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NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4 - Surface Area and Volumes

Edited By Vishal kumar | Updated on May 05, 2025 02:24 PM IST

In the previous exercise 11.3, we discussed the volume of the cone, and in this exercise, we solve the questions related to the volume of a sphere and a hemisphere as per the NCERT book for class 9. A sphere is a three-dimensional object that is circular. 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. 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 11 exercise 11.3 consist of 10 questions regarding the volume of a sphere and a hemisphere.

This Story also Contains
  1. Download the PDF of NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Exercise 11.4
  2. Access Surface Area and Volumes Class 9 Maths Chapter 11 Exercise: 11.4
  3. Q1 (i) Find the volume of a sphere whose radius is 7 cm
  4. Topics covered in Chapter 11 Surface Area and Volumes: Exercise 11.4
  5. NCERT Solutions of Class 9 Subject Wise
  6. NCERT Subject-Wise Exemplar Solutions
NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4 - Surface Area and Volumes
NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.4 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Exercise 11 - This 9th class maths exercise 11.4 solution 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 11 exercise 11.4 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 11.4 class 9 maths solutions in PDF format, enabling offline access and convenient learning.


**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 11 Surface Areas and Volumes Exercise 11.4

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Access Surface Area and Volumes Class 9 Maths Chapter 11 Exercise: 11.4

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

Answer:

Given,

The radius of the sphere = r=7 cm

We know, Volume of a sphere = 43πr3

The required volume of the sphere = 43×227×(7)3

=43×22×7×7

=43123

=143713 cm3

Q1 (ii) Find the volume of a sphere whose radius is 0.63m

Answer:

Given,

The radius of the sphere = r=0.63 m

We know, Volume of a sphere = 43πr3

The required volume of the sphere = 43×227×(0.63)3

=4×22×0.03×0.63×0.63

=1.048 m3

=1.05 m3   (approx.)

Q2 (i) Find the amount of water displaced by a solid spherical ball of diameter 28 cm

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = r=282 cm=14 cm

We know, Volume of a sphere = 43πr3

The required volume of the sphere = 43×227×(14)3

=43×22×2×14×14

=344693=1148923 cm3

Therefore, the amount of water displaced will be 1148923 cm3

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter 0.21m

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = r=0.212 m

We know, Volume of a sphere = 43πr3

The required volume of the sphere = 43×227×(0.212)3

=4×22×0.01×0.21×0.218

=11×0.01×0.21×0.21

=0.004851 m3

Therefore, the amount of water displaced will be 0.004851 m3

Q3 The diameter of a metallic ball is 4.2cm . What is the mass of the ball, if the density of the metal is 8.9g per cm3 ?

Answer:

Given,

The radius of the metallic sphere = r=4.22 cm=2.1 cm

We know, Volume of a sphere = 43πr3

The required volume of the sphere = 43×227×2.13

=4×22×0.1×2.1×2.1

=38.808 cm3

Now, the density of the metal is 8.9g per cm3 ,which means,

Mass of 1 cm3 of the metallic sphere = 8.9g

Mass of 38.808 cm3 of the metallic sphere = (8.9×38.808) g

345.39 g

Q4 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?

Answer:

Given,

Let de be the diameter of Earth

The diameter of the Moon = dm=14de

We know, Volume of a sphere =

43πr3=43π(d2)3=16πd3

The ratio of the volumes = Volume of the EarthVolume of the Moon

=16πde316πdm3=de3(de4)3=64:1

Therefore, the required ratio of the volume of the moon to the volume of the earth is 1:64

Q5 How many litres of milk can a hemispherical bowl of diameter 10.5cm hold?

Answer:

The radius of the hemispherical bowl = r=10.52 cm

We know, Volume of a hemisphere = 23πr3

The volume of the given hemispherical bowl = 23×227×(10.52)3

=23×8×22×1.5×10.5×10.5

=303.1875 cm3

The capacity of the hemispherical bowl = =303.187510000.303 litres   (approx.)

Q6 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.

Answer:

Given,

Inner radius of the hemispherical tank = r1=1 m

Thickness of the tank = 1 cm=0.01 m

Outer radius = Internal radius + thickness = r2=(1+0.01) m=1.01 m

We know, Volume of a hemisphere = 23πr3

Volume of the iron used = Outer volume - Inner volume

=23πr2323πr13

=23×227×(1.01313)

=4421×0.030301

=0.06348 m3  (approx)

Q7 Find the volume of a sphere whose surface area is 154cm2 .

Answer:

Given,

The surface area of the sphere = 154cm2

We know, Surface area of a sphere = 4πr2

4πr2=154

4×227×r2=14×11r2=7×74r=72r=3.5 cm

The volume of the sphere = 43πr3

=43×227×(3.5)3

=17923 cm3

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs4989.60 . If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

Rs20 is the cost of white-washing 1 m2 of the inside area

Rs4989.60 is the cost of white-washing 120×4989.60 m2=249.48 m2 of inside area

(i) Therefore, the surface area of the inside of the dome is 249.48 m2

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs4989.60 . If the cost of white-washing is Rs 20 per square metre, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be r m

Inside the surface area of the dome = 249.48 m2

We know, Surface area of a hemisphere = 2πr2

2πr2=249.48r2=249.48×72×22r=6.3 m

The volume of the hemisphere = 23πr3

=23×227×(6.3)3

=523.908 m3

Q9 (i) 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 radius r of the new sphere

Answer:

Given,

The radius of a small sphere = r

The radius of the bigger sphere = r

The volume of each small sphere= 43πr3

And, Volume of the big sphere of radius r = 43πr3

According to question,

27×43πr3=43πr3

r3=27×r3r=3×r

r=3r

Q9 (ii) 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 ratio of S and S .

Answer:

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 = 4πr2

The ratio of their surface areas = 4πr24πr2

=(3r)2r2=9

Therefore, the required ratio is 1:9

Q10 A capsule of medicine is in the shape of a sphere of diameter 3.5mm . How much medicine (in mm3 ) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule = r=3.52

The volume of the capsule = 43πr3

=43×227×(3.52)3

=43×22×0.5×3.5×3.58

=22.458 mm322.46 mm3  (approx)

Therefore, 22.46 mm3  (approx) of medicine is needed to fill the capsule.


Also Read:

Surface Area and Volumes Exercise 11.1
Surface Area and Volumes Exercise 11.2
Surface Area and Volumes Exercise 11.3

Topics covered in Chapter 11 Surface Area and Volumes: Exercise 11.4

To find the volume of a sphere in the NCERT solutions for Class 9 Maths exercise 11.4, 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 ) by 43π. Thus, the final answer will be the volume of the sphere. Similarly, the volume of a hemisphere is given by 23πr3 cubic units.

Also See:

NCERT Solutions of Class 9 Subject Wise

Students must check the NCERT solutions for Class 9 Maths and Science given below:

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NCERT Subject-Wise Exemplar Solutions

Students must check the NCERT exemplar solutions for Class 9 Maths and Science given below:

Frequently Asked Questions (FAQs)

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.

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