# 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.

**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 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 cm ^{2}. (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 cm ^{2}. (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).**

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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