 NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes in English and Hindi Medium updated for session 2023-24. Please note that the effectiveness of Tiwari Academy or any educational platform can vary depending on individual learning styles and needs. According to NCERT book for CBSE 2023-24 and new syllabus of class 9 Maths, chapter 11 has only four exercises.
Class 9 Maths Chapter 11 Solutions in English Medium
Class 9 Maths Exercise 11.1 in English
Class 9 Maths Exercise 11.2 in English
Class 9 Maths Exercise 11.3 in English
Class 9 Maths Exercise 11.4 in English
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Class 9 Maths Chapter 11 Solutions in Hindi Medium
Class 9 Maths Exercise 11.1 in Hindi
Class 9 Maths Exercise 11.2 in Hindi
Class 9 Maths Exercise 11.3 in Hindi
Class 9 Maths Exercise 11.4 in Hindi

## NCERT Solutions for Class 9 Maths Chapter 11

 Class: 9 Mathematics Chapter 11: Surface Areas and Volumes Number of Exercises: 4 (Four) Content: NCERT Exercises Solutions Content Type: Online Videos and Text Format Academic Session: CBSE 2023-24 Medium: English and Hindi Medium UP Board Students of Class 9 (High School) are using NCERT Book in session 2023-24, so they can also using these textbooks solutions as UP Board Solutions for Class 9 Maths Chapter 11. Here, they can download Prashnavali 11.1, Prashnavali 11.2, Prashnavali 11.3 and Prashnavali 11.4 in Hindi Medium to use offline for session 2023-24. NCERT Solutions for class 9 chapter 11 is applicable for CBSE Delhi Board, MP Board, UP Board – High School, UK Board, Gujrat Board and other board using NCERT Books. NCERT Solutions for other are also given in downloadable format.

### What are the formulae for Cuboid?

Cuboid: A cuboid is a solid bounded by six rectangular plane surfaces for example match box, brick, books, etc. are cuboid.
1. Surface area (or total surface area) of cuboid = 2 (lb + bh + hl) square units
2. Lateral surface area of cuboid = 2(l + b)h square units
3. Diagonal of a cuboid = √[l² + b² + h²] units
4. Total length of a edges of a cuboid = 4 (l + b + h) units
5. Volume of cuboid = lbh cubic units

### What are the formulae for Cube?

Cube: A cuboid whose length, breadth and height are equal, is called a cube.
1. Surface area (or total surface area) of cuboid = 6a² square units
2. Lateral surface area of cuboid = 4a² square units
3. Diagonal of a cuboid = √3a units
4. Total length of a edges of a cuboid = 12a units
5. Volume of cuboid = a³ cubic units

### What is meant by a Cylinder? Write its formulae?

Cylinder: A solid generated by the revolution of a rectangle about one of its sides which is kept fixed is called right circular cylinder.
Curved Surface Area (CSA) = 2πrh square units
Total Surface Area (TSA) = 2πr(r + h) square units
Volume = πr²h cubic units

### What do you understand by a Cone? Write its Formulae?

Cone: A right circular cone is solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with a fixed line.
Slant Height = √[r² + h²]
Curved Surface Area (CSA) = πrl square units
Total Surface Area (TSA) = πr(r + l) square units
Volume = 1/3 πr²h cubic units

### What are the basic formulae of Sphare?

Sphere: A sphere is three dimensional figure which is made up of all points in the space, which lie at a constant distance, form a fixed point called the centre of the sphere and the constant distant is called its radius.
Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr² square units
Volume = 4/3 πr³ cubic units

### 9th Maths Chapter 11 Solutions in English & Hindi Medium

Download free NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes all exercises which are given below in PDF format to download for new academic session 2023-24. NCERT Solutions as well as NCERT Solutions Offline Apps are updated for new academic session 2023-24 based on latest CBSE Syllabus 2023-24. NCERT Books in English and Hindi are now implemented in Uttar Pradesh also. So, students can download UP Board solutions for class 9 Maths chapter 11 all exercises from here.

### The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Inner radius of cylindrical pipe r = 24/2 = 12 cm,
outer radius R = 28/2 = 14 cm and
length h = 35 m
Volume of cylindrical wooden pipe
= π(R²-r²)h
= 22/7 × (14²-12²)×35
= 22 × (196 – 144) × 5
= 22 × 52 × 5
= 5720 cm³
Mass of cylindrical wooden pipe
= 5720 × 0.6 g
= 3432 g
= 3.432 kg [∵1 cm³ of wood has a mass of 0.6 g]
Hence, the volume of cylindrical wooden pipe is 3.432 kg.

### A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Length of tin can l = 5 cm, breadth b = 4 cm and height h = 15 cm
Volume of tin can = lbh = 5 × 4 × 15 = 300 cm³
Radius of plastic cylinder r = 7/2 = 3.5 cm and height H = 10 cm
Volume of plastic cylinder
= πr²H = 22/7 × 3.5 × 3.5 × 10
= 22 × 0.5 × 3.5 × 10
= 385 cm³
Difference between capacities of two packs = 385 – 300 = 85 cm³
Hence, the capacity of plastic cylindrical pack is greater than tin can by 85 cm³.

### If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find radius of its base.

Lateral surface area of cylinder C = 94.2 cm² and height h = 5 cm.
Let, the radius of cylinder = r cm
Lateral surface area of cylinder C = 2πrh
⇒ 94.2 = 2 × 3.14 × r × 5
⇒ r = 94.2/(3.14×10) = 3 cm
Hence, the radius of base is 3 cm.

### It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find inner curved surface area of the vessel.

Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200 and
height h = 10 m.
Let, the inner radius of cylindrical vessel = r m
The inner curved surface area of cylindrical vessel = 2πrh
The cost of painting is at the rate of ₹ 20 per m² = ₹ 20 × 2πrh
According to question,
₹ 20 × 2πrh = ₹ 2200
⇒ 2πrh = 2200/20 = 110
Hence, the inner curved surface area is 110 m².

### A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Radius of cylindrical bowl r = 7/2 = 3.5 and
height of soup inside the cylindrical bowl h = 4 cm
Volume of cylindrical bowl
= πr²h = 22/7 × (3.5)² × 4
= 22/7× 3.5 × 3.5 × 4
= 22 × 0.5 × 3.5 × 4 = 154 cm³
Therefore, the volume of soup per day for 250 patient
= 250 × 154
= 38500 cm³
Hence, hospital has to prepare 38500 cm³ soup daily to serve 250 patients.

### The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Curved surface area of cylinder 88 cm² and height h = 14 cm
Let, the radius of base of cylinder = r cm
Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × 22/7 × r × 14
⇒ 88 = 88r
⇒ r = 1 cm
Hence, the diameter of base of cylinder
= 2r = 2 × 1 = 2 cm

### The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Radius of roller r = 84/2 = 42 cm = 0.42 m and
length h = 120 cm = 1.2 m
Outer curved surface area of roller
= 2πrh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²
Area of ground levelled in on revolution = 3.168 m²
Therefore, area of ground levelled in 500 revolutions
= 500 × 3.168 = 1584 m²
Hence, the area of playground is 1584 m².

### Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Curved surface area of cylinder 4.4 m² and radius r = 0.7 m
Let, the height of cylinder = h m
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × 22/7 × 0.7 × h
⇒ 4.4 = 4.4h
⇒ h = 1 m
Hence, the height of the cylinder is 1 m.

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

Radius of hemisphere r = 10 cm
Surface area of hemisphere = 3πr²
= 3 × 3.14 × 10 × 10
= 942 cm²
Hence, the total surface area of hemisphere is 942 cm².

Surface Area:
Surface area of a solid body is the area of all of its surface together and it is always measured in square units. Surface area is also known as total surface area (TSA).
Volume:
Space occupied by an object (solid body) is called the volume of the object. Volume is always measured in cubic units.

### How many exercises are there in chapter 11 of 9th Maths?

According to new syllabus, there are total 4 exercises in chapter 11 surface areas and volumes of 9th mathematics. In all exercises only 8 or 9 questions are given based on separate sections. Only 1 or 2 questions in exercise are tricky.

### Is chapter 11 of class 9 Maths easy?

Class 9 Maths chapter 11 Surface Areas and Volumes, is easy to solve and understand but need to learn formulae. If a student learn formulae based on areas and volumes of 3-dimensional figures, the questions in each exercise look simple.

Last Edited: September 18, 2023