NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Areas and Volumes-Term 2 CBSE Board Exam - Future Study Point

NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Areas and Volumes-Term 2 CBSE Board Exam

class 9 maths ex.13.4

NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Areas and Volumes-Term 2 CBSE Board Exam

NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13-Surface Areas and Volumes created here for the purpose of boosting the preparation of class 9 students in maths paper of term 2 CBSE Board exam.NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13-Surface Areas and Volumes will help you in clearing the basic concepts of maths that require solving the questions of maths paper in term 2 CBSE Board exam, although chapter 13 is based on problems related to Surface areas and Volumes of three-dimensional figures here this exercise accommodates the surface area of the sphere only.

Students can practice and improve their math abilities by tackling the NCERT Solutions.NCERT Solutions is one of the greatest insightful study materials for Class 9 Maths that is created here. These NCERT Solutions incorporate questions from the chapters given in the NCERT Course study according to the CBSE norms. The primary point in solving these questions is to empower the students to score well in Class 9 first and second term CBSE Board exams. NCERT Solutions for Class 9 Maths Chapter13–Surface Areas and Volumes Exercise 13.4 assists students with scoring great and furthermore to confront the exams all the more unhesitatingly, as they gain confidence in tackling these exercises.

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We present to you an itemized assortment of questions and solutions from the exercises with applicable responses, made by our subject specialists and experienced teachers. NCERT solutions plan is to assist students with scoring high in the first and second-term exams. We give appropriate logic and clarifications in the solution of the questions, so students can comprehend the ideas in a superior way.

class 9 maths ex.13.4

 

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

Class 9 Maths Exercise -13.8

Class 9 Maths Exercise 13.9

NCERT Solutions Class 9 Maths-All Chapters

NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13-Surface Areas and Volumes

Q1. Find the surface area of a sphere of radius:

(i) 10.5cm (ii) 5.6cm (iii) 14cm

(Assume π=22/7)

Ans. Surface area of the sphere = 4πr²

(i) Radius (r) of the sphere = 10.5 cm

Surface area of the sphere

= 4 ×(22/7) ×10.5 ×10.5

= 4 ×22 ×1.5 ×10.5

=1386

Hence surface area of the sphere is 1386 cm²

(ii) Radius (r) of the sphere = 5 .6cm

Surface area of the sphere

= 4 ×(22/7) ×5.6 ×5.6

= 4 ×22 ×.8 ×5.6

=394.24

Hence surface area of the sphere is 394.24 cm²

(iii) Radius (r) of the sphere = 14cm

Surface area of the sphere

= 4 ×(22/7) ×14 ×14

= 4 ×22 ×2 ×14

=2464

Hence surface area of the sphere is 2464 cm²

Q2.Find the surface area of a sphere of diameter:

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

Ans.(i) Diameter of the sphere is 14 cm

Therefore radius(r) of the sphere =14/2 = 7 cm

Surface area of the sphere

= 4πr²

Surface area of the sphere with radius 7 cm

= 4 ×(22/7) ×7 ×7

= 4 ×22  ×7

=616

Hence surface area of the sphere is 616 cm²

(ii) Diameter of the sphere is 21 cm

Therefore radius(r) of the sphere =21/2 = 10.5 cm

The surface area of the sphere

= 4πr²

The surface area of the sphere

= 4 ×(22/7) ×10.5 ×10.5

= 4 ×22 ×1.5 ×10.5

=1386

Hence the surface area of the sphere is 1386 cm²

(iii) Diameter of the sphere is 3.5cm

Therefore radius(r) of the sphere =3.5/2 = 1.75 cm

The surface area of the sphere

= 4πr²

The surface area of the sphere

= 4 ×(22/7) ×1.75 ×1.75

= 4 ×22 ×0.25 ×1.75

=38.5

Hence the surface area of the sphere is 38.5 cm²

Q3.Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

Radius(r) of the hemisphere =10 cm

The total surface area of the hemisphere

= 3πr²

The surface area of the sphere

= 3 ×3.14 ×10 ×10

= 3×314

=942

Hence the surface area of the sphere is 942 cm²

Q4.The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans. The surface area of the spherical balloon is =4πr²

When the radius (r) of the balloon was 7 cm then

The surface area of the balloon was = 4π×7×7

When the radius (r) of the balloon was 14 cm then

The surface area of the balloon was = 4π×14×14

The ratio of surface areas of the balloon in the two cases

= (4π×7×7)/(4π×14×14)

= 1/4

Hence the ratio of surface areas of the balloon in the two cases is 1 : 4

Q5.A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)

Ans.The inner diameter of the hemispherical bowl is given 10.5 cm,therefore the radius(r) =10.5/2 =5.25 cm

Inner curved surface area of the hemispherical bowl 44×0.75 ×5.25

=2πr²

=2×(22/7) ×5.25 ×5.25

=44×0.75 ×5.25

= 173.25

Therefore inner curved surface area of the bowl is 173.25 cm²

The rate of tin plating is Rs 16 per 100 cm²

The cost of tinplating inside hemispherical bowl is

=(16/100) × 173.25

= 16×1.7325

=27.72

Hence  the cost of tinplating inside hemispherical bowl is Rs. 27.72 

Q6.Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Ans.Surface area of the sphere = 4πr²

The given surface area of the sphere is 154 cm²

4πr² = 154

4×(22/7) ×r²= 154

r² = (154 ×7)/(4×22)

r² = (7×7)/4

r = 7/2 =3.5

Hence the radius of the sphere is 3.5 cm

Q7.The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.

Ans. Let the diameter of the earth be x, then the radius of the earth is x/2

The diameter of the moon is

=(1/4) of the earth’s diameter

=(1/4) of x = x/4⇒radius of the moon is x/8

Surface area  of the earth is

= 4π(x/2)²=πx²

Surface area of the  moon= 4πr² =4π(x/8)² =πx²/16

Ratio of the surface area of the  moon and the earth

= πx²/16 ÷ πx²

= 1/16

Hence the ratio of surface areas between moon and the earth is 1 :16

Q8.A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans. The inner radius of a hemispherical bowl is 5 cm

The thickness of the hemispherical bowl is 0.25 cm

The radius of the outer curved surface of the hemispherical bowl is = 5 + o.25 =5.25 cm

The outer curved surface area of the bowl

= 2π(Radius of the outer curved surface)²

=2(22/7)×5.25²

=2(22/7)×5.25×5.25

=44×0.75×5.25

=173.25

Hence outer curved surface area of the bowl is 173.25 cm²

Q9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii).

Q9 ex.13.4 class 9 maths

Ans. It is given that a right circular cylinder just encloses a sphere of radius r

The height(h) of the sphere is = diameter of the sphere =2r

(i) Surface area of the sphere,S =4πr²

(ii) Radius(r) of the sphere = radius of the cylinder(r)

Curved surface area of the cylinder,S’

= 2πrh

=2πr(2r)

S’=4πr²

(iii) The ratio between surface areas of sphere(S) and cylinder(S’)

S/S’ = 4πr²/4πr² =1/1

Hence the ratio between the surface areas of sphere and the curved surface area of the cylinder

S : S’ = 1 : 1

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