NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2 - Future Study Point

# NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2

NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2 will help you to boost your preparation for CBSE Term 2 examination.NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2 is helpful for the preparation of competitive entrance exam also. The NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2 are the solutions of unsolved questions of the Class 9 NCERT maths textbook exercise 13.8 Surface area and Volumes prescribed by CBSE for Class 9 students.NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2 is the most important study material for the preparation of the CBSE Term 2 examination.

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## NCERT Solutions for Class 9 Maths  Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam

Class 9 Maths Exercise-13.1

Class 9 Maths Exercise-13.2

Class 9 Maths Exercise-13.3

Class 9 Maths Exercise -13.5

Class 9 Maths Exercise -13.7

NCERT Solutions Class 9 Maths-All Chapters

## NCERT Solutions for Class 9 Maths of Exercise 13.8 Chapter 13 Surface areas and Volumes-Term 2

Q1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

(Assume π =22/7)

Ans.(i) The radius(r) of sphere is 7 cm

Volume of the sphere

= (4/3) πr³

=(4/3) (22/7)×7×7×7

=4312/3

The volume of the sphere is 4312/3 cm³

Ans.(ii) The radius(r) of sphere is 0.63 m

The volume of the sphere

= (4/3) πr³

=(4/3) (22/7)×0.63 ×0.63 ×0.63

=88×0.09 ×0.21×0.63

=1.0478

The volume of the sphere is 1.0478 m³

Q2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

(Assume π =22/7)

Ans.(i) The diameter of the ball is 28 cm, therefore, radius(r) of the ball is 28/2 =14 cm

According to Archimedes Principle

The amount of water displaced by a solid spherical ball

= Volume of the spherical ball

= (4/3) πr³

=(4/3) (22/7)×14 ×14 ×14

=34496/3

Hence the amount of water displaced by a solid spherical ball is 34496/3 cm³

(ii)The diameter of the ball is 0.21 m, therefore, radius(r) of the ball is 0.21/2 =0.105 m

According to Archimedes Principle

The amount of water displaced by a solid spherical ball

= Volume of the spherical ball

= (4/3) πr³

=(4/3) (22/7)×0.105 ×0.105 ×0.105

=88×0.15×0.35×0.105

=0.004851

Hence the amount of water displaced by a solid spherical ball is 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 8.9 g per cm3? (Assume π=22/7)

Ans. The diameter of the ball is 4.2cm, therefore, radius(r) of the ball is 4.2/2 =2.1 cm

The mass of the ball = Volume of the ball×density of the ball

Volume of the ball

=(4/3) πr³

=(4/3) (22/7)×2.1×2.1 ×2.1

=88×0.7×0.3×0.105

=38.808

The volume of the sphere is 38.808 cm³

The density of the metal is 8.9 g per cm³

The mass of the ball = (38.808×8.9 )g=345.3912g

Hence mass of the ball is 345.3912g

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?

Ans. Let the diameter of the earth is 2r then its radiuus(r) is r

The diameter of the moon is 2r/4 =r/2 and its radius r’ =r/4

The volume(V) of the earth

V=(4/3) πr³…….(i)

The volume(V’) of the earth

V’=(4/3) πr’³

V’=(4/3) π(r/4)³

V’ =(1/3) πr³/16

V’ = (1/48)πr³ …….(ii)

Dividig equation (ii) from (i)

V’/V = (1/48)πr³ ÷(4/3) πr³

V’/V =1/48 ÷ 4/3

V’/V =1/48 ×3/4 =1/64

Hence the volume of the moon is (1/64) portion of the earth’s volume

Q5.How many liters of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Ans. The diameter of a hemispherical bowl is 10.5cm then the radius(r) of the bowl is 10.5/2 =5.25 cm

The volume of the hemispherical bowl is

= (2/3)πr³

= (2/3)(22/7)×5.25 ×5.25 ×5.25

=44×1.75×0.75×5.25

=303.1875

The volume of the hemispherical bowl is 303.1875 cm³

1 litre = 1000 cm³

303.1875 cm³ = 303.1875/1000 =0.303 l

Hence hemispherical bowl can hold 0.303 l milk

Q6.A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)

Ans. The inner radius(r) of the hemispherical tank is 1 cm and its thickness is 1 cm =0.01m

Outer radius(R) of the tank is =1+0.01 =1.01 m

Volume of the iron used in the tank

=(2/3)(22/7)(R³-r³)

=(2/3)(22/7)(1.01³-1³)

=(44/21)(1.030301-1)

=(44/21)(.030301)

=0.06348 m³

Hence the volume of the iron  used to make the tank is 0.06348 m³

Q7.Find the volume of a sphere whose surface area is 154 cm². (Assume π = 22/7)

Ans. The surface area of the sphere is 154 cm²

Surface area of the sphere = 4πr²(where r is the radius )

4πr² = 154

4(22/7) r² = 154

r² = (154×7)/(4×22) = (7×7)/4

r = 7/2 = 3.5

Volume of the sphere is

=(4/3)πr³

=(4/3)(22/7)×3.5×3.5×3.5

=(88/3) ×0.75×3.5×3.5

=179.67

Hence the volume of the sphere is 179.67 cm³

Q8.A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the

(i) inside surface area of the dome (ii) volume of the air inside the dome

(Assume π = 22/7)

Ans.Total cost of white washig inner curved surface area of the dome is = Rs 4989.60

The rate of whitewashing is =Rs20 per square meter

Inner curved surface area of the dome

=Total cost of white washig CSA of the dome/Rate of whitewashing

=4989.60/20

=249.48

Hence inner curved surface area of the dome is 249.48 m²

(ii) Inner curved surface area of the dome =2πr²

2πr² = 249.48

2(22/7)r² = 249.48

r² = (249.48×7)/44 =39.69

r = 6.3

Volume of the air inside the dome

= (2/3)πr³

=(2/3)(22/7)×6.3×6.3×6.3

=44×2.1×0.09×6.3

=523.9

Hence the volume of the air inside the dome is 523.9 m³

Q9.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 Sand S’.

Ans. (i) Radius of the smaller sphere is r and radius of the new sphere is r’

Since twenty-seven solid iron spheres are melted to form a sphere

∴ Volume of  27 smaller spheres = Volume of new sphere

27(4/3)πr³ = (4/3)πr’³

27r³= r’³

r’ = 3r

Hence radius r’ of the new sphere is 3r

(ii) Surface area of the smaller sphere is

S = 4πr²

S’ = 4πr’²

S/S’ = 4πr²/4πr’²

S/S’ = r²/r’²= (r/r’)² =(r/3r)² =1/9

Therefore the ratio of Sand S’ 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? (Assume π = 22/7)

Ans.The diameter of a spherical capsule is 3.5mm then the radius(r) of the capsle is 3.5/2 =1.75 mm

The volume of the spherical capsle is

= (4/3)πr³

= (4/3)(22/7)×1.75 ×1.75 ×1.75

=(88/3)×0.25×1.75×1.75

=22.46

The volume of the hemispherical bowl is 22.46 mm³

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