**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**

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**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 cm ^{3}? (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 mm ^{3}) 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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