Important Questions CBSE Class 9 Maths Chapter 13 Surface Areas Volumes

Important questions for CBSE Class 9 Maths Chapter 13 Surface Areas and Volumes are given here for students who are preparing for their final exams. All these questions are prepared by our subject experts based on the CBSE syllabus (2022-2023) and are relevant to the NCERT book. The important questions are formulated as per the latest exam pattern.  Practising these questions will help students to score well in their exams. Students can revise by accessing the Class-wise important questions for CBSE Class 9 Maths at BYJU’S.

Chapter 13 of Class 9 maths deals with various solid shapes such as cubes, cuboids, cones, cylinders and spheres. In this chapter, students will learn how to find the surface area and volumes of all these solids and how we can apply these formulas in word problems based on real-world scenarios.

Also Check:

• Important 2 Marks Questions for CBSE 9th Maths
• Important 3 Marks Questions for CBSE 9th Maths
• Important 4 Marks Questions for CBSE 9th Maths

Important Questions For CBSE Class 9 Surface Areas and Volumes  (Chapter 13)

Q.1: Hameed has built a cubical water tank with a lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see in the figure below). Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.

Solution:

Given,

Edge of the cubical tank (a) = 1.5 m = 150 cm 

So, surface area of the tank = 5 × 150 × 150 cm²

The measure of side of a square tile = 25 cm

Area of each square tile = side × side = 25 × 25 cm²

Required number of tiles = (Surface area of the tank)/(area of each tile)

= (5 × 150 × 150)/(25 × 25)

= 180

Also, given that the cost of the tiles is Rs. 360 per dozen.

Thus, the cost of each tile = Rs. 360/12 = Rs. 30

Hence, the total cost of 180 tiles = 180 × Rs. 30 = Rs. 5400

Q.2: The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Solution:

Given,

Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm

Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm

Surface area of 1 brick  = 2(lb + bh + hl) 

= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²

= 2(225 + 75 + 168.75) cm² 

= 2 x 468.75 cm² 

= 937.5 cm² 

Area that can be painted by the container = 9.375 m² (given)

= 9.375 × 10000 cm²

= 93750 cm²

Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick)

= 93750/937.5

= 937500/9375

= 100

Q.3: The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of  Rs.7.50 per sq.m.

Solution:

Given,

Length of the room (l) = 5 m

Breadth of the room (b) = 4 m

Height of the room (h) = 3 m

Area of walls of the room = Lateral surface area of cuboid

= 2h(l + b)

= 2 × 3(5 + 4)

= 6 × 9

= 54 sq.m

Area of ceiling = Area of base of the cuboid

= lb

= 5 × 4

= 20 sq.m

Area to be white washed = (54 + 20) sq.m = 74 sq.m

Given that, the cost of white washing 1 sq.m = Rs. 7.50

Therefore, the total cost of white washing the walls and ceiling of the room = 74 × Rs. 7.50 = Rs. 555

Q.4: The curved surface area of a right circular cylinder of height 14 cm is 88 sq.cm. Find the diameter of the base of the cylinder.

Solution:

Let d be the diameter and r be the radius of a right circular cylinder.

Given,

Height of cylinder (h) = 14 cm 

Curved surface area of right circular cylinder = 88 cm²

⇒ 2πrh = 88 cm²

⇒ πdh = 88 cm² (since d = 2r) 

⇒ 22/7 x d x 14 cm = 88 cm²

⇒ d = 2 cm 

Hence, the diameter of the base of the cylinder is 2 cm.

Q.5: Curved surface area of a right circular cylinder is 4.4 sq.m. If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:

Let h be the height of the cylinder.

Given,

Radius of the base of the cylinder (r) = 0.7 m 

Curved surface area of cylinder = 4.4 m²

Thus,

2πrh = 4.4  

2 × 3.14 × 0.7 × h = 4.4

4.4 × h = 4.4 

h = 4.4/4.4

h = 1  

Therefore, the height of the cylinder is 1 m.

Q.6: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Given,

Length of the cylindrical pipe = h = 28 m

Diameter of the pipe = 5 cm

Now, the radius of piper (r) = 5/ 2 cm = 2.5 cm = 0.025 m

Total radiating surface in the system = Total surface area of the cylinder  

= 2πr(h + r)  

= 2 × (22/7) × 0.025 (28 + 0.025) m²  

= (44 x 0.025 x 28.025)/7 m²

= 4.4 m² (approx)

Q.7: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Take π = 3.14)

Solution:

Given

Height of a cone (h) = 16 cm

Radius of the base (r) = 12 cm

Now,

Slant height of cone (l) = √(r² + h²)

= √(256 + 144)

= √400

= 20 cm

Curved surface area of cone = πrl

= 3.14 × 12 × 20 cm²

= 753.6 cm²

Total surface area = πrl + πr²

= (753.6 + 3.14 × 12 × 12) cm²

= (753.6 + 452.16) cm²

= 1205.76 cm²

Q.8: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

Given,

Diameter of the cone = 24 m 

Radius of the cone (r) = 24/2 = 12 m 

Slant height of the cone (l) = 21 m 

Total surface area of a cone = πr(l + r) 

= (22/7) × 12 × (21 + 12) 

= (22/7) × 12 × 33 

= 1244.57 m²

Q.9: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs.210 per 100 sq.m.

Solution:

Given,

Slant height of a cone (l) = 25 m  

Diameter of the base of cone = 2r = 14 m  

∴ Radius = r = 7 m  

Curved Surface Area = πrl 

= (22/7) x 7 x 25 

= 22 × 25  

= 550 sq.m  

Also, given that the cost of white-washing 100 sq.m = Rs. 210 

Hence, the total cost of white-washing for 550 sq.m = (Rs. 210 × 550)/100 = Rs. 1155

Q.10: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.

Solution:

Given,

Diameter of the sphere = 7 m

Radius (r) = 7/2 = 3.5 m 

Now, the riding space available for the motorcyclist = Surface area of the sphere

= 4πr²

= 4 × (22/7) × 3.5 × 3.5

= 154 m²

Q.11: 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.

Solution:

Given,

Radius of balloon = r =  7 cm

Radius of pumped balloon = R = 14 cm

Ratio of surface area = (TSA of balloon with r = 7 cm)/(TSA of balloon with R = 14 cm)

= (4πr²)/(4πR²)

= r²/R²

= (7)²/(14)²

= 49/196

= 1/4

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

Q.12: A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Solution:

Given,

Depth of the river (h) = 3 m

Width of the river (w) = 40 m

Flow rate of water = 2 km/hr

i.e. Flow of water in 1 hour = 2 km = 2000 m

Flow of water in 1 minute = 2000/60 = 100/3 m

Thus, length (l) = 100/3 m

Volume of water falling into the sea in 1 minute = Volume of cuboid with dimension l, w, h

= l × w × h

= (100/3) × 40 × 3

= 4000 m³

= 4000 x 1000 L 

= 4000000 L

Q.13: A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:

Given,

Diameter of the pencil = 7  mm

Radius of the pencil (R) = 7/2 mm

Diameter of the graphite cylinder = 1 mm

Radius of the graphite (r) = 1/2 mm

Height (h) = 14 cm = 140 mm (since 1 cm = 10 mm)

Volume of a cylinder = πr²h

Volume of graphite cylinder = πr2h 

= (22/7) × (1/2) × (1/2) × 140

= 110 mm³

Volume of pencil = πR²h

= (22/7) × (7/2) × (7/2) × 140

= 490 × 11 

= 5390 mm²

Volume of wood = Volume of pencil – Volume of graphite

= 5390- 110 = 5280 mm³

= 5280/1000 (since 1 mm³ = 1/1000 cm³)

= 5.28 cm³

Q.14: Meera has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m², find the volume of the tent that can be made with it.

Solution:

Given,

Area of the canvas = 551 m²

Area of the canvas lost in wastage = 1 m²

Thus, the area of canvas available for making the tent = (551 – 1) m² = 550 m²

Now, the surface area of the tent = 550 m²

The required base radius of the conical tent = 7 m

Curved surface area of tent = 550 m²

That means,

πrl = 550 

(22/7) × 7 × l = 550

l = 550/22

l = 25 m

Now, l² = h² + r²

h² = l² – r² = (25)² – (7)² = 625 – 49 = 576

h = 24 m

So, the volume of the conical tent = (1/3)πr²h

= (1/3) × (22/7) × 7 × 7 × 24

= 1232 m³

Q.15: A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Solution:

Given,

Diameter of capsule = 3.5 mm

Radius of capsule = (r) = 3.5/2 = 1.75 mm

Volume of spherical capsule = (4/3)πr³

= (4/3) × (22/7) × 1.75 × 1.75 × 1.75

= 22.458 mm³

Therefore, the volume of the capsule is 22.46 mm³ approx.

Q.16: Calculate the amount of ice-cream that can be put into a cone with base radius 3.5 cm and height 12 cm.

Solution:

Given,

Base radius = r = 3.5 cm

Height = h = 12 cm

The amount of ice-cream that can be put into a cone = Volume of cone

= (1/3)πr²h

= (1/3) × (22/7) × 3.5 × 3.5 × 12

= 154 cm³

Q.17: A spherical ball is divided into two equal halves. Given that the curved surface area of each half is 56.57 cm, what will be the volume of the spherical ball?

Solution:

Given,

Curved surface area of of half of the spherical ball = 56.57 cm²

(1/2) 4πr² = 56.57

2 × 3.14 × r² = 56.57

r² = 56.57/6.28

r² = 9 (approx)

r = 3 cm

Now,

Volume of spherical ball = (4/3)πr³

= (4/3) × 3.14 × 3 × 3 × 3

= 113.04 cm³

Articles for Class 9

• Algebraic Identities For Class 9
• Areas Of Parallelograms And Triangles Class 9 Notes
• Circles Class 9
• Class 9 Maths MCQs
• Congruence of Triangles Class 9
• Constructions Class 9
• Coordinate Geometry Class 9 Notes
• Heron’s Formula Class 9 Notes
• Linear Equations in Two Variables Class 9 Notes
• Lines and Angles Class 9
• Maths Formulas for Class 9
• Number System for Class 9
• Polynomial Class 9 Notes
• Probability Class 9 Notes
• Quadrilaterals Class 9 Notes
• Surface Area and Volume Class 9

Practice Problems for Surface Areas and Volumes Class 9 Maths

1. How many shots, each having a diameter 3 cm, can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?
2. Two identical cubes, each of volume 64 cm³ are joined together end to end. What is the surface area of the resulting cuboid?
3. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?
4. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
5. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m³?
6. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm. 
7. A joker’s cap is in the form of a right circular cone with a base radius of 7 cm and a height of 24 cm. Find the area of the sheet required to make 10 such caps. 
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.