling with. In the NCERT Solutions for Maths Class 9 Chapter 13 Exercise 13.8, students may find a vast combination of knowledge that they can use in various ways to upgrade their execution. In order to concentrate on their revision efforts, these solutions can assist students in identifying their areas of weakness. Use of the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 is a great choice for students who hope to get better in Areas and Volumes of different shapes. Students may evaluate their academic progress and identify their areas for improvement with the aid of these solutions.

CBSE Study Materials for Class 9

There are many accomplishments of availing the CBSE Study Materials for Class 9 or the NCERT Solutions to get high marks in the examinations. One of the most obvious benefits is that students can get help with their homework or assignments. If students are struggling with a question, they can go through the solutions to see how it can be solved. This can help students understand the material better and become more focused on their studies. Another benefit of using the NCERT Solutions and CBSE materials is that students can review the subject matter and understand the topics, as well as questions that they are not able to solve. Finally, using the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 can help students examine the study material. The solutions are typically well-written and easy to grasp. To access the Class 9 Maths Chapter 13 Exercise 13.8 NCERT Solutions, students can visit the website of Extramarks. Therefore, if Class 9 students are looking for a way to strengthen their Mathematics examination performance, they can use the solutions for Class 9 Chapter 13 Maths Exercise 13.8, which are quickly accessible at the online platform of Extramarks.

CBSE Study Materials

One of the easiest ways to prepare for the CBSE test is to use the study materials for the Board exams that are provided by Extramarks website or mobile app. These materials must be used by students to study the curriculum and arrange the thoughts and concepts for the examinations. Students can operate this support emphasized by Extramarks to achieve their best results on the CBSE Board exam. Students can download and make use of these NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 assisted by Extramarks, when the Mathematics exam is going to take place.

Q.1 Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m.

Ans

$\begin{array}{l} Volume of sphere = \frac{4}{3} π r^{3} \\ (i) W h e n radius = 7 cm \\ Volume of sphere = \frac{4}{3} π {(7)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \\ = 1437.33 c m^{3} \\ (i i) When radius = 0 .63 cm \\ Volume of sphere = \frac{4}{3} π {(0 .63)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 0 .63 \times 0 .63 \times 0 .63 \\ = 4 \times 22 \times 0 .03 \times 0 .63 \times 0 .63 \\ = 1.047816 c m^{3} \\ = 1.05 c m^{3} (a p p r o x .) \end{array}$

Q.2 Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m, when the solid spherical ball is placed in a large tank containing water.

Ans

$\begin{array}{l} Volume of sphere = \frac{4}{3} π r^{3} \\ Water displaced by a spherical ball is equal to the volume of the sphere. \\ (i) When diameter of sphere = 28 cm \\ Radius of sphere = 14 c m \\ Volume of sphere = \frac{4}{3} π {(14)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 \\ = 11498.64 c m^{3} \\ (ii) When diameter of sphere = 0. 21 m \\ R a d i u s of sphere = \frac{0.21}{2} m \\ Volume of sphere = \frac{4}{3} π {(\frac{0.21}{2})}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{0.21}{2} \times \frac{0.21}{2} \times \frac{0.21}{2} \\ = 11 \times 0 .01 \times 0 .21 \times 0 .21 \\ = 0.004851 m^{3} \end{array}$

Q.3 The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Ans

$\begin{array}{l} D i a m e t e r of a metallic ball = 4.2 c m \\ Radius of a metallic ball = 2.1 c m \\ Volume of metallic ball = \frac{4}{3} π r^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \\ = 4 \times 22 \times 0.1 \times 2.1 \times 2.1 \\ = 38.808 c m^{3} \\ Density of metal = 8.9 g p e r c m^{3} \\ Mass of metallic ball = V o l u m e \times D e n s i t y \\ = 38.808 \times 8.9 \\ = 345.39 grams . \end{array}$

Q.4 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?

Ans

$\begin{matrix} Radius of earth (R_{e}) = \frac{Diameter of earth}{2} \\ = \frac{D_{e}}{2} \\ Radius of moon (R_{m}) = \frac{\frac{1}{4} \times D_{e}}{2} \\ = \frac{D_{e}}{8} \\ \frac{Volume of moon}{Volume of earth} = \frac{\frac{4}{3} π {(R_{m})}^{3}}{\frac{4}{3} π {(R_{e})}^{3}} \\ = \frac{\frac{4}{3} π {(\frac{D_{e}}{8})}^{3}}{\frac{4}{3} π {(\frac{D_{e}}{2})}^{3}} \\ = \frac{\frac{1}{8} \times \frac{1}{8} \times \frac{1}{8}}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \\ = \frac{1}{64} \\ Volume of moon = \frac{1}{64} \times Volume of earth \end{matrix}$

Q.5 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Ans

$\begin{array}{l} Diameter of hemispherical bowl = 10.5 c m \\ Radius of hemispherical bowl = \frac{10.5}{2} c m \\ Volume of hemispherical bowl = \frac{2}{3} π r^{3} \\ = \frac{2}{3} \times \frac{22}{7} \times \frac{10.5}{2} \times \frac{10.5}{2} \times \frac{10.5}{2} \\ = 11 \times \frac{5}{20} \times \frac{105}{10} \times \frac{105}{10} \\ = 11 \times \frac{1}{4} \times \frac{21}{2} \times \frac{21}{2} \\ = 303.1875 c m^{3} \\ = \frac{303.1875}{1000} l i t r e s (a p p r o x .) \\ [∵ 1 c m^{3} = \frac{1}{1000} l i t r e s] \\ = 0.3031875 l i t r e s \\ Capacity of hemispherical bowl = 0.303 l i t r e s . \end{array}$

Q.6 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.

Ans

$\begin{array}{l} The inner radius of hemispherical tank (r_{2}) = 1 m \\ Thickness of hemispherical tank = 1 c m \\ = 0.01 m \\ Oute r radius of spherical tank (r_{1}) = 1 + 0.01 \\ = 1.01 m \\ Volume of iron used to make the tank = \frac{2}{3} π (r_{1}^{3} - r_{2}^{3}) \\ = \frac{2}{3} \times \frac{22}{7} {{(1.01)}^{3} - {(1)}^{3}} \\ = \frac{44}{21} (1.030301 - 1) \\ = \frac{44}{21} (0.030301) \\ = 0.06348 m^{3} \\ Thus, the volume of iron used to make the tank is 0.06348 m^{3} . \end{array}$

Q.7 Find the volume of a sphere whose surface area is 154 cm².

Ans

$\begin{array}{l} Surface area of a sphere = 154 c m^{2} \\ 4 \times \frac{22}{7} \times r^{2} = 154 c m^{2} \\ r^{2} = \frac{154 \times 7}{4 \times 22} \\ = \frac{7 \times 7}{4} \\ r = \sqrt{\frac{7 \times 7}{4}} \\ = \frac{7}{2} c m \\ Volume of a sphere = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\ = \frac{11 \times 49}{3} = \frac{539}{3} \\ = 179 \frac{2}{3} c m^{3} \\ Thus, volume of sphere is 179 \frac{2}{3} c m^{3} . \end{array}$

Q.8 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 498.96. If the cost of white-washing is ₹ 2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Ans

$\begin{array}{l} (i) \\ Cost of white-washing the hemisphere = ₹ 498 .96 \\ Rate of white-washing the hemisphere = ₹ 2 .00 / m^{2} \\ Therefore, Surface area of the hemisphere = \frac{498 .96}{2} \\ = 249 .48 m^{2} \\ Thus, inside surface area of the dome is 249 .48 m^{2} . \\ (ii) ∵ Surface area of the hemisphere = 249 .48 m^{2} \\ 2 {πr}^{2} = 249 .48 m^{2} \\ 2 \times \frac{22}{7} \times r^{2} = 249 .48 m^{2} \\ r = \sqrt{\frac{249 .48 \times 7}{2 \times 22}} \\ r = 6 .3 m \\ Volume of air inside the tomb = \frac{4}{3} {πr}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 6 .3 \times 6 .3 \times 6 .3 \\ = 88 \times 0 .3 \times 6 .3 \times 6 .3 \\ = 1047 .816 m^{3} \\ Thus, the volume of air inside the tomb is 1047 .816 m^{3} . \end{array}$

Q.9 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r ′ of the new sphere,
(ii) ratio of S and S′.

Ans

$\begin{array}{l} (i) Volume of new sphere = 27 \times volume of solid iron sphere \\ \frac{4}{3} π {(r ‘)}^{3} = 27 \times \frac{4}{3} {πr}^{3} \\ \Rightarrow r ‘ = \sqrt{27 r^{3}} \\ \Rightarrow r ‘ = 3 r \\ Thus, the radius (r ‘) of new sphere is 3r. \\ (ii) Ratio of S and S’ = \frac{4 {πr}^{2}}{4 πr ‘^{2}} \\ = \frac{r^{2}}{{(3 r)}^{2}} [∵ r ‘ = 3 r] \\ = \frac{1}{9} \\ Thus, ration of S and S’ is 1:9. \end{array}$

Q.10 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine material (in mm³) is needed to fill this capsule?

Ans

$\begin{array}{l} Diameter of spherical capsule = 3 . 5 mm \\ Radius of spherical capsule = \frac{3.5}{2} mm \\ Volume of spherical capsule = \frac{4}{3} π r^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times \frac{3.5}{2} \\ = \frac{11}{3} \times 0.5 \times 3.5 \times 3.5 \\ = 22.458 m m^{3} \\ = 22.46 m m^{3} (A p p r o x .) \\ Thus, 22.46 m m^{3} medicine materials are required to fill the capsule . \end{array}$

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The NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 are effortlessly available on Extramarks website. Sample or intext questions and papers from past years are incorporated in the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8. On Extramarks website and smartphones application, students may access unanswered questions from NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8.

Students simply need to log into Extramarks and view the syllabus to acquire the latest syllabus for Class 9 Mathematics. Students may find it helpful to have an opportunity of accessing Extramarks websites and application for the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8. However, the curriculum helps students gain an abundance of information about the discipline of Mathematics as well as for the other chapters and subjects they are related to.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

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

Chapter wise NCERT Solutions for Class 9 Maths

Class 9 Maths Chapter 13 Includes:

NCERT Solutions for Class 9

CBSE Study Materials for Class 9

CBSE Study Materials

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

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

Chapter wise NCERT Solutions for Class 9 Maths

Class 9 Maths Chapter 13 Includes:

NCERT Solutions for Class 9

CBSE Study Materials for Class 9

CBSE Study Materials

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