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NCERT Solutions
Class 9
Maths
Chapter 11 Surface Area and Volume
Exercise 11.2

NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.2 guide students through the calculation of surface areas of right circular cylinders. This exercise introduces the formulas for curved surface area and total surface area, helping students connect geometry to practical situations like finding the area of pipes, cans, or pillars. Clear examples and solved questions ensure learners understand how to apply these formulas step by step.

These solutions follow the latest NCERT Solution syllabus and are designed to strengthen conceptual understanding and improve problem-solving skills. With detailed explanations, they help students prepare confidently for exams and tackle complex questions with ease. Practicing Exercise 11.2 is crucial for mastering the geometry of cylinders and laying a strong foundation for higher classes.

1.0Download NCERT Solutions of Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.2 Free PDF

Find solutions to all questions in NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.2 which are acceptable, accurate and systematic. Perfect your problem-solving skills and studying capabilities with this downloadable PDF.

NCERT Solutions Class 9 Maths Chapter 11 Ex 11.2

Key Concepts of Exercise 11.2

  • Right circular cylinder
  • Curved surface area (CSA) = 2πrh
  • Total surface area (TSA) = 2πr(h + r)
  • Description of CSA and TSA in word problems
  • Unit conversions related to and beyond the topic with real life knowledge of the cylindrical shape.

2.0NCERT Exercise Solutions Class 9 Chapter 11 Surface Area and Volume: All Exercises 

Here is a quick overview and access to solutions for all exercises from Chapter 11, Surface Area and Volume.

NCERT Solutions Class 9 Maths Chapter 11 Exercise 11.1

NCERT Solutions Class 9 Maths Chapter 11 Exercise 11.2

NCERT Solutions Class 9 Maths Chapter 11 Exercise 11.3

NCERT Solutions Class 9 Maths Chapter 11 Exercise 11.4

3.0NCERT Class 9 Maths Chapter 11 Exercise 11.2 : Detailed Solutions

(Assume π=22/7 unless stated otherwise)

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

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Sol.

The surface area of a sphere is given by the formula A = 4πr².

(i) Radius (r) = 10.5 cm

Surface area = 4 × (22/7) × (10.5)² cm²

= 4 × (22/7) × 110.25 cm²

= 4 × 22 × 15.75 cm²

= 1386 cm².

(ii) Radius (r) = 5.6 cm

Surface area = 4 × (22/7) × (5.6)² cm²

= 4 × (22/7) × 31.36 cm²

= 4 × 22 × 4.48 cm²

= 394.24 cm².

(iii) Radius (r) = 14 cm

Surface area = 4 × (22/7) × (14)² cm²

= 4 × (22/7) × 196 cm²

= 4 × 22 × 28 cm²

= 2464 cm².


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

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Sol.

The surface area of a sphere is given by A = 4πr², where r is the radius (diameter/2).

(i) Diameter = 14 cm

Radius (r) = 14/2 cm = 7 cm.

Surface area = 4 × (22/7) × (7)² cm²

= 4 × (22/7) × 49 cm²

= 4 × 22 × 7 cm²

= 616 cm².

(ii) Diameter = 21 cm

Radius (r) = 21/2 cm.

Surface area = 4 × (22/7) × (21/2)² cm²

= 4 × (22/7) × (441/4) cm²

= 22 × 63 cm²

= 1386 cm².

(iii) Diameter = 3.5 cm

Radius (r) = 3.5/2 cm.

Surface area = 4 × (22/7) × (3.5/2)² cm²

= 4 × (22/7) × (12.25/4) cm²

= 22 × (12.25/7) cm²

= 22 × 1.75 cm²

= 38.5 cm².


3. Find the total surface area of a hemisphere of radius 10 cm. (Use π=3.14)

Sol.

Radius (r) = 10 cm.

Total surface area (TSA) of the hemisphere = 3πr²

= 3 × 3.14 × (10)² cm²

= 3 × 3.14 × 100 cm²

= 3 × 314 cm²

= 942 cm².


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

Sol.

Initial radius (r₁) = 7 cm

Final radius (r₂) = 14 cm

Let S₁ and S₂ be the surface areas of the balloon in the two cases.

Surface area of a sphere = 4πr².

S₁ = 4πr₁²

S₂ = 4πr₂²

Ratio S₁/S₂ = (4πr₁²) / (4πr₂²) = r₁² / r₂² = (r₁/r₂)²

= (7/14)² = (1/2)² = 1/4.

The ratio of surface areas S₁:S₂ is 1:4.


5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm².

Sol.

Inner diameter = 10.5 cm

Inner radius (r) = 10.5/2 cm = 5.25 cm.

Inner curved surface area (CSA) of the hemispherical bowl = 2πr²

= 2 × (22/7) × (5.25)² cm²

= 2 × (22/7) × 27.5625 cm²

= 44 × 3.9375 cm²

= 173.25 cm².

Cost of tin-plating for 100 cm² = Rs. 16.

Cost of tin-plating per cm² = Rs. 16/100.

Cost of tin-plating for 173.25 cm² = 173.25 × (16/100) Rs.

= (173.25 × 16) / 100 Rs.

= 2772 / 100 Rs.

= Rs. 27.72.


6. Find the radius of a sphere whose surface area is 154 cm².

Sol.

Surface area of a sphere = 154 cm².

We know the formula: 4πr² = 154.

4 × (22/7) × r² = 154

(88/7) × r² = 154

r² = (154 × 7) / 88

r² = (7 × 7) / 4

r² = 49/4

r = √(49/4)

r = 7/2 cm, which is 3.5 cm.


7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Sol.

Let d_moon be the diameter of the moon and d_earth be the diameter of the earth.

Given d_moon = (1/4) d_earth.

Let r_moon and r_earth be their respective radii. So, r_moon = (1/4) r_earth.

Let S_moon and S_earth be their respective surface areas.

Surface area of a sphere = 4πr².

S_moon = 4π(r_moon)²

S_earth = 4π(r_earth)²

Ratio S_moon / S_earth = (4π(r_moon)²) / (4π(r_earth)²) = (r_moon / r_earth)²

Substitute r_moon = (1/4) r_earth:

= ((1/4) r_earth / r_earth)² = (1/4)² = 1/16.

The ratio of their surface areas (Moon : Earth) is 1:16.


8. 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.

Sol.

Inner radius (r_inner) = 5 cm.

Thickness of steel sheet = 0.25 cm.

Outer radius (r_outer) = Inner radius + Thickness = 5 cm + 0.25 cm = 5.25 cm.

Outer curved surface area (CSA) of the hemispherical bowl = 2π(r_outer)²

= 2 × (22/7) × (5.25)² cm²

= 2 × (22/7) × 27.5625 cm²

= 44 × 3.9375 cm²

= 173.25 cm².


9. A right circular cylinder just encloses a sphere of radius r. Find

(i) Surface area of the sphere,

(ii) Curved surface area of the cylinder,

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

Sol.

When a right circular cylinder just encloses a sphere of radius r:

The radius of the cylinder's base will be equal to the radius of the sphere (r).

The height of the cylinder will be equal to the diameter of the sphere (2r).

(i) Surface Area of the Sphere = 4πr².

(ii) Curved Surface Area (CSA) of the Cylinder = 2πrh

Substitute h = 2r:

CSA of Cylinder = 2πr(2r) = 4πr².

(iii) Ratio of the areas obtained in (i) and (ii):

(Surface Area of Sphere) / (CSA of Cylinder) = (4πr²) / (4πr²) = 1/1.

The ratio is 1:1.

4.0Key Features and Benefits Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.2

  • Clear Explanations: Each question is broken down into conceptual clarity.  
  • Formula Recap with Examples: Helps reinforce understanding of surface area formulas. 
  • Useful for Exams: Helps build a strong foundation for geometry based questions on exams.  
  • Application-Based Study: Helps students apply geometry to real life applications.  
  • Accurate and Reliable: Prepared according to CBSE guidelines and updated NCERT syllabus.

NCERT Class 9 Maths Ch. 7 Surface Areas and Volumes Other Exercises:-

Exercise 11.1

Exercise 11.2

Exercise 11.3

Exercise 11.4


NCERT Solutions for Class 9 Maths Other Chapters:-

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics

Frequently Asked Questions

Curved Surface Area (CSA) of a Cylinder: 2πrh Total Surface Area (TSA) of a Cylinder: 2πr(h+r) where rr is the radius and hh is the height of the cylinder.

It strengthens understanding of 3D geometry and helps students solve practical problems involving cylindrical shapes, which are common in real life. It also builds the base for more advanced geometry topics in higher classes.

Yes! Many questions involve real-life scenarios like calculating the cost of painting a cylinder, finding the area of labels for cans, or wrapping cylindrical objects.

Yes, some problems require converting units like cm to m or vice versa before calculating areas. It’s important to keep units consistent for accurate answers.

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