NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space: Surface Area and Volume
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 – Math of Space: Surface Area and Volume help students understand how to calculate the surface area and volume of different three-dimensional shapes. The chapter covers important concepts related to cubes, cuboids, cylinders, cones, spheres and other solids, along with their practical applications.
The step-by-step solutions make it easier for students to apply the correct formulas, solve numerical problems accurately and strengthen their understanding of mensuration. Students can also refer to the NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space: Surface Area and Volume PDF for convenient practice, revision and exam preparation.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.1
1. The volume of a cube is 64 cm³. What is its total surface area?
Answer:
For a cube,
Volume = a³
Given:
a³ = 64
So,
a = 4 cm
Total surface area of a cube:
TSA = 6a²
= 6 × 4²
= 6 × 16
= 96 cm²
Therefore, the total surface area of the cube is 96 cm².
2. How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Answer:
Side of the cubical box:
2 m = 200 cm
Side of each small cube = 20 cm
Number of cubes along one edge:
200 ÷ 20 = 10
Therefore, total number of small cubes:
10 × 10 × 10 = 1000
Hence, 1000 small cubes can be packed tightly in the cubical box.
3. The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.
Answer:
Volume of the godown:
40 × 25 × 10 = 10,000 m³
Volume of one box:
2 × 1.25 × 1 = 2.5 m³
Number of boxes:
10,000 ÷ 2.5 = 4000
Therefore, the godown can hold 4000 boxes.
4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Side of each cube = ∛125 = 5 cm. The cuboid is 10 cm × 5 cm × 5 cm.
TSA = 2(10 × 5 + 5 × 5 + 10 × 5) = 2(50 + 25 + 50) = 250 cm².
(Check: two cubes have 2 × 150 = 300 cm², minus the two joined faces, 2 × 25 = 50 cm².)
5. A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes?
Answer:
Original cube: 6 × 4² = 96 cm².
Number of 1 cm cubes = 4³ = 64; total surface area = 64 × 6 = 384 cm².
Ratio = 96 : 384 = 1 : 4.
6. The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Answer:
Let the edges be
,
,
, with
,
,
. Multiplying the three:
∴ The volume is 30 cm³. (The edges are 2 cm, 3 cm and 5 cm.)
7. A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
Answer:
A cube of side 5 cm is divided into 1 cm³ cubes.
So, the number of small cubes along each edge is:
5 ÷ 1 = 5
(i) Exactly three faces painted
These are the cubes at the 8 corners.
Number of cubes = 8
(ii) Exactly two faces painted
These lie along the edges, excluding the corner cubes.
Each edge has:
5 − 2 = 3 cubes
There are 12 edges.
Number of cubes = 12 × 3 = 36
(iii) Exactly one face painted
These lie in the interior of each face, excluding edges.
On each face:
(5 − 2)² = 3² = 9 cubes
There are 6 faces.
Number of cubes = 6 × 9 = 54
(iv) No face painted
These are completely inside the cube.
Number of cubes = (5 − 2)³ = 3³ = 27
Therefore:
(i) 8
(ii) 36
(iii) 54
(iv) 27
8. Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
Answer:
Let the integer edge lengths of the cuboid be a, b, c cm.
Total surface area of a cuboid is:
2(ab + bc + ca) = 100
So,
ab + bc + ca = 50 ...(1)
Assume a ≤ b ≤ c.
(i) Is there more than one such cuboid?
Yes. There are two different cuboids with integer edge lengths.
(ii) Find all such cuboids
Since a ≤ b ≤ c,
ab + ac + bc ≥ 3a²
Using (1),
3a² ≤ 50
So,
a ≤ 4
Therefore, we only need to check a = 1, 2, 3, 4.
For a = 1:
b + c + bc = 50
Adding 1 to both sides in factorised form:
(b + 1)(c + 1) = 51
Since 51 = 3 × 17,
b + 1 = 3, c + 1 = 17
So,
b = 2, c = 16
Hence, one cuboid is:
1 cm × 2 cm × 16 cm
For a = 2:
2b + 2c + bc = 50
Adding 4:
(b + 2)(c + 2) = 54
Using 54 = 6 × 9,
b + 2 = 6, c + 2 = 9
So,
b = 4, c = 7
Hence, another cuboid is:
2 cm × 4 cm × 7 cm
For a = 3:
bc + 3b + 3c = 50
So,
(b + 3)(c + 3) = 59
Since 59 is prime, there is no valid integer solution.
For a = 4:
bc + 4b + 4c = 50
So,
(b + 4)(c + 4) = 66
There is no factor pair of 66 satisfying 4 ≤ b ≤ c, so there is no valid solution.
(iii) Show that all have been found
We proved that a ≤ 4, so every possible cuboid must occur among the four cases above. Only two cases give positive integer dimensions.
Therefore, all possible cuboids are:
1 cm × 2 cm × 16 cm
and
2 cm × 4 cm × 7 cm
Checking:
For 1 × 2 × 16:
TSA = 2(2 + 16 + 32) = 2 × 50 = 100 cm²
For 2 × 4 × 7:
TSA = 2(8 + 14 + 28) = 2 × 50 = 100 cm²
Hence, there are exactly two such cuboids.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.2
1. Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.
Answer:
Let the radius and height of cylinder A be r and h.
Then for cylinder B:
Radius = 2r
Height = h/2
Curved surface area
Curved surface area of a cylinder:
CSA = 2πrh
For cylinder A:
CSAₐ = 2πrh
For cylinder B:
CSAᵦ = 2π(2r)(h/2) = 2πrh
Therefore,
CSA of A : CSA of B = 1 : 1
Volume
Volume of a cylinder:
V = πr²h
For cylinder A:
Vₐ = πr²h
For cylinder B:
Vᵦ = π(2r)²(h/2)
= 2πr²h
Therefore,
Volume of A : Volume of B = 1 : 2
Hence:
Curved surface area ratio = 1 : 1
Volume ratio = 1 : 2
2. The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas.
Let the radii of the two cylinders be 2r and 3r, and their heights be 3h and 2h.
(a) Ratio of their volumes
Volume of a cylinder:
V = πr²h
So,
V₁ : V₂ = π(2r)²(3h) : π(3r)²(2h)
= 4 × 3 : 9 × 2
= 12 : 18
= 2 : 3
Therefore, the ratio of their volumes is:
2 : 3
(b) Ratio of their curved surface areas
Curved surface area of a cylinder:
CSA = 2πrh
So,
CSA₁ : CSA₂ = 2π(2r)(3h) : 2π(3r)(2h)
= 6 : 6
= 1 : 1
Therefore:
(a) Volume ratio = 2 : 3
(b) Curved surface area ratio = 1 : 1
3. The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?
Answer:
The largest possible right circular cylinder cut from a cube of edge r cm will have:
Height = r cm
and its circular base will be inscribed in a square of side r cm.
So, the diameter of the cylinder is r, hence:
Radius = r/2
Volume of a cylinder:
V = πR²h
Substituting:
V = π(r/2)²(r)
= πr³/4
Therefore, the volume of the cylinder is:
πr³/4 cm³
4. The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.
Answer:
Let the original radius and height of the cylinder be r and h.
Original volume:
V = πr²h
The radius is increased by 10%, so the new radius is:
1.1r
The height is decreased by x%, so the new height is:
h(1 − x/100)
Since the volume remains unchanged:
π(1.1r)²h(1 − x/100) = πr²h
Cancelling πr²h:
1.21(1 − x/100) = 1
So,
1 − x/100 = 1/1.21 = 100/121
Therefore,
x/100 = 1 − 100/121 = 21/121
x = 2100/121 ≈ 17.36
Hence, the height must be decreased by approximately 17.36%
5. A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find:
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (π ≈ 22/7)
Answer:
Given:
- Side of cube = 12 cm
- Radius of each cylindrical rod = 2 cm
- Height of each rod = 12 cm
- π ≈ 22/7
(i) Volume of the cube
Volume = side³
= 12³
= 1728 cm³
Therefore, the volume of the cube is 1728 cm³.
(ii) Volume of one cylindrical rod
Volume of a cylinder:
V = πr²h
= 22/7 × 2² × 12
= 22/7 × 48
= 1056/7
≈ 150.86 cm³
Therefore, the volume of one cylindrical rod is approximately 150.86 cm³.
(iii) Number of complete cylindrical rods
Number of rods = Volume of cube ÷ Volume of one rod
= 1728 ÷ 150.86
≈ 11.45
Only complete rods can be formed.
Therefore, approximately 11 complete cylindrical rods can be made.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.3
1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
Given:
- Slant height, l = 21 m
- Diameter = 24 m, so radius r = 12 m
Total surface area of a cone:
TSA = πr(r + l)
Substituting the values:
TSA = π × 12 × (12 + 21)
= 12π × 33
= 396π m²
Using π ≈ 22/7:
TSA ≈ 1244.57 m²
Therefore, the total surface area of the cone is 396π m² or approximately 1244.57 m².
2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Answer:
Given:
- Slant height, l = 10 cm
- Radius, r = 7 cm
Curved surface area of a cone:
CSA = πrl
Substituting the values:
CSA = π × 7 × 10
= 70π cm²
Using π = 22/7:
CSA = 70 × 22/7 = 220 cm²
Therefore, the curved surface area of the cone is 220 cm².
3. 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.
Given:
- Height, h = 16 cm
- Radius, r = 12 cm
First, find the slant height:
l = √(r² + h²)
= √(12² + 16²)
= √(144 + 256)
= √400
= 20 cm
Curved Surface Area
CSA = πrl
= π × 12 × 20
= 240π cm²
Using π = 22/7:
CSA ≈ 754.29 cm²
Total Surface Area
TSA = πr(l + r)
= π × 12 × (20 + 12)
= 384π cm²
Using π = 22/7:
TSA ≈ 1206.86 cm²
Therefore:
- Curved surface area = 240π cm² ≈ 754.29 cm²
- Total surface area = 384π cm² ≈ 1206.86 cm².
4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
Answer:
Given:
- Height, h = 15 cm
- Volume, V = 1570 cm³
Volume of a cone:
V = (1/3)πr²h
Substituting the values:
1570 = (1/3) × π × r² × 15
1570 = 5πr²
Using π = 3.14:
1570 = 15.7r²
r² = 100
r = 10 cm
Therefore, the radius of the base is 10 cm
5. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
Answer:
Given:
- Curved surface area, CSA = 308 cm²
- Slant height, l = 14 cm
(i) Radius of the base
Curved surface area of a cone:
CSA = πrl
Substituting the values:
308 = (22/7) × r × 14
308 = 44r
r = 7 cm
Therefore, the radius of the base is 7 cm.
(ii) Total surface area of the cone
TSA = πr(l + r)
= (22/7) × 7 × (14 + 7)
= 22 × 21
= 462 cm²
Therefore:
(i) Radius = 7 cm
(ii) Total surface area = 462 cm²
6. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:
Given:
- Radius, r = 7 cm
- Height, h = 24 cm
First, find the slant height:
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625
= 25 cm
Since a joker’s cap is open at the base, only the curved surface area is required.
Area of one cap:
CSA = πrl
= (22/7) × 7 × 25
= 550 cm²
For 10 caps:
10 × 550 = 5500 cm²
Therefore, the area of sheet required to make 10 caps is 5500 cm².
7. What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
Answer:
Given:
- Height of tent, h = 8 m
- Base radius, r = 6 m
- Width of tarpaulin = 3 m
First, find the slant height:
l = √(r² + h²)
= √(6² + 8²)
= √(36 + 64)
= 10 m
Since the tent is open at the bottom, only the curved surface area is required.
CSA = πrl
= π × 6 × 10
= 60π m²
Using π = 22/7:
CSA = 60 × 22/7 = 1320/7 m²
Let the required length of tarpaulin be L.
3 × L = 1320/7
L = 440/7 ≈ 62.86 m
Extra material required = 20 cm = 0.20 m
Therefore,
Total length = 62.86 + 0.20 = 63.06 m
Hence, approximately 63.06 m of tarpaulin is required.
8. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
When the right triangle with sides 6 cm, 8 cm and 10 cm is rotated about the side of 8 cm, it forms a right circular cone.
For the cone:
- Height, h = 8 cm
- Radius, r = 6 cm
- Slant height, l = 10 cm
Volume of the cone
V = (1/3)πr²h
= (1/3) × π × 6² × 8
= 96π cm³
Using π = 22/7:
V ≈ 301.71 cm³
Curved surface area
CSA = πrl
= π × 6 × 10
= 60π cm²
Using π = 22/7:
CSA ≈ 188.57 cm²
Therefore:
Volume = 96π cm³ ≈ 301.71 cm³
Curved surface area = 60π cm² ≈ 188.57 cm²
9. Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
Let the full cone have height h and radius r.
If the water is filled to half the depth, then the water forms a smaller similar cone with:
Height = h/2
Since the cones are similar:
Radius = r/2
Volume of full cup:
V = (1/3)πr²h
Volume of water:
V₁ = (1/3)π(r/2)²(h/2)
= (1/3)πr²h × 1/8
So,
V₁ = V/8
Therefore, the water occupies:
1/8 of the volume of the cup.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.4
1. A ball bearing has a radius of 0.7 cm. Find its surface area.
Given:
Radius, r = 0.7 cm
Surface area of a sphere:
SA = 4πr²
Substituting the value:
SA = 4 × 22/7 × (0.7)²
= 4 × 22/7 × 0.49
= 6.16 cm²
Therefore, the surface area of the ball bearing is 6.16 cm².
2. Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Answer:
Since both spheres are made of the same metal, their weights are proportional to their volumes.
So,
Volume ratio = Weight ratio
= 5920 : 740 = 8 : 1
For spheres:
Volume ∝ r³
Therefore,
R³ : r³ = 8 : 1
So,
R : r = 2 : 1
Diameter of the smaller sphere = 5 cm
Hence, its radius:
r = 2.5 cm
Therefore,
R = 2 × 2.5 = 5 cm
Hence, the radius of the larger sphere is 5 cm.
3. The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.
Answer:
The diameter of the moon is 1/4 the diameter of the earth.
Therefore, their radii are also in the ratio:
Moon : Earth = 1 : 4
Surface area of a sphere is proportional to the square of its radius:
Surface area ∝ r²
So,
Surface area of moon : Surface area of earth
= 1² : 4²
= 1 : 16
Therefore, the ratio of their surface areas is 1 : 16.
4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Answer:
Given:
Radius, r = 21 cm
(i) Curved surface area of the hemisphere
CSA = 2πr²
= 2 × 22/7 × 21²
= 2772 cm²
(ii) Total surface area of the hemisphere
TSA = 3πr²
= 3 × 22/7 × 21²
= 4158 cm²
Therefore:
(i) Curved surface area = 2772 cm²
(ii) Total surface area = 4158 cm²
5. Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
Answer:
Volume of the rectangular box:
V = 16 × 8 × 8 = 1024 cm³
Radius of each sphere:
r = 2 cm
Volume of one sphere:
V = (4/3)πr³
= (4/3) × π × 2³
= 32π/3 cm³
Volume of 16 spheres:
16 × 32π/3 = 512π/3
Using π ≈ 3.14:
Volume of 16 spheres ≈ 535.89 cm³
Therefore, volume of preservative liquid:
= 1024 − 535.89
≈ 488.11 cm³
Rounded to the nearest integer:
488 cm³
Hence, the volume of the preservative liquid is approximately 488 cm³.
6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Answer:
Let the original radius of the sphere be r.
Original volume:
V = (4/3)πr³
If the radius is increased by 10%, the new radius is:
1.1r
New volume:
V' = (4/3)π(1.1r)³
= (4/3)π × 1.331r³
= 1.331V
Increase in volume:
V' − V = 1.331V − V = 0.331V
Percentage increase:
0.331 × 100 = 33.1%
Therefore, the volume increases by approximately 33.1%.
7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Answer:
Let the original radius of the sphere be r.
Original volume:
V = (4/3)πr³
If the radius is increased by x%, the new radius is:
r(1 + x/100)
Given that the volume increases by 72.8%, the new volume is:
1.728V
Therefore,
(1 + x/100)³ = 1.728
Taking cube roots:
1 + x/100 = 1.2
So,
x/100 = 0.2
x = 20
Therefore, the radius is increased by 20%.
8. The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹ 10 per 100 cm².
Answer:
Given:
- Circumference of the base of the hemispherical dome = 35.2 m
- Painting cost = ₹10 per 100 cm²
First, find the radius of the dome.
Circumference = 2πr
35.2 = 2 × 22/7 × r
So,
r = 5.6 m
The area to be painted is the curved surface area of the hemisphere:
CSA = 2πr²
= 2 × 22/7 × (5.6)²
= 197.12 m²
Convert into square centimetres:
197.12 m² = 197.12 × 10,000 = 19,71,200 cm²
Cost of painting 100 cm² = ₹10.
Therefore,
Cost = (19,71,200 / 100) × 10
= ₹1,97,120
Hence, the cost of painting the hemispherical dome is ₹1,97,120.
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.