## NCERT Solutions for Class 9 Maths Chapter 13 (Ex 13.8)

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

## FAQs on NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.8

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 $\dfrac{2}{3}) \pi r^3$

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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.ย 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.