• NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • Class 6-10
      • Class 6th
      • Class 7th
      • Class 8th
      • Class 9th
      • Class 10th
    • View All Options
      • Online Courses
      • Distance Learning
      • Hindi Medium Courses
      • International Olympiad
    • NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE (Main+Advanced)
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE Main
      • Class 11th
      • Class 12th
      • Class 12th Plus
  • Classroom
  • NEW
    • NEET
      • 2025
      • 2024
      • 2023
      • 2022
    • JEE
      • 2025
      • 2024
      • 2023
      • 2022
    • Class 6-10
    • JEE Main
      • Previous Year Papers
      • Sample Papers
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • JEE Advanced
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NEET
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
      • College Predictor
      • Counselling
    • NCERT Solutions
      • Class 6
      • Class 7
      • Class 8
      • Class 9
      • Class 10
      • Class 11
      • Class 12
    • CBSE
      • Notes
      • Sample Papers
      • Question Papers
    • Olympiad
      • NSO
      • IMO
      • NMTC
    • TALLENTEX
    • AOSAT
  • ALLEN E-Store
    • ALLEN for Schools
    • About ALLEN
    • Blogs
    • News
    • Careers
    • Request a call back
    • Book home demo
NCERT Solutions
Class 9
Maths
Chapter 11 Surface Area and Volume
Exercise 11.3

NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3

The NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3 focus on understanding the curved surface area and total surface area of cylinders. This exercise helps students apply formulas to calculate areas of real-life objects like pipes, tins, and pillars. By solving these problems, learners enhance their ability to visualize three-dimensional shapes and develop practical skills essential for geometry and mensuration topics.

The step-by-step solutions provided here follow the latest NCERT guidelines and are designed to clear doubts quickly. These solutions not only help students secure good marks in exams but also strengthen their problem-solving skills and boost confidence in handling complex questions related to surface area and volume.

1.0Download NCERT Solutions of Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3 Free PDF

Obtain NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3  in PDF format for download. These solutions have been developed by subject experts and follow the latest guidelines of the CBSE.

NCERT Solutions Class 9 Maths Chapter 11 Ex 11.3

Key Concepts of Exercise 11.3

  • Surface Curved Area:
    • CSA= 𝜋rl
  • Total surface area:
    • TSA= 𝜋r(l + r)
  • Application-based questions include finding CSA and TSA given a radius and slant height.
  • The concept of the Pythagorean Theorem is used to arrive at the missing dimension if required.

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 11.3 : Detailed Solutions

(Assume π=22/7 unless stated otherwise)

1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

Sol.

The volume of a cone is given by V = (1/3)πr²h.

(i) Radius (r) = 6 cm, Height (h) = 7 cm.

Volume = (1/3) × (22/7) × (6)² × 7 cm³

= (1/3) × 22 × 36 cm³

= 22 × 12 cm³

= 264 cm³.

(ii) Radius (r) = 3.5 cm = 7/2 cm, Height (h) = 12 cm.

Volume = (1/3) × (22/7) × (7/2)² × 12 cm³

= (1/3) × (22/7) × (49/4) × 12 cm³

= (1/3) × 22 × 7 × 3 cm³

= 22 × 7 cm³

= 154 cm³.


2. Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm.

(ii) height 12 cm, slant height 13 cm.

Sol.

(i) Radius (r) = 7 cm, Slant height (ℓ) = 25 cm.

First, find the height (h):

r² + h² = ℓ²

(7)² + h² = (25)²

49 + h² = 625

h² = 625 − 49

h² = 576

h = √576 = 24 cm.

Volume of cone = (1/3)πr²h

= (1/3) × (22/7) × (7)² × 24 cm³

= (1/3) × 22 × 7 × 24 cm³

= 22 × 7 × 8 cm³

= 1232 cm³.

Capacity in litres: 1 L = 1000 cm³.

Capacity = 1232/1000 L = 1.232 L.

(ii) Height (h) = 12 cm, Slant height (ℓ) = 13 cm.

First, find the radius (r):

r² + h² = ℓ²

r² + (12)² = (13)²

r² + 144 = 169

r² = 169 − 144

r² = 25

r = √25 = 5 cm.

Volume of cone = (1/3)πr²h

= (1/3) × (22/7) × (5)² × 12 cm³

= (1/3) × (22/7) × 25 × 12 cm³

= (22 × 25 × 4)/7 cm³

= 2200/7 cm³.

Capacity in litres:

Capacity = (2200/7) cm³ = (2200/7)/1000 L = 2200/7000 L = 22/70 L = 11/35 L.


3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π=3.14)

Sol.

Height (h) = 15 cm.

Volume = 1570 cm³.

We know Volume = (1/3)πr²h.

1570 = (1/3) × 3.14 × r² × 15

1570 = 3.14 × r² × 5

1570 = 15.70 × r²

r² = 1570 / 15.70

r² = 100

r = √100 = 10 cm.


4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Sol.

Height (h) = 9 cm.

Volume = 48π cm³.

We know Volume = (1/3)πr²h.

48π = (1/3) × π × r² × 9

48 = (1/3) × r² × 9 (Cancel π from both sides)

48 = 3r²

r² = 48/3

r² = 16

r = √16 = 4 cm.

Diameter (d) = 2r = 2 × 4 = 8 cm.


5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Sol.

For conical pit:

Diameter = 3.5 m.

Radius (r) = 3.5/2 m = 1.75 m.

Depth (h) = 12 m.

Capacity of the conical pit = (1/3)πr²h

= (1/3) × (22/7) × (1.75)² × 12 m³

= (1/3) × (22/7) × 3.0625 × 12 m³

= 22 × 3.0625 × (12/21) m³ (Incorrect calculation in original, 12/3 = 4, then divide by 7)

Let's recalculate:

= (1/3) × (22/7) × (1.75 × 1.75) × 12 m³

= (1/3) × (22/7) × (7/4 × 7/4) × 12 m³

= (1/3) × (22/7) × (49/16) × 12 m³

= 22 × (7/16) × 4 m³

= 22 × (7/4) m³ = 11 × 3.5 m³ = 38.5 m³.

Capacity in kilolitres: 1 m³ = 1 kilolitre (kl).

Capacity = 38.5 kl.


6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

Sol.

Given Volume = 9856 cm³.

Diameter of base = 28 cm.

Radius (r) = 28/2 = 14 cm.

(i) To find the height (h):

Volume = (1/3)πr²h

9856 = (1/3) × (22/7) × (14)² × h

9856 = (1/3) × (22/7) × 196 × h

9856 = (22 × 28) × h / 3

9856 = 616h / 3

h = (9856 × 3) / 616

h = 48 cm.

(ii) To find the slant height (ℓ):

ℓ² = h² + r²

ℓ² = (48)² + (14)²

ℓ² = 2304 + 196

ℓ² = 2500

ℓ = √2500 = 50 cm.

(iii) To find the curved surface area (CSA):

CSA = πrℓ

= (22/7) × 14 × 50 cm²

= 22 × 2 × 50 cm²

= 22 × 100 cm²

= 2200 cm².


7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Sol.

When a right triangle is revolved about one of its perpendicular sides, the solid formed is a cone.

The side about which it revolves becomes the height (h).

The other perpendicular side becomes the radius (r) of the base.

The hypotenuse becomes the slant height (ℓ).

Here, the triangle is revolved about the side 12 cm.

So, Height (h) = 12 cm.

Radius (r) = 5 cm.

Slant height (ℓ) = 13 cm. (This is consistent with 5-12-13 being a Pythagorean triplet).

Volume of the cone = (1/3)πr²h

= (1/3) × π × (5)² × 12 cm³

= (1/3) × π × 25 × 12 cm³

= π × 25 × 4 cm³

= 100π cm³.


8. If the triangle ABC in the question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Question 7 and 8.

Sol.

When the triangle ABC (sides 5 cm, 12 cm, 13 cm) is revolved about the side 5 cm:

Height (h) = 5 cm.

Radius (r) = 12 cm.

Slant height (ℓ) = 13 cm.

Volume of the new cone = (1/3)πr²h

= (1/3) × π × (12)² × 5 cm³

= (1/3) × π × 144 × 5 cm³

= π × 48 × 5 cm³

= 240π cm³.

Ratio of the volumes of the two solids (Volume from Q7 : Volume from Q8):

Volume (Q7) = 100π cm³

Volume (Q8) = 240π cm³

Ratio = (100π) / (240π) = 100/240 = 10/24 = 5/12.

The ratio is 5:12.


9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Sol.

For the conical heap of wheat:

Diameter = 10.5 m.

Base Radius (r) = 10.5/2 m = 5.25 m.

Height (h) = 3 m.

First, find its volume:

Volume = (1/3)πr²h

= (1/3) × (22/7) × (5.25)² × 3 m³

= (1/3) × (22/7) × 27.5625 × 3 m³

= 22 × (27.5625/7) m³

= 22 × 3.9375 m³

= 86.625 m³.

Next, find the area of the canvas required. This is the curved surface area (CSA) of the cone.

We need the slant height (ℓ):

ℓ = √(r² + h²) = √((5.25)² + (3)²) m

ℓ = √(27.5625 + 9) m

ℓ = √36.5625 m

ℓ ≈ 6.0467 m (approximately 6.05 m as in original solution)

Area of the canvas required (CSA) = πrℓ

= (22/7) × 5.25 × 6.05 m²

= 22 × 0.75 × 6.05 m²

= 16.5 × 6.05 m²

= 99.825 m².

Thus, the required area of the canvas is 99.825 m².

4.0Key Features and Benefits Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3

  • There are step by step solutions to each question that will assist with learning
  • There are diagrams and breaks on formulas to assist with visualization
  • The work is based on the recent CBSE Grade 9 curriculum
  • Suitable for homework help and learning support in preparation for exams and tests.
  • Will increase your ability to extend geometry formulas to answer real world problems. 

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

Exercise 11.3 deals with calculating the surface area and volume of right circular cones, including practical word problems.

Curved Surface Area (CSA) = πrl Total Surface Area (TSA) = πr(l + r)

Use the Pythagoras Theorem: l = √(r² + h²)

Yes! Students learn the formula: Volume of cone = (1/3)πr²h

Join ALLEN!

(Session 2025 - 26)


Choose class
Choose your goal
Preferred Mode
Choose State
  • About
    • About us
    • Blog
    • News
    • MyExam EduBlogs
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET UG
    • CBSE
    • NCERT Solutions
    • Olympiad
    • NEET 2025 Results
    • NEET 2025 Answer Key
    • NEET College Predictor
    • NEET 2025 Counselling

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO