NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume End of Chapter Exercises

End-of-Chapter Exercises

Question 1. Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
Answer:
Guess: about 50 scoops.

Ice cream in the container = 10 × 15 × 20 = 3000 cm³
Assumption: one scoop is roughly a sphere of diameter 5 cm (radius 2.5 cm).
Volume of one scoop = \( \frac{4}{3} \) πr³ ≈ \( \frac{4}{3} \) × 3.14 × 2.5³ = \( \frac{4}{3} \) × 3.14 × 15.625 ≈ 65.4 cm³
Number of scoops ≈ 3000 ÷ 65.4 ≈ 45.9
About 45 full scoops can be obtained. (With a smaller scoop of radius 2 cm, each holds about 33.5 cm³ and we get about 89 scoops; the estimate depends on the scoop size assumed.)

Question 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.
Answer:
A cube of side a has a³ unit cubes and a cube of side a + 1 has (a + 1)³.
Number to be added = (a + 1)³ − a³
= (a² + 2a + 1)(a + 1) − a³
= a³ + 3a² + 3a + 1 − a³
= 3a² + 3a + 1

Seeing it in the figure: to grow the cube by 1 in every direction we add 3 slabs a × a × 1 on three faces (3a² cubes), 3 rods a × 1 × 1 along the edges where two slabs meet (3a cubes) and 1 corner cube.
Example: for a = 2, 3(4) + 3(2) + 1 = 19, and indeed 27 − 8 = 19.
Number of unit cubes to be added = 3a² + 3a + 1.

Teacher's Note:
1. The algebra (a + 1)³ − a³ and the picture of slabs, rods and a corner give the same answer; showing both builds understanding.
2. Test the formula with a small value such as a = 1 (7 cubes) or a = 2 (19 cubes).

Question 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:
(i) Guess: probably not.
Assumptions: classroom 8 m × 6 m × 3.5 m; a person drinks about 2.5 litres of water a day for about 80 years.
Volume of the classroom = 8 × 6 × 3.5 = 168 m³ = 168 × 1000 = 1,68,000 litres (1 m³ = 1000 litres).
Water drunk in a lifetime ≈ 2.5 × 365 × 80 = 73,000 litres
73000 ÷ 168000 ≈ 0.43
No. A lifetime's drinking water (about 73,000 litres) fills less than half (about 43%) of such a classroom; even 5 litres a day would give only about 1,46,000 litres.
(ii) Guess: several thousand bricks.
Assumptions: walls of an 8 m × 6 m room, 3.5 m high and 20 cm thick; one door 1 m × 2.1 m and three windows 1.5 m × 1.2 m; one brick with its share of mortar takes up 20 cm × 10 cm × 10 cm.
Area of the four walls = 2(8 + 6) × 3.5 = 98 m²
Door and windows = 1 × 2.1 + 3 × (1.5 × 1.2) = 2.1 + 5.4 = 7.5 m²
Brickwork area = 98 − 7.5 = 90.5 m²; brickwork volume = 90.5 × 0.2 = 18.1 m³
One brick with mortar = 0.2 × 0.1 × 0.1 = 0.002 m³
Number of bricks ≈ 18.1 ÷ 0.002 = 9050
About 9,000 bricks (slightly fewer, because the corners are shared by two walls).

Question 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:

Option A: 50³ = 50 × 50 × 50 = 1,25,000 mm³
Option B: 100 × 10³ = 100 × 1000 = 1,00,000 mm³
Since 125000 > 100000, option A has more chocolate. In fact, one 50 mm cube equals \( \frac{125000}{1000} \) = 125 small cubes of side 10 mm.
Option (A), the single 50 mm cube, gives more chocolate — 25% more.

Question 5. If all the people in the world are crowded together in one location, how much area would it cover?
Answer:
Guess: a few thousand square kilometres.
Assumptions: world population about 8 billion = 8 × 10⁹ people; in a tightly packed standing crowd each person needs about 0.5 m × 0.5 m = 0.25 m².
Area ≈ 8 × 10⁹ × 0.25 m² = 2 × 10⁹ m²
Since 1 km² = 1000 × 1000 = 10⁶ m², the area is \( \frac{2 \times 10^9}{10^6} \) = 2000 km².
This is a square of side about √2000 ≈ 45 km.
About 2000 km² — roughly 1.35 times the area of Delhi (about 1,480 km²). All of humanity could stand in a square about 45 km on each side.

Question 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:

Radius of a glass r = 7 ÷ 2 = 3.5 cm; height of milk h = 12 cm.
Milk in one glass = πr²h = \( \frac{22}{7} \) × 3.5 × 3.5 × 12 = 22 × 0.5 × 3.5 × 12 = 462 cm³
Milk for 1600 students = 1600 × 462 = 7,39,200 cm³
In litres: 739200 ÷ 1000 = 739.2 litres
739.2 litres of milk are needed.

Question 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:

Surface area of the sphere = 4πr² = 4π × 5² = 100π cm²
Let the slant height of the cone be l; its curved surface area is π × 4 × l = 4πl.
100π = 5 × 4πl = 20πl, so l = 5 cm.
Height: h = √(l² − r²) = √(5² − 4²) = √(25 − 16) = √9 = 3 cm
Volume: V = \( \frac{1}{3} \) πr²h = \( \frac{1}{3} \) × π × 4² × 3 = 16π ≈ 16 × \( \frac{22}{7} \) ≈ 50.29 cm³
Height of the cone = 3 cm; volume = 16π ≈ 50.29 cm³.

Teacher's Note:
1. The given relation leads to the slant height first; the height then comes from l² = h² + r².
2. Keeping π as a symbol until the last step makes the equation 100π = 20πl easy to solve.

Question 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 two spheres of radii R₁ and R₂, \( \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3} = \left(\frac{R_1}{R_2}\right)^3 \).
(i) \( \frac{V_J}{V_E} = \left(\frac{69900}{6370}\right)^3 \approx (10.973)^3 \approx 1321 \)
So Jupiter : Earth ≈ 1321 : 1 — about 1300 Earths would fit inside Jupiter.
(ii) \( \frac{V_S}{V_E} = \left(\frac{695700}{6370}\right)^3 \approx (109.22)^3 \approx 1302700 \)
So Sun : Earth ≈ 13,02,700 : 1, i.e. about 13 lakh Earths would fit inside the Sun.

Question 9. Show that the volume of a sphere is equal to \( \frac{2}{3} \) of the volume of the smallest cylinder which encloses it.
Answer:

Let the sphere have radius r. The smallest cylinder that encloses it touches it all round and at the top and bottom, so
radius of the cylinder = r, height of the cylinder = 2r (the diameter of the sphere).
Volume of the cylinder = πr² × 2r = 2πr³
Volume of the sphere = \( \frac{4}{3} \) πr³
\( \frac{\text{Volume of sphere}}{\text{Volume of cylinder}} = \frac{\frac{4}{3}\pi r^3}{2\pi r^3} = \frac{4}{3} \times \frac{1}{2} = \frac{2}{3} \)
Hence the volume of the sphere is \( \frac{2}{3} \) of the volume of the smallest enclosing cylinder.

Question 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. 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 metres and let the second string be x metres above the ground all round. Then the second string is a circle of radius R + x.
(length of second string) − (length of first string) = 1
2π(R + x) − 2πR = 1
2πx = 1, so x = \( \frac{1}{2\pi} \) ≈ \( \frac{1}{6.283} \) ≈ 0.159 m
So the second string is about 15.9 cm above the ground.
R has cancelled out, so the answer does not depend on the size of the sphere:
(i) Moon: ≈ 15.9 cm; (ii) Jupiter: ≈ 15.9 cm; (iii) volleyball: ≈ 15.9 cm.
Most people are surprised: they expect 1 extra metre spread round a huge planet to lift the string by a tiny amount, but round a small volleyball by a lot. However, circumference = 2π × radius, so adding 1 m to the length always adds exactly \( \frac{1}{2\pi} \) m to the radius, whatever the original radius.

Teacher's Note:
1. Ask students to guess first; almost everyone guesses a tiny gap for the Earth.
2. The key point is that the change in circumference depends only on the change in radius, not on the radius itself.

Question 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 corners A, B, C, D of the square base lie on the circle of radius r, so the diagonal AC of the square is a diameter: AC = 2r.
If each side of the square is s, then in right triangle ABC, s² + s² = (2r)², so 2s² = 4r² and s² = 2r².
So the base of the pyramid has area 2r² cm². The apex is on the top of the cylinder, so the pyramid's height is h.
Volume of pyramid = \( \frac{1}{3} \) × area of base × height = \( \frac{1}{3} \) × 2r² × h = \( \frac{2r^2h}{3} \)
Volume of cylinder = πr²h
\( \frac{\text{Volume of pyramid}}{\text{Volume of cylinder}} = \frac{\frac{2r^2h}{3}}{\pi r^2 h} = \frac{2}{3\pi} \)
The ratio is 2 : 3π, about 0.21 — the pyramid takes up roughly 21% of the cylinder.

Teacher's Note:
1. The key observation is that the diagonal of the square is a diameter of the circle.
2. Both r and h cancel, so the ratio is the same for every such cylinder.

Question 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) Change = (l + 1)wh − lwh = lwh + wh − lwh = wh
The volume grows by wh cubic units — one extra slab of thickness 1 on the face of area wh. Correct option: (c).
(ii) Change = π(r + 1)²h − πr²h = πh[(r + 1)² − r²] = πh(r² + 2r + 1 − r²) = πh(2r + 1) = 2πrh + πh
The volume grows by (2πrh + πh) cubic units. Correct option: (e).
(iii) Using (r − 1)³ = r³ − 3r² + 3r − 1,
Change = \( \frac{4}{3} \) πr³ − \( \frac{4}{3} \) π(r − 1)³ = \( \frac{4}{3} \) π[r³ − (r³ − 3r² + 3r − 1)] = \( \frac{4}{3} \) π(3r² − 3r + 1)
= 4πr² − 4πr + \( \frac{4}{3} \) π
The volume decreases by \( \frac{4}{3} \) π(3r² − 3r + 1) = (4πr² − 4πr + \( \frac{4}{3} \) π) cubic units.

Teacher's Note:
1. Option (d) in part (ii) is a common trap: (r + 1)² − r² is 2r + 1, not 2r + 2.
2. Write 'new volume − old volume' and expand; do not guess from the shape.

Question 13. Given a cube with volume V, express its total surface area S in terms of V.
Answer:
Let the side of the cube be a. Then V = a³, so a = ∛V = \( V^{\frac{1}{3}} \).
S = 6a² = 6\( \left(V^{\frac{1}{3}}\right)^2 \) = 6\( V^{\frac{2}{3}} \)
S = \( 6V^{\frac{2}{3}} = 6\sqrt[3]{V^2} \).
Check: for V = 8, a = 2 and S = 24; also 6 × ∛64 = 6 × 4 = 24 ✓

Question 14. Given a cube with total surface area S, express its volume V in terms of S.
Answer:
Let the side of the cube be a. Then S = 6a², so a² = \( \frac{S}{6} \) and a = \( \sqrt{\frac{S}{6}} \).
V = a³ = \( \left(\sqrt{\frac{S}{6}}\right)^3 = \left(\frac{S}{6}\right)^{\frac{3}{2}} \)
V = \( \left(\frac{S}{6}\right)^{\frac{3}{2}} = \frac{S}{6}\sqrt{\frac{S}{6}} = \frac{S\sqrt{S}}{6\sqrt{6}} \).
Check: for S = 24, S ÷ 6 = 4, a = 2 and V = 4 × 2 = 8 ✓

Question 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?
Answer:

The marbles sink and push the water up; the volume of the marbles equals the volume by which the water rises.
Rise needed = 20 − 16 = 4 cm (the glass is 25 cm tall, so nothing overflows).
Volume of the rise = πr²h = π × 4² × 4 = 64π cm³
Volume of one marble = \( \frac{4}{3} \) π × 1³ = \( \frac{4}{3} \) π cm³
Number of marbles = \( \frac{64\pi}{\frac{4}{3}\pi} = 64 \times \frac{3}{4} = 48 \)
The crow should drop 48 marbles.

Teacher's Note:
1. Only the rise of 4 cm matters, not the full 20 cm of water.
2. Keeping π as a symbol lets it cancel, so no approximation is needed.

Question 16. Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
Answer:
Guess: less than one millilitre.

Sample measurements: the ink is in the refill, a thin cylindrical tube. In a new pen the ink column is about 9 cm long and the inner diameter of the tube is about 2 mm, so r = 1 mm = 0.1 cm.
V = πr²h ≈ 3.14 × 0.1² × 9 = 3.14 × 0.01 × 9 ≈ 0.28 cm³
Since 1 cm³ = 1 millilitre, this is about 0.28 mL.
A new ball point pen holds roughly 0.3 cm³ (about a quarter of a millilitre) of ink. Your answer will depend on the pen you measure.

Question 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.
Answer:
(i) Volume of dough = \( \frac{4}{3} \) πr³ = \( \frac{4}{3} \) π × 6³ = 288π cm³ ≈ 904.8 cm³
Assumption: a rolled chapati is a thin disc (cylinder) of diameter 16 cm (radius 8 cm) and thickness 2 mm = 0.2 cm.
Dough in one chapati = π × 8² × 0.2 = 12.8π cm³ ≈ 40.2 cm³
Number of chapatis = \( \frac{288\pi}{12.8\pi} \) = 22.5
About 22 chapatis can be made.
(ii) Coconut (sample measurements): outer radius of the shell about 6 cm, shell about 0.4 cm thick, white flesh about 1 cm thick. The flesh is a thick spherical shell between radii 6 − 0.4 = 5.6 cm and 5.6 − 1 = 4.6 cm.
V = \( \frac{4}{3} \) π(5.6³ − 4.6³) = \( \frac{4}{3} \) × 3.14 × (175.616 − 97.336) = \( \frac{4}{3} \) × 3.14 × 78.28 ≈ 328 cm³ (about 164 cm³ in each half).
Muskmelon (sample measurements): radius about 7 cm, rind about 0.5 cm thick, hollow seed cavity of radius about 3 cm.
V = \( \frac{4}{3} \) π(6.5³ − 3³) = \( \frac{4}{3} \) × 3.14 × (274.625 − 27) ≈ 1037 cm³
Roughly 330 cm³ of coconut flesh, or about 1000 cm³ (1 litre) of muskmelon fruit. The answers depend on the fruit you measure.

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:
Assumptions: the book has about 160 pages, i.e. 80 sheets (leaves); each page is about 21 cm × 27.5 cm; the closed book is about 1 cm thick; the classroom floor is about 8 m × 6 m.
(i) Area of one page = 21 × 27.5 = 577.5 cm² = 0.05775 m²
80 sheets laid side by side cover 80 × 0.05775 ≈ 4.6 m². Even if all 160 pages were laid separately (as photocopies), the area would be only about 9.2 m².
Area of the classroom floor = 8 × 6 = 48 m²
No. The pages would cover only about 4.6 m² (at most about 9.2 m²), less than one-fifth of the floor.
(ii) 15 ft ≈ 457.2 cm, 20 ft ≈ 609.6 cm, 30 ft ≈ 914.4 cm.
Upper bound from volumes: \( \frac{457.2 \times 609.6 \times 914.4}{21 \times 27.5 \times 1} \) ≈ \( \frac{254850000}{577.5} \) ≈ 4,41,300
Actual packing: books cannot be cut, so count how many fit along each direction. The best arrangement is thickness (1 cm) along the 15 ft side, 21 cm along the 20 ft side and 27.5 cm along the 30 ft side:
along 15 ft: 457.2 ÷ 1 → 457 books; along 20 ft: 609.6 ÷ 21 ≈ 29.03 → 29 books; along 30 ft: 914.4 ÷ 27.5 ≈ 33.25 → 33 books
Number of textbooks = 457 × 29 × 33 = 4,37,349
About 4.37 lakh textbooks (4,37,349 in this arrangement; the volume upper bound is about 4.41 lakh).

Question 19. If the entire human population decided to climb into one giant cube, how long would the side have to be?
Answer:
Guess: about a kilometre.
Assumptions: population about 8 × 10⁹ people; the average human body has a volume of about 0.065 m³ (mass about 65 kg, density close to that of water, 1000 kg/m³).
Packed with no gaps at all: total volume ≈ 8 × 10⁹ × 0.065 = 5.2 × 10⁸ m³; side = ∛(5.2 × 10⁸) ≈ 804 m.
More realistically, people stand in floors: each person needs about 0.5 m × 0.5 m of floor and a layer about 2 m high, i.e. 0.5 m³ of space.
Total volume ≈ 8 × 10⁹ × 0.5 = 4 × 10⁹ m³; side = ∛(4 × 10⁹) ≈ 1587 m ≈ 1.6 km.
The cube would need a side of roughly 0.8 km (tightly packed) to about 1.6 km (people standing on floors) — only about a kilometre or so!

Question 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.
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
Answer:

Let R = 6371 km be the Earth's radius, V = 1.38 × 10⁹ km³ the volume of water and t the thickness of the layer.
(i) The water fills the space between two spheres of radii R and R + t:
\( \frac{4}{3}\pi (R + t)^3 - \frac{4}{3}\pi R^3 = V \;\Rightarrow\; t = \sqrt[3]{R^3 + \frac{3V}{4\pi}} - R \)
Because the layer is very thin compared with the Earth, it is almost a flat sheet spread over the surface area 4πR², so approximately
volume ≈ surface area × thickness, i.e. \( t \approx \frac{V}{4\pi R^2} \)
(ii) 4πR² = 4 × 3.1416 × 6371² ≈ 4 × 3.1416 × 4,05,89,641 ≈ 5.1006 × 10⁸ km²
t ≈ \( \frac{1.38 \times 10^9}{5.1006 \times 10^8} \) ≈ 2.71 km (the exact expression gives t ≈ 2.70 km, almost the same).
The water layer would be about 2.7 km thick.

Teacher's Note:
1. For a very thin shell, volume ≈ surface area × thickness; this avoids working with cube roots.
2. Comparing the exact and approximate answers shows why the approximation is good here.

Question 21. Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Answer:
We need a cuboid P and a reshaped cuboid Q with twice the surface area and half the volume. To gain surface while losing volume, Q must be much flatter.
Try P = 8 cm × 14 cm × 14 cm and Q = 1 cm × 28 cm × 28 cm (height cut from 8 to 1, the other two edges doubled).

Cuboid P: V = 8 × 14 × 14 = 1568 cm³; S = 2(8 × 14 + 14 × 14 + 14 × 8) = 2(112 + 196 + 112) = 840 cm²
Cuboid Q: V = 1 × 28 × 28 = 784 cm³ = 1568 ÷ 2; S = 2(1 × 28 + 28 × 28 + 28 × 1) = 2(28 + 784 + 28) = 1680 cm² = 2 × 840
A cuboid of 8 cm × 14 cm × 14 cm works: reshaped to 1 cm × 28 cm × 28 cm, its surface area doubles (840 → 1680 cm²) while its volume halves (1568 → 784 cm³).
(Other pairs also work, for example 8 cm × 10 cm × 24 cm becoming 1 cm × 24 cm × 40 cm: V 1920 → 960 cm³, S 1024 → 2048 cm².)

Teacher's Note:
1. This is an open question with many answers; any pair that is checked for both conditions is correct.
2. Making the cuboid thinner and wider raises surface area while lowering volume.

Question 22. Project: Find the volume of your house making necessary approximations. Present how you solved it.
Answer:
Method: treat each room as a cuboid, measure (or estimate) its length, width and height, find V = l × w × h for each, and add. Ignore wall thickness and small items.
Sample for a two-bedroom flat with 3 m high ceilings:
1. Living room: 5 m × 4 m × 3 m = 60 m³
2. Bedroom 1: 4 m × 3.5 m × 3 m = 42 m³
3. Bedroom 2: 3.5 m × 3 m × 3 m = 31.5 m³
4. Kitchen: 3 m × 2.5 m × 3 m = 22.5 m³
5. Two bathrooms: 2 × (2 m × 1.5 m × 3 m) = 2 × 9 = 18 m³
6. Passage: 4 m × 1 m × 3 m = 12 m³
Volume of the house ≈ 60 + 42 + 31.5 + 22.5 + 18 + 12 = 186 m³
Since 1 m³ = 1000 litres, the house encloses about 1,86,000 litres of space.
Presentation: draw a rough floor plan, write the measured dimensions of each room, show the calculation list as above, and state the approximations (rooms treated as cuboids; walls, cupboards and the balcony ignored).
Sample answer: about 186 m³. Your own measurements will give a different value.