Class 9 Maths Ganita Manjari Chapter 14 End of Chapter Exercises Math of Space: Surface Area and Volume
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 End of Chapter Exercises Math of Space: Surface Area and Volume provides detailed, step-by-step solutions to help students revise the important concepts covered in this chapter.
The solutions explain how to calculate the surface area and volume of three-dimensional shapes using relevant formulas and methods in simple language. Students can access the solutions in PDF format for easy download, offline study, and revision. Practising these end-of-chapter exercises helps students strengthen their mathematical concepts, improve problem-solving skills, and prepare effectively for exams.
NCERT Class 9 Maths Ganita Manjari Chapter 14 End-of-Chapter Exercises
1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
Assume the container is completely full, there is no wastage, and each scoop is approximately a sphere of radius 2 cm.
Volume of the cuboidal container:
V = 10 × 15 × 20 = 3000 cm³
Volume of one scoop:
V = (4/3)πr³
= (4/3) × π × 2³
= 32π/3
≈ 33.52 cm³
Number of scoops:
3000 ÷ 33.52 ≈ 89.5
Therefore, the container can provide approximately 90 scoops of ice cream.
The actual number may vary depending on the size of each scoop.
2. A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
A cube of side length a contains:
a³ unit cubes
A cube of side length a + 1 contains:
(a + 1)³ unit cubes
Therefore, the number of unit cubes to be added is:
(a + 1)³ − a³
Expanding:
= a³ + 3a² + 3a + 1 − a³
= 3a² + 3a + 1
Hence, the required number of unit cubes is:
3a² + 3a + 1.
3. Solve the following:
(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom
Answer:
3. Solve the following:
(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
This is an estimation question, so assume a person drinks about 2 litres of water per day for 70 years.
Total water drunk:
= 2 × 365 × 70
= 51,100 litres
= 51.1 m³
Assume the classroom measures:
8 m × 6 m × 3 m
Volume of the classroom:
= 8 × 6 × 3
= 144 m³
= 1,44,000 litres
Since:
51,100 < 1,44,000
a person would not drink enough water in a lifetime to fill a classroom of this size.
The water consumed would fill approximately:
(51,100 / 1,44,000) × 100 ≈ 35.5%
of the classroom.
Therefore, No, a person would not drink enough water in a lifetime to fill a normal-sized classroom.
(ii) Estimate the number of bricks used to build the walls of your classroom.
Assume:
- Internal length = 8 m
- Internal breadth = 6 m
- Wall height = 3 m
- Wall thickness = 0.20 m
Outer dimensions:
8.4 m × 6.4 m
Volume of all four walls before subtracting openings:
= [(8.4 × 6.4) − (8 × 6)] × 3
= 17.28 m³
Assume one door of size 1 m × 2 m and two windows of size 1.5 m × 1.2 m.
Volume of openings:
= [1 × 2 + 2(1.5 × 1.2)] × 0.20
= 1.12 m³
Net volume of the walls:
= 17.28 − 1.12
= 16.16 m³
Assume one brick with its share of mortar occupies approximately:
0.20 × 0.10 × 0.10 = 0.002 m³
Number of bricks:
= 16.16 / 0.002
= 8080
Therefore, approximately 8,100 bricks are required to build the walls of the classroom.
Note: Since this is an estimation problem, the answer may vary depending on the assumed classroom size, wall thickness, openings, and brick dimensions.
4. You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
Answer:
Volume of a cube = side³
Option A
One cube of side 50 mm:
Volume = 50³ = 125,000 mm³
Option B
100 cubes of side 10 mm:
Volume of one cube:
10³ = 1,000 mm³
Volume of 100 cubes:
100 × 1,000 = 100,000 mm³
Since:
125,000 mm³ > 100,000 mm³
Therefore, Option A gives more chocolate.
It gives 25,000 mm³ more chocolate than Option B.
5. If all the people in the world are crowded together in one location, how much area would it cover?
This is an estimation question, so we need to make a reasonable assumption.
Assume the world population is about 8 billion people and, when crowded together, each person occupies about 0.25 m² of ground area.
Total area required:
= 8,000,000,000 × 0.25
= 2,000,000,000 m²
Since:
1 km² = 1,000,000 m²
Area = 2,000 km²
Therefore, all the people in the world crowded together would occupy roughly 2,000 km².
This is approximately the area of a square about 45 km × 45 km.
The estimate will change depending on the assumed world population and how tightly people are packed.
6. A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
Answer:
Diameter of each cylindrical glass = 7 cm
So, radius:
r = 7/2 = 3.5 cm
Height of milk:
h = 12 cm
Volume of milk in one glass:
V = πr²h
= 22/7 × 3.5 × 3.5 × 12
= 462 cm³
For 1600 students:
Total volume = 462 × 1600
= 739,200 cm³
Since:
1000 cm³ = 1 litre
739,200 cm³ = 739.2 litres
Therefore, the school needs 739.2 litres of milk.
7. The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.
Answer:
Given:
- Radius of sphere = 5 cm
- Radius of cone = 4 cm
Surface area of sphere:
4πr² = 4π(5²) = 100π cm²
It is five times the curved surface area of the cone.
So,
Curved surface area of cone = 100π / 5 = 20π cm²
For a cone:
CSA = πrl
So,
π × 4 × l = 20π
l = 5 cm
Now,
l² = r² + h²
5² = 4² + h²
25 = 16 + h²
h² = 9
h = 3 cm
Volume of the cone
V = (1/3)πr²h
= (1/3)π × 4² × 3
= 16π cm³
Therefore:
Height of the cone = 3 cm
Volume of the cone = 16π cm³ ≈ 50.27 cm³
8. Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 6,95,700 km.
Compute approximately:
(i) the ratio of the volume of Jupiter to the volume of the Earth;
(ii) the ratio of the volume of the Sun to the volume of the Earth.
(Calculator can be used.)
Answer:
For a sphere,
Volume = (4/3)πr³
Therefore, the ratio of the volumes of two spheres is equal to the cube of the ratio of their radii.
(i) Ratio of the volume of Jupiter to the volume of Earth
V₍Jupiter₎ / V₍Earth₎ = (69900 / 6370)³
≈ 1321.34
Therefore,
Volume of Jupiter : Volume of Earth ≈ 1321 : 1
(ii) Ratio of the volume of the Sun to the volume of Earth
V₍Sun₎ / V₍Earth₎ = (695700 / 6370)³
≈ 13,02,709.9
Therefore,
Volume of Sun : Volume of Earth ≈ 13,02,710 : 1
Hence:
(i) Jupiter : Earth ≈ 1321 : 1
(ii) Sun : Earth ≈ 13,02,710 : 1
9. Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder which encloses it.
Answer:
Let the radius of the sphere be r.
The smallest cylinder that can enclose the sphere will have:
Radius = r
Height = 2r
Volume of the cylinder:
V₍cylinder₎ = πr²h
= πr²(2r)
= 2πr³
Volume of the sphere:
V₍sphere₎ = (4/3)πr³
Now,
V₍sphere₎ / V₍cylinder₎ = [(4/3)πr³] / [2πr³]
= 2/3
Therefore,
Volume of the sphere = 2/3 × Volume of the smallest enclosing cylinder.
Hence proved.
10. Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere.
Answer:
Let the radius of the Earth be r metres, and let the second string be h metres above the ground.
The first string has circumference:
2πr
The second string is 1 metre longer:
2π(r + h) = 2πr + 1
So,
2πr + 2πh = 2πr + 1
2πh = 1
Therefore,
h = 1/(2π)
h ≈ 0.159 m
≈ 15.9 cm
So, the second string is about 15.9 cm above the ground.
The same calculation applies to any sphere because the original radius cancels out.
Therefore:
- Earth: about 15.9 cm
- Moon: about 15.9 cm
- Jupiter: about 15.9 cm
- Volleyball: about 15.9 cm
Yes, this may seem surprising. The answer is the same in every case because increasing the circumference by 1 metre increases the radius by exactly 1/(2π) metre, regardless of the original size of the sphere.
11. We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
Answer:
The square base of the pyramid is inscribed in the circular base of the cylinder.
So, the diagonal of the square equals the diameter of the circle:
Diagonal = 2r
If the side of the square is a, then:
a√2 = 2r
Therefore,
a = r√2
Area of the square base:
a² = 2r²
Volume of the square pyramid:
Vₚ = (1/3) × 2r² × h
= (2/3)r²h
Volume of the cylinder:
V꜀ = πr²h
Therefore,
Vₚ : V꜀ = (2/3)r²h : πr²h
Cancelling r²h:
Vₚ : V꜀ = 2 : 3π
Hence, the ratio of the volume of the pyramid to the volume of the cylinder is:
2 : 3π.
12. What is the change in volume when:
(i) the length of a cuboid with dimensions l, w, h is increased by 1 unit?
(a) 1 cubic unit
(b) (lwh + 1) cubic unit
(c) wh cubic units
(d) lw cubic units
(e) lh cubic units
(ii) the radius of a cylinder with dimensions r, h is increased by 1 unit?
(a) 1 cubic unit
(b) πr²h cubic units
(c) πr² cubic units
(d) 2πrh + 2πh cubic units
(e) 2πrh + πh cubic units
(iii) the radius of a sphere is decreased by 1 unit?
Answer:
(i) The length of a cuboid with dimensions \(l, w, h\) is increased by 1 unit?
Original volume:
V = lwh
New volume:
V' = (l + 1)wh
Change in volume:
V' − V = (l + 1)wh − lwh
= wh
Therefore, the correct option is:
(c) wh cubic units
(ii) The radius of a cylinder with dimensions \(r, h\) is increased by 1 unit?
Original volume:
V = πr²h
New radius = r + 1
New volume:
V' = π(r + 1)²h
Change in volume:
V' − V = πh[(r + 1)² − r²]
= πh(r² + 2r + 1 − r²)
= 2πrh + πh
Therefore, the correct option is:
(e) 2πrh + πh cubic units
(iii) The radius of a sphere is decreased by 1 unit
Let the original radius be r.
Original volume:
V = (4/3)πr³
New radius = r − 1
New volume:
V' = (4/3)π(r − 1)³
Decrease in volume:
V − V' = (4/3)π[r³ − (r − 1)³]
Since,
(r − 1)³ = r³ − 3r² + 3r − 1
Therefore,
V − V' = (4/3)π(3r² − 3r + 1)
= 4πr² − 4πr + 4π/3
Hence, the volume decreases by:
4πr² − 4πr + 4π/3 cubic units.
13. Given a cube with volume V, express its total surface area S in terms of V.
Let the side of the cube be a.
Volume of a cube:
V = a³
So,
a = V^(1/3)
Total surface area of a cube:
S = 6a²
Substituting \(a = V^{1/3}\):
S = 6(V^(1/3))²
Therefore,
S = 6V^(2/3)
Hence, the total surface area in terms of volume is:
S = 6V^(2/3).
14. Given a cube with total surface area S, express its volume V in terms of S.
Let the side of the cube be a.
Total surface area of a cube is:
S = 6a²
So,
a² = S/6
Therefore,
a = √(S/6)
Volume of the cube:
V = a³
Substituting the value of a:
V = (√(S/6))³
Hence,
V = (S/6)^(3/2)
Therefore, the volume of the cube in terms of its total surface area is:
V = (S/6)^(3/2).
15. A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?
The water level must rise from 16 cm to 20 cm, so the required increase in height is:
20 − 16 = 4 cm
Radius of the cylindrical glass = 4 cm
Volume of water displaced to raise the level by 4 cm:
V = πr²h
= π × 4² × 4
= 64π cm³
Each marble is a sphere of radius 1 cm.
Volume of one marble:
V = (4/3)πr³
= (4/3)π × 1³
= 4π/3 cm³
Number of marbles required:
= 64π ÷ (4π/3)
= 64π × 3/(4π)
= 48
Therefore, the crow should drop 48 marbles into the glass.
16. Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
This is an estimation-based question, so the answer depends on the dimensions of the ink tube.
Assume the ink inside the ball-point pen forms a small cylinder of:
- Length = 10 cm
- Internal radius = 1 mm = 0.1 cm
Volume of ink:
V = πr²h
= π × (0.1)² × 10
= 0.1π
≈ 0.314 cm³
Since 1 cm³ = 1 mL,
Volume of ink ≈ 0.314 mL
Therefore, a new ball-point pen may contain approximately 0.3 mL of ink.
The actual value may vary depending on the length and internal diameter of the pen refill.
17. Solve:
(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?
(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
es, the answer is correct based on the assumptions stated. Since both parts are estimation-based, clearly stating the assumed dimensions is important.
For (i):
Volume of dough ball = (4/3)π(6³) = 288π cm³
Volume of one chapati:
= π(6²)(0.2) = 7.2π cm³
Number of chapatis:
= 288π / 7.2π = 40
So, about 40 small chapatis is correct for the assumed radius and thickness.
For (ii):
Volume of edible flesh in one half:
V = (2/3)π(R³ − r³)
= (2/3)π(5³ − 4³)
= (2/3)π(125 − 64)
= 122π/3
Using π = 22/7:
V ≈ 127.81 cm³
So the estimate is approximately 128 cm³ for one half, or 256 cm³ for the whole coconut if both halves are approximately equal.
Your solution is mathematically consistent and suitable as a final answer for an estimation question.
18. Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders:
(i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions 15 ft × 20 ft × 30 ft?
Answer:
(i) If all the pages of the textbook were laid side-by-side on the floor, would they cover the entire classroom floor?
Assume each paper sheet measures 25 cm × 20 cm and the book has 200 numbered pages, i.e. 100 physical sheets, ignoring the covers.
Area of one sheet:
= 25 × 20
= 500 cm²
Area of 100 sheets:
= 100 × 500
= 50,000 cm²
= 5 m²
Assume the classroom floor measures:
8 m × 6 m = 48 m²
Since:
5 m² < 48 m²
the sheets would not cover the entire classroom floor.
Therefore, No, the pages would not cover the classroom floor.
(ii) What is the maximum number of textbooks that can fit in a storeroom of dimensions 15 ft × 20 ft × 30 ft?
Assume one closed textbook measures:
25 cm × 20 cm × 2 cm
Using:
1 ft ≈ 30 cm
Storeroom dimensions are approximately:
450 cm × 600 cm × 900 cm
Arrange the books with:
- 25 cm side along 450 cm
- 20 cm side along 600 cm
- 2 cm thickness along 900 cm
Number of books:
= (450 ÷ 25) × (600 ÷ 20) × (900 ÷ 2)
= 18 × 30 × 450
= 2,43,000
Therefore, approximately 2,43,000 textbooks, or 2.43 lakh textbooks, can fit in the storeroom under these idealised assumptions.
19. If the entire human population decided to climb into one giant cube, how long would the side have to be?
Assume the world population is approximately 8 billion and allow about 1 m³ of space per person, including some empty space.
Total volume required:
V = 8,000,000,000 m³
Let the side of the cube be a metres.
For a cube:
a³ = 8,000,000,000
a = ∛(8 × 10⁹)
a = 2000 m
= 2 km
Therefore, the giant cube would need to have a side of approximately 2 km.
20. The Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~6371 km.
(ii) Simplify the expression in (i) using a calculator.
Let the Earth’s radius be R = 6371 km and the thickness of the water layer be t km.
The total volume of water is:
W = 1.38 × 10⁹ km³
The outer radius of the water-covered Earth is:
R + t
(i) Expression for the thickness of the water layer
Volume of the spherical water layer:
W = (4/3)π[(R + t)³ − R³]
So,
(R + t)³ = R³ + 3W/(4π)
Therefore,
t = ∛[R³ + 3W/(4π)] − R
Substituting the given values:
t = ∛[6371³ + (3 × 1.38 × 10⁹)/(4π)] − 6371 km
(ii) Using a calculator
t ≈ 2.704 km
Therefore, the thickness of the water layer is approximately:
2.70 km
or
2704 m.
21. Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Answer:
One possible example is to take the original cuboid as:
8 cm × 14 cm × 14 cm
Original cuboid
Volume:
V₁ = 8 × 14 × 14
= 1568 cm³
Total surface area:
S₁ = 2(lb + bh + hl)
= 2(8×14 + 14×14 + 8×14)
= 2(112 + 196 + 112)
= 840 cm²
Now take the new cuboid as:
1 cm × 28 cm × 28 cm
New cuboid
Volume:
V₂ = 1 × 28 × 28
= 784 cm³
Since:
784 = 1568/2
the volume is halved.
Total surface area:
S₂ = 2(1×28 + 28×28 + 1×28)
= 2(28 + 784 + 28)
= 1680 cm²
Since:
1680 = 2 × 840
the surface area is doubled.
Therefore, one valid pair of cuboid dimensions is:
8 cm × 14 cm × 14 cm and 1 cm × 28 cm × 28 cm.
The dimensions are not unique; this is one possible example.
22. Project: Find the volume of your house making necessary approximations. Present how you solved it.
22. Project: Find the volume of your house making necessary approximations.
Aim: To estimate the internal volume of a single-storey house.
Assume the house is divided into separate rooms and passages. Treat each space as a cuboid and calculate its volume using:
Volume = length × breadth × height
Sample measurements:
| Space | Dimensions | Volume |
|---|---|---|
| Living room | 4 m × 3 m × 3 m | 36 m³ |
| Bedroom 1 | 4 m × 3 m × 3 m | 36 m³ |
| Bedroom 2 | 3 m × 3 m × 3 m | 27 m³ |
| Kitchen | 3 m × 2 m × 3 m | 18 m³ |
| Bathroom | 2 m × 1.5 m × 3 m | 9 m³ |
| Passage | 2 m × 3 m × 3 m | 18 m³ |
Total internal volume:
= 36 + 36 + 27 + 18 + 9 + 18
= 144 m³
Therefore, the estimated internal volume of the sample house is:
144 m³
Assumptions: The ceiling height is taken as 3 m throughout, all rooms are approximated as cuboids, wall thickness is ignored, and furniture is not subtracted from the room volume.