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# NCERT Solutions for Class 9 Maths Chapter 13 - Exercise Last updated date: 29th Nov 2023
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## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.8) Exercise 13.8

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NCERT Solutions for Class 9 Maths Chapter 13 - Exercise Surface Area and Volume L-1 | Surface Area & Volume of Cuboid & Cube | CBSE 9 Maths Ch 13 | Term 2
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## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

Important Topics Covered in Exercise 13.8 of Class 9 Maths NCERT Solutions

Chapter 13 - Surface Areas and Volumes is a very important chapter in Class 9 Maths. Exercise 13.8 of Class 9 Maths NCERT Solutions is mainly based on the concept of the volume of a sphere.

Below are the important formulas discussed in this chapter.

• Volume of a Sphere = (4/3)πr3

• Volume of a Hemisphere = (2/3) πr3

This exercise consists of questions based on finding the volume of a sphere. The students are advised to practise the questions present in this exercise as it will help them to have a better understanding of the concepts and they will be able to score high in the exam.

Opting for the NCERT solutions for Ex 13.8 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.8 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

### Class 9 Maths Chapter 13 Includes:

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Surface Areas and Volumes textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 13 Exercise 13.8 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.8, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.8 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 9 Maths Chapter 13 - Exercise

1. How many questions are there in NCERT Solutions Class 9 Maths Chapter 13 Exercise 13.8?

The NCERT Solutions Class 9 Maths Chapter 13 Exercise 13.8 contain 10 questions. In addition to these, this chapter contains a good number of exercises that are packed with questions. Subject matter experts from Vedantu have resolved or responded to all of these queries. Don't put it off any longer. For better exam preparation, download the NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.8 from the Vedantu website right away.

2. Which formulas are crucial for NCERT Class 9 Maths Chapter 13 Exercise 13.8?

The idea of the volume of a sphere is mostly used in NCERT Solutions Class 9 Maths Chapter 13 Exercise 13.8. The key formulas mentioned in this chapter are listed below.

• The volume of a Sphere is $(\dfrac{4}{3}) \pi r^3$

• The volume of a Hemisphere is $\dfrac{2}{3}) \pi r^3$

Utilize the free Vedantu NCERT Solutions PDFs to get started learning! You can get the same through the Vedantu app if you have it on your phone. The ability of these solutions to be used both online and offline is their most appealing feature.

3. What is the Volume of a hollow Sphere?

If the inner sphere's radius is r, the outer sphere's radius is R, and the sphere's volume is V, The sphere's volume is then determined by:

The volume of the required sphere, V is given by, the volume of the outer sphere minus the volume of the inner sphere

$V = (\dfrac{4}{3}) \pi R^3 - (\dfrac{4}{3}) \pi r^3 = (\dfrac{4}{3}) \pi (R^3-r^3)$

4. How Does a Halfing of the Radius of a Sphere Affect the Volume of the Sphere?

The volume of a sphere gets one-eighth when the radius is halved as r = r/2. As the volume of the sphere is equal to $(\dfrac{4}{3}) \pi r^3 = (\dfrac{4}{3}) \pi (\dfrac{r^3}{2}) = (\dfrac{4}{3}) \pi (\dfrac{r^3}{8}) = volume/8$ . Thus, the volume of a sphere gets cut by one-eighth as soon as its radius gets halved.

5.  What is the ratio of the Surface Area to the Volume of a sphere with a certain radius?

Let us assume the radius of the given sphere to be r, as we know a sphere's volume is calculated using the formula $(\dfrac{4}{3})\pi r^3$ while its surface area is calculated using the formula $4 \pi r^2$ . Thus, $\dfrac{(\dfrac{4}{3})}{4}$ = 1:3 is the ratio of the sphere's surface area to its volume.