Math of Space: Surface Area and Volume

Think and Reflect

Q1. Try to work out for yourself why this model explains the formula for the volume of a cuboid, i.e., volume = area of base × height. Write the formula for the surface area and the volume of a cube.

Answer:

Why the card model works:

• Each playing card in the stack is a thin rectangle of length l and width w. Its area is lw — which is exactly the area of the base of the cuboid.
• The stack has h such identical layers piled one on top of the other.
• So the total volume = (area of one layer) + (area of one layer) + … repeated h times = lw × h.
• Hence Volume = area of base × height, and for a cuboid this becomes V = lwh.

Formulas for a cube of side a:

• Total surface area = 6a²
• Volume = a³

Q2. Suppose a swimming pool is being built in your school. Two choices are available for the dimensions of the pool: (A) 8 ft (depth) × 50 ft × 20 ft and (B) 6 ft (depth) × 100 ft × 15 ft. Which option would be suitable/appropriate for your school? Why? What parameters/aspects did you consider to make the choice? [1 cubic foot ≈ 28.3 litres]

Answer:

Conclusion: Option A is more suitable for a school because it needs less land, uses less water, costs less to build and maintain, and is safer for students (deeper pools need trained lifeguards and stricter supervision).

Aspects considered: safety of students, available land in the school, volume of water and its recurring cost, construction and tiling cost, ease of cleaning and maintenance, and the purpose of the pool (splash pool vs. swimming practice).

Exercise Set 14.1

Q1. The volume of a cube is 64 cm³. What is its total surface area?

Answer:

Given: Volume of cube = 64 cm³
To find: Total surface area (TSA)

Solution: a³ = 64 ⇒ a = 4 cm
TSA = 6a² = 6 × 4² = 6 × 16 = 96 cm²

Final answer: TSA = 96 cm²

Q2. How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?

Answer:

Given: Side of box = 2 m = 200 cm; side of small cube = 20 cm

Solution: Along each edge of the box, number of small cubes = 200 ÷ 20 = 10

Total number of cubes = 10 × 10 × 10 = 1000 cubes

Q3. 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:

Given: Godown = 40 m × 25 m × 10 m; Box = 2 m × 1.25 m × 1 m

Solution:
Volume of 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 boxes

Q4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Given: Volume of each cube = 125 cm³

Solution: a³ = 125 ⇒ a = 5 cm

On joining end to end, the cuboid formed has dimensions l = 5 + 5 = 10 cm, w = 5 cm, h = 5 cm.

Surface area = 2(lw + wh + lh) = 2[(10 × 5) + (5 × 5) + (5 × 10)] = 2[50 + 25 + 50] = 2 × 125 = 250 cm²

Q5. 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? (Note that there is no change in volume but a big change in the surface area. This property has major consequences in the biological world.)

Answer:

Solution:
Surface area of original cube = 6 × 4² = 96 cm²

Number of 1 cm cubes = 4 × 4 × 4 = 64
Surface area of one small cube = 6 × 1² = 6 cm²
Total surface area of all small cubes = 64 × 6 = 384 cm²

Ratio = 96 : 384 = 1 : 4

Note: The volume stays the same (64 cm³) but the surface area becomes four times. This is why cells, leaves and lungs are finely divided — to expose a large surface area for exchange of gases and nutrients.

Q6. 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:

Given: lw = 6 cm², wh = 15 cm², lh = 10 cm² (areas of the three faces meeting at a corner)

Solution: Multiplying all three,
(lw) × (wh) × (lh) = 6 × 15 × 10
(lwh)² = 900
lwh = √900 = 30

Final answer: Volume = 30 cm³ (the cuboid is 2 cm × 3 cm × 5 cm)

Q7. 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:

Given: Cube of side 5 cm cut into 1 cm³ cubes ⇒ 5 × 5 × 5 = 125 small cubes

• (i) Exactly three faces painted: these are the corner cubes = 8
• (ii) Exactly two faces painted: cubes along the edges, excluding corners = 12 × (5 − 2) = 12 × 3 = 36
• (iii) Exactly one face painted: cubes on the faces, leaving the border = 6 × (5 − 2)² = 6 × 9 = 54
• (iv) No face painted: cubes hidden inside = (5 − 2)³ = 3³ = 27

Check: 8 + 36 + 54 + 27 = 125 ✔

Q8. 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:

Given: TSA = 2(lw + wh + lh) = 100 ⇒ lw + wh + lh = 50

Let l ≥ w ≥ h (whole numbers). Since 3h² ≤ 50, we get h ≤ 4. So h can be 1, 2, 3 or 4.

• h = 1: lw + w + l = 50 ⇒ (l + 1)(w + 1) = 51 = 17 × 3 ⇒ l = 16, w = 2 → 16 × 2 × 1
• h = 2: lw + 2w + 2l = 50 ⇒ (l + 2)(w + 2) = 54 = 9 × 6 ⇒ l = 7, w = 4 → 7 × 4 × 2
• h = 3: (l + 3)(w + 3) = 59, which is prime and both factors must be ≥ 4 → no solution
• h = 4: (l + 4)(w + 4) = 66; factor pairs (1,66), (2,33), (3,22), (6,11) — none has both numbers ≥ 8 → no solution

(i) Yes, there is more than one.
(ii) The cuboids are 16 cm × 2 cm × 1 cm and 7 cm × 4 cm × 2 cm (and their rearrangements).
(iii) Since h cannot exceed 4 (otherwise 3h² > 50), checking h = 1, 2, 3, 4 covers every possibility, and only two cases gave whole-number solutions. Hence the list is complete.

Exercise Set 14.2

Q1. 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:

Given: Let cylinder A have radius r and height h. Then cylinder B has radius 2r and height h/2.

Curved surface area:
CSA of A = 2πrh; CSA of B = 2π(2r)(h/2) = 2πrh
Ratio = 1 : 1

Volume:
V of A = πr²h; V of B = π(2r)²(h/2) = 2πr²h
Ratio = πr²h : 2πr²h = 1 : 2

Q2. 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.

Answer:

Given: r₁ : r₂ = 2 : 3 and h₁ : h₂ = 3 : 2

(a) Volumes: V₁ : V₂ = r₁²h₁ : r₂²h₂ = (4 × 3) : (9 × 2) = 12 : 18 = 2 : 3

(b) Curved surface areas: CSA₁ : CSA₂ = r₁h₁ : r₂h₂ = (2 × 3) : (3 × 2) = 6 : 6 = 1 : 1

Q3. 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:

Given: Cube edge = r cm

The largest cylinder that fits inside has diameter equal to the edge of the cube, so its radius = r/2 and its height = r.

Volume = π(radius)² × height = π(r/2)² × r = πr³/4 cm³

Q4. 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:

Given: New radius = 1.1r; new height = h(1 − x/100); volume unchanged.

Solution:
πr²h = π(1.1r)² × h(1 − x/100)
1 = 1.21 × (1 − x/100)
1 − x/100 = 1/1.21 = 0.82645
x/100 = 0.17355 ⇒ x ≈ 17.36

Final answer: x ≈ 17.4%

Q5. 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:

Solution:

• (i) Volume of cube = 12³ = 1728 cm³
• (ii) Volume of one rod = πr²h = (22/7) × 2² × 12 = (22/7) × 48 = ≈ 150.86 cm³
• (iii) Number of rods = 1728 ÷ 150.86 ≈ 11.45

Since only complete rods are counted, the number is 11 rods (some metal is left over).

Exercise Set 14.3

Q1. 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: l = 21 m, diameter = 24 m ⇒ r = 12 m

TSA = πr(l + r) = (22/7) × 12 × (21 + 12) = (22/7) × 12 × 33 = 8712/7

Final answer: TSA ≈ 1244.57 m²

Q2. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.

Answer:

CSA = πrl = (22/7) × 7 × 10 = 220 cm²

Q3. 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.

Answer:

Given: h = 16 cm, r = 12 cm

Slant height: l = √(h² + r²) = √(16² + 12²) = √(256 + 144) = √400 = 20 cm

CSA = πrl = (22/7) × 12 × 20 = 5280/7 ≈ 754.29 cm²
TSA = πr(l + r) = (22/7) × 12 × 32 = 8448/7 ≈ 1206.86 cm²

Q4. A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.

Answer:

Given: h = 15 cm, V = 1570 cm³ (take π = 3.14)

V = (1/3)πr²h ⇒ 1570 = (1/3) × 3.14 × r² × 15 = 15.7 r²

r² = 1570 ÷ 15.7 = 100 ⇒ r = 10

Final answer: radius = 10 cm

Q5. 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: CSA = 308 cm², l = 14 cm

• (i) πrl = 308 ⇒ (22/7) × r × 14 = 308 ⇒ 44r = 308 ⇒ r = 7 cm
• (ii) TSA = πr(l + r) = (22/7) × 7 × 21 = 462 cm²

Q6. 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: r = 7 cm, h = 24 cm

l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm

Sheet required for one cap = CSA = πrl = (22/7) × 7 × 25 = 550 cm²

Sheet required for 10 caps = 550 × 10 = 5500 cm²

Q7. 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: h = 8 m, r = 6 m, width of tarpaulin = 3 m, wastage = 20 cm = 0.2 m

Slant height l = √(8² + 6²) = √100 = 10 m

Curved surface area of tent = πrl = (22/7) × 6 × 10 = 1320/7 ≈ 188.57 m²

Length of tarpaulin = Area ÷ width = 188.57 ÷ 3 ≈ 62.86 m

Adding wastage: 62.86 + 0.2 = 63.06 m

Final answer: about 63.1 m of tarpaulin is required.

Q8. 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.

Answer:

Given: 6² + 8² = 36 + 64 = 100 = 10², so it is right-angled with the 10 cm side as hypotenuse.

Rotating about the 8 cm side gives a cone with r = 6 cm, h = 8 cm, l = 10 cm.

Volume = (1/3)πr²h = (1/3) × 3.14 × 36 × 8 ≈ 301.44 cm³

CSA = πrl = 3.14 × 6 × 10 = 188.4 cm²

[Diagram: a right triangle with legs 6 cm and 8 cm and hypotenuse 10 cm, rotated about the 8 cm leg to form a cone of radius 6 cm and height 8 cm]

Q9. 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?

Answer:

The water forms a smaller cone similar to the whole cone, with every length scaled down by the factor 1/2.

Volume of a similar solid scales as the cube of the length ratio:

V_water = (1/2)³ × V_cup = (1/8) V_cup

Final answer: The water occupies 1/8 (one-eighth) of the cup's volume.

[Diagram: a cone with water filled to half its height, showing the small similar cone at the bottom]

Exercise Set 14.4

Q1. A ball bearing has a radius of 0.7 cm. Find its surface area.

Answer:

Surface area = 4πr² = 4 × (22/7) × 0.7 × 0.7 = 4 × 1.54 = 6.16 cm²

Q2. 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:

Given: Since the metal is the same, weights are proportional to volumes.

Ratio of volumes = 5920 : 740 = 8 : 1

Ratio of radii = ∛8 : ∛1 = 2 : 1

Radius of smaller sphere = 5/2 = 2.5 cm

Final answer: Radius of the larger sphere = 2 × 2.5 = 5 cm

Q3. 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:

Given: d_moon : d_earth = 1 : 4 ⇒ r_moon : r_earth = 1 : 4

Surface area = 4πr², so the ratio of surface areas = 1² : 4² = 1 : 16

Q4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.

Answer:

Given: r = 21 cm

• (i) CSA = 2πr² = 2 × (22/7) × 21 × 21 = 2772 cm²
• (ii) TSA = 3πr² = 3 × (22/7) × 21 × 21 = 4158 cm²

Q5. 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:

Given: Box = 16 cm × 8 cm × 8 cm; 16 spheres of radius 2 cm

Volume of box = 16 × 8 × 8 = 1024 cm³

Volume of one sphere = (4/3)πr³ = (4/3) × (22/7) × 8 = 704/21 ≈ 33.52 cm³

Volume of 16 spheres ≈ 16 × 33.52 = 536.38 cm³

Volume of liquid = 1024 − 536.38 = 487.62 cm³

Final answer: ≈ 488 cm³ of preservative liquid

Q6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.

Answer:

New radius = 1.1r

New volume = (4/3)π(1.1r)³ = (4/3)πr³ × 1.331 = 1.331 × original volume

Increase = 1.331 − 1 = 0.331 = 33.1%

Hence the volume increases by about 33.1%.

Q7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.

Answer:

New volume = 1.728 × original volume

So (1 + x/100)³ = 1.728

1 + x/100 = ∛1.728 = 1.2

x/100 = 0.2 ⇒ x = 20

Q8. 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 it ₹10 per 100 cm².

Answer:

Given: Circumference = 35.2 m = 2πr

r = (35.2 × 7)/(2 × 22) = 246.4/44 = 5.6 m

Curved surface area of dome = 2πr² = 2 × (22/7) × 5.6 × 5.6 = 197.12 m²

Convert to cm²: 197.12 m² = 197.12 × 10,000 = 19,71,200 cm²

Cost = (19,71,200 ÷ 100) × ₹10 = 19,712 × 10 = ₹1,97,120

Final answer: The cost of painting the dome is ₹1,97,120.

Activities and Hands-on Work

Q1. Activity (Section 14.2.1): Explain how stacking thin circular khakhras helps us understand the curved surface area and volume of a cylinder.

Answer:

• Take a large number of very thin circular khakhras, each of radius r and thickness t.
• Stack them one above the other to form a cylinder of height h. The number of khakhras is h/t.
• Volume: volume of one khakhra = area of base × thickness = πr²t. Total volume = πr²t × (h/t) = πr²h.
• Curved surface: the rim (edge) of every khakhra contributes a thin strip of length 2πr and height t. Adding all the rims gives 2πr × h = 2πrh.

Q2. Activity (Section 14.3): Verify experimentally that the volume of a cone is one-third the volume of a cylinder of the same base radius and height.

Answer:

What to do:

1. Take a cylindrical tin and a paper cone with the same base radius and the same height.
2. Fill the cone to the brim with fine dry salt (or sand) and empty it into the cylinder.
3. Repeat this two more times.

Observation: The cylinder becomes completely full only after the third pouring.

Conclusion: Volume of cylinder = 3 × volume of cone, so V_cone = (1/3)πr²h.

Alternative (clay activity): Mould modelling clay into a solid cylinder, measure r and h, then reshape the same clay into cones of the same r and h. Exactly three cones can be made — again showing V_cone = (1/3)πr²h.

[Diagram: a cone full of salt being poured into a cylindrical tin; three such pourings fill the tin]

Q3. Activity (Section 14.5.1): Verify the formula for the surface area of a sphere using a string and four circles.

Answer:

What to do:

1. Take a rubber ball and drive a nail into it. Wind a string tightly around the ball, using pins to hold the string at the fullest part, until the whole surface is covered.
2. Mark the starting and finishing points and unwind the string.
3. Measure the diameter of the ball and calculate its radius r.
4. Draw four circles of radius r on a sheet of paper and fill them completely with the same string.

Observation: The string that covered the whole sphere exactly fills the four circles.

Conclusion: Surface area of sphere = 4 × area of a circle of radius r = 4πr².

[Diagram: a ball completely wound with string, and four circles of the same radius filled with the same string]

Q4. Activity (Section 14.6): Find the approximate volumes of a gas cylinder, a bucket, any cylindrical vessel, a cupboard and a water tank that you can find near you (use 1000 cm³ = 1 litre).

Answer:

Method: Measure the dimensions with a tape and use the correct formula for each shape.

Note: Answers will vary from student to student because the objects measured are different — what matters is the correct choice of formula and correct unit conversion.

Q5. Activity (Example 6): For packing tennis balls in a room, make an estimate based on the second and third configurations shown in Fig. 14.17.

Answer:

• Upper bound (no gaps): 33,45,132 balls — impossible in practice.
• Configuration A (simple cubic stacking): 150 × 150 × 75 = 16,87,500 balls.
• Configurations B and C (staggered / close packing): these pack spheres with about 74% efficiency. So the estimate becomes roughly 0.74 × 33,45,132 ≈ 24,75,000 balls (about 25 lakh).

Conclusion: The staggered arrangements fit about 1.5 times as many balls as the simple cubic arrangement, because they leave less empty space.

End-of-Chapter Exercises

Q1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.

Answer:

Guess first: about 50 scoops.

Volume of container = 10 × 15 × 20 = 3000 cm³

Assumption: an average scoop is a sphere of radius about 2.5 cm.

Volume of one scoop = (4/3)πr³ = (4/3) × 3.14 × (2.5)³ ≈ 65.4 cm³

Number of scoops = 3000 ÷ 65.4 ≈ 45.9

Final answer: about 45–46 scoops (if smaller scoops of radius 2 cm are used, the number rises to about 90).

Q2. 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.

Answer:

Unit cubes in the cube of side a = a³
Unit cubes in the cube of side (a + 1) = (a + 1)³

Number to be added = (a + 1)³ − a³ = a³ + 3a² + 3a + 1 − a³

= 3a² + 3a + 1 unit cubes

Q3. 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:

(i) No. Take a classroom of 10 m × 8 m × 4 m = 320 m³ = 3,20,000 litres.
A person drinks about 2.5 litres a day ⇒ about 2.5 × 365 ≈ 912 litres a year.
In 70 years ≈ 64,000 litres ≈ 64 m³.
This is only about one-fifth of the classroom's volume, so a person cannot drink enough water in a lifetime to fill a classroom.

(ii) Assume the classroom walls are 10 m × 8 m with height 4 m and thickness 0.23 m.
Perimeter = 2(10 + 8) = 36 m; wall area = 36 × 4 = 144 m²; subtract about 20 m² for doors and windows ⇒ 124 m².
Volume of brickwork = 124 × 0.23 ≈ 28.5 m³.
A standard brick with mortar occupies about 0.002 m³ (1 m³ needs about 500 bricks).
Number of bricks ≈ 28.5 × 500 ≈ 14,000 bricks.

Q4. 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:

Option A: side = 50 mm = 5 cm ⇒ volume = 5³ = 125 cm³

Option B: side = 10 mm = 1 cm ⇒ volume of one cube = 1 cm³; for 100 cubes = 100 cm³

Final answer: Option A gives more chocolate (125 cm³ > 100 cm³).

Q5. If all the people in the world are crowded together in one location, how much area would it cover?

Answer:

Guess first: a large city.

World population ≈ 8 billion (8 × 10⁹).

Assume a tightly packed crowd gives each person about 0.25 m² (roughly a 0.5 m × 0.5 m square).

Total area = 8 × 10⁹ × 0.25 = 2 × 10⁹ m² = 2000 km²

Final answer: about 2000 km² — roughly the area of a large Indian city such as Delhi (about 1500 km²). If people stand more comfortably (1 m² each), the area becomes about 8000 km².

Q6. 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:

Given: diameter = 7 cm ⇒ r = 3.5 cm, h = 12 cm

Volume of one glass = πr²h = (22/7) × 3.5 × 3.5 × 12 = 462 cm³

Milk for 1600 students = 462 × 1600 = 7,39,200 cm³

In litres = 7,39,200 ÷ 1000 = 739.2 litres

Q7. 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: r_sphere = 5 cm, r_cone = 4 cm

Surface area of sphere = 4π(5)² = 100π cm²

This is 5 times the CSA of the cone ⇒ CSA of cone = 100π ÷ 5 = 20π cm²

πrl = 20π ⇒ 4l = 20 ⇒ l = 5 cm

Height: h = √(l² − r²) = √(25 − 16) = √9 = 3 cm

Volume = (1/3)πr²h = (1/3)π(16)(3) = 16π ≈ 50.3 cm³

Q8. 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.

Answer:

Volume of a sphere ∝ r³, so the ratio of volumes = (ratio of radii)³

(i) (69,900 ÷ 6370)³ = (10.97)³ ≈ 1321

(ii) (6,95,700 ÷ 6370)³ = (109.2)³ ≈ 13,00,000 (about 1.3 million)

Final answer: Jupiter is about 1300 times and the Sun about 13 lakh times the volume of the Earth.

Q9. Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder which encloses it.

Answer:

The smallest cylinder that encloses a sphere of radius r has radius r and height 2r.

Volume of cylinder = πr²(2r) = 2πr³

Volume of sphere = (4/3)πr³

Required fraction = (4/3)πr³ ÷ 2πr³ = (4/3) ÷ 2 = 2/3

Hence the volume of the sphere is 2/3 of the volume of the smallest enclosing cylinder.

[Diagram: a sphere of radius r fitting exactly inside a cylinder of radius r and height 2r]

Q10. 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. How high above the ground is the second string? Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?

Answer:

Let the Earth's radius be R and the height of the second string be h.

First string length = 2πR; second string length = 2π(R + h) = 2πR + 2πh

Given: 2πh = 1 m ⇒ h = 1/(2π) ≈ 0.159 m ≈ 16 cm

The same calculation holds for the Moon, Jupiter and a volleyball — the answer is about 16 cm in every case.

Why this is not really surprising: the extra 1 metre is spread over a circumference, and the increase in radius depends only on the extra length (h = extra length ÷ 2π), not on the size of the sphere. It is a striking but genuine result.

Q11. 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 is inscribed in a circle of radius r, so its diagonal = 2r.

Side of the square = 2r/√2 = r√2, so area of base = (r√2)² = 2r²

Height of the pyramid = h (same as the cylinder)

Volume of pyramid = (1/3) × base area × height = (1/3)(2r²)h = (2/3)r²h

Volume of cylinder = πr²h

Ratio = (2/3)r²h : πr²h = 2 : 3π

Final answer: 2 : 3π (about 1 : 4.7)

Q12. What is the change in volume when: (i) the length of a cuboid with dimensions l, w, h is increased by 1 unit? (ii) the radius of a cylinder with dimensions r, h is increased by 1 unit? (iii) the radius of a sphere is decreased by 1 unit?

Options for (i):

• (a) 1 cubic unit
• (b) (lwh + 1) cubic unit
• (c) wh cubic units
• (d) lw cubic units
• (e) lh cubic units

Options for (ii):

• (a) 1 cubic unit
• (b) πr²h cubic units
• (c) πr² cubic units
• (d) 2πrh + 2πh cubic units
• (e) 2πrh + πh cubic units

Answer:

(i) Correct option: (c) wh cubic units.
Reason: New volume = (l + 1)wh = lwh + wh, so the increase is wh.

(ii) Correct option: (e) 2πrh + πh cubic units.
Reason: New volume = π(r + 1)²h = πr²h + 2πrh + πh, so the increase is 2πrh + πh = πh(2r + 1).

(iii) Decrease = 4πr² − 4πr + (4/3)π cubic units.
Reason: New volume = (4/3)π(r − 1)³. Change = (4/3)π[r³ − (r − 1)³] = (4/3)π(3r² − 3r + 1) = 4πr² − 4πr + (4/3)π.

Q13. Given a cube with volume V, express its total surface area S in terms of V.

Answer:

V = a³ ⇒ a = V^1/3

S = 6a² = 6(V^1/3)²

S = 6V^2/3

Q14. Given a cube with total surface area S, express its volume V in terms of S.

Answer:

S = 6a² ⇒ a² = S/6 ⇒ a = √(S/6)

V = a³ = (S/6)^3/2

V = (S/6)^3/2 cubic units

Q15. 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?

Answer:

Given: Glass radius r = 4 cm; water must rise from 16 cm to 20 cm, i.e., by 4 cm.

Volume of water that must be displaced = πr² × rise = π × 4² × 4 = 64π cm³

Volume of one marble of radius 1 cm = (4/3)π(1)³ = (4/3)π cm³

Number of marbles = 64π ÷ (4π/3) = 64 × 3/4 = 48

Final answer: 48 marbles

[Diagram: a cylindrical glass with water at 16 cm and a heap of 1 cm marbles beside it]

Q16. Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.

Answer:

Measurements (sample): length of the ink column in the refill ≈ 12 cm, inner diameter of the refill ≈ 1.5 mm ⇒ radius ≈ 0.075 cm.

Volume = πr²h = 3.14 × (0.075)² × 12 ≈ 3.14 × 0.005625 × 12 ≈ 0.21 cm³

Final answer: about 0.2 cm³ (≈ 0.2 mL) of ink.

Note: Different pens will give slightly different values; what matters is the correct method and sensible approximations.

Q17. 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.

Answer:

(i) Volume of dough ball = (4/3)πr³ = (4/3) × 3.14 × 216 ≈ 904.8 cm³

Assume each chapati is a disc of radius 8 cm and thickness 0.2 cm (2 mm).

Volume of one chapati = πr²t = 3.14 × 64 × 0.2 ≈ 40.2 cm³

Number of chapatis = 904.8 ÷ 40.2 ≈ 22.5

So about 22 chapatis can be made.

(ii) Sample answer for a muskmelon: outer radius = 6 cm, inner (seed cavity) radius = 4.5 cm.

Volume of edible flesh = (4/3)π(R³ − r³) = (4/3) × 3.14 × (216 − 91.125) = (4/3) × 3.14 × 124.875 ≈ 523 cm³

[Diagram: a sphere of radius 6 cm with an inner empty sphere of radius 4.5 cm, showing the edible shell in between]

Q18. 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) One page ≈ 28 cm × 22 cm = 616 cm². With about 200 pages, total area ≈ 200 × 616 = 1,23,200 cm² ≈ 12.3 m².

A classroom floor is about 10 m × 8 m = 80 m².

No, one textbook's pages would cover only about one-seventh of the floor.

(ii) Volume of storeroom = 15 × 20 × 30 = 9000 ft³

Using 1 ft³ ≈ 28317 cm³, volume ≈ 9000 × 28317 ≈ 25,48,53,000 cm³

Volume of one textbook ≈ 28 × 22 × 1.5 = 924 cm³

Maximum number = 25,48,53,000 ÷ 924 ≈ 2,75,800

Final answer: about 2.7 lakh textbooks (an upper bound; in practice fewer, because gaps are left between books).

Q19. If the entire human population decided to climb into one giant cube, how long would the side have to be?

Answer:

World population ≈ 8 billion = 8 × 10⁹ people.

An average person's body volume is about 0.07 m³ (70 litres), and with some empty space between people take about 0.1 m³ each.

Total volume = 8 × 10⁹ × 0.1 = 8 × 10⁸ m³

Side of the cube = ∛(8 × 10⁸) = 928 m

Final answer: a cube of side about 900 m to 1 km — smaller than most people expect!

[Diagram: a huge cube drawn to scale next to a tall building to show how big a 1 km cube would be]

Q20. 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.

Answer:

Surface area of the Earth = 4πr² = 4 × 3.14 × (6371)² ≈ 5.10 × 10⁸ km²

Volume of water = 1.38 × 10⁹ km³

Thickness = Volume ÷ Surface area = (1.38 × 10⁹) ÷ (5.10 × 10⁸) ≈ 2.7 km

Final answer: the water layer would be about 2.7 km thick.

Q21. Give the dimension of a cuboid whose volume is halved when its surface area is doubled.

Answer:

Changing only the size (scaling) cannot do this, because scaling by k multiplies the volume by k³ but the surface area only by k². So we must change the shape.

Example: A cube of side 1 cm has V = 1 cm³ and SA = 6 cm².

Take a flat cuboid of dimensions about 2.36 cm × 2.36 cm × 0.09 cm:

• V = 2.36 × 2.36 × 0.09 ≈ 0.5 cm³ (half of 1 cm³)
• SA = 2[(2.36 × 2.36) + (2.36 × 0.09) + (0.09 × 2.36)] ≈ 2(5.57 + 0.42) ≈ 12 cm² (double of 6 cm²)

Hence the required cuboid is roughly 2.4 cm × 2.4 cm × 0.1 cm.

Idea: a very flat, spread-out cuboid has a large surface area but a small volume.

Q22. Project: Find the volume of your house making necessary approximations. Present how you solved it.

Answer:

Sample project plan:

1. Divide the house into simple solids: each room as a cuboid, a water tank as a cylinder, a dome or sloped roof as a hemisphere or prism.
2. Measure the length, breadth and height of every room with a tape (in metres).
3. Compute each part: Volume of a room = l × b × h. For a cylindrical tank, V = πr²h. For a hemispherical dome, V = (2/3)πr³.
4. Subtract nothing for walls — we are finding the volume of space enclosed, so we simply add the volumes of all the parts.
5. Add up all the parts to get the total volume in m³. Convert to litres if needed (1 m³ = 1000 litres).

Sample result: 4 rooms × (4 m × 3.5 m × 3 m) = 4 × 42 = 168 m³, plus a hall of 6 m × 5 m × 3 m = 90 m³ ⇒ total ≈ 258 m³.

Presentation: Draw a rough plan of the house, label the measurements, show the formula used for each part and present a table of volumes with the final total.

[Diagram: a rough floor plan of a house divided into labelled cuboids with their measurements]