CBSE Class 9 Maths Ganita Manjari Part 2 Ch 14 Math of Space Surface Area and Volume Short Answer Questions with Answers

Short Answer Questions

Question 1. A closed wooden box measures 15 cm by 12 cm by 10 cm. Find its total surface area, its volume, and the cost of painting the outside at ₹8 per 100 cm².
Answer:
TSA = 2(180 + 120 + 150) = 900 cm².
V = 15 × 12 × 10 = 1800 cm³.
Cost = (900 ÷ 100) × 8 = ₹72.
In simple words: Find the area, then count how many 100 cm² units it has.

Teacher's Note:
1. Rate is per 100 cm², not per cm².
2. Painting uses area, not volume.

Question 2. A godown measures 30 m by 20 m by 8 m. It is to be filled with cuboidal cartons, each measuring 1.5 m by 1 m by 0.8 m. Find how many cartons it will hold.
Answer:
Godown volume = 4800 m³; carton volume = 1.2 m³.
4800 ÷ 1.2 = 4000 cartons.
Check by edges: 20 × 20 × 10 = 4000; they fit exactly.
In simple words: Because the cartons fit exactly, dividing volumes gives the right count.

Teacher's Note:
1. If dimensions do not divide exactly, count along edges.
2. Volume division then gives only an upper bound.

Question 3. A solid cube of side 4 cm is cut into cubes of side 1 cm. Find the total surface area of the original cube and the combined total surface area of all the small cubes, and comment on what has happened to the volume.
Answer:
Original TSA = 6 × 16 = 96 cm².
64 small cubes × 6 cm² = 384 cm² (four times as much).
Volume is unchanged: 64 cm³ before and after.
In simple words: Cutting adds surface but never changes volume.

Teacher's Note:
1. New faces come from the cuts.
2. Volume is conserved.

Question 4. A cube of side 4 cm is painted on all six faces and then cut into cubes of side 1 cm, as shown.
Find how many of the small cubes have exactly three painted faces, exactly two, exactly one, and none at all. Check that your four numbers add up correctly.
Answer:
Three faces: the corners = 8.
Two faces: along edges, not corners: 12 × 2 = 24.
One face: inside each face: 6 × 4 = 24.
None: inner 2 × 2 × 2 core = 8.
Check: 8 + 24 + 24 + 8 = 64 = 4³ ✓.
In simple words: Corners, edges, faces and the hidden core.

Teacher's Note:
1. General: 8, 12(n − 2), 6(n − 2)², (n − 2)³.
2. Always check the total.

Question 5. A closed cylindrical drum has base radius 10.5 cm and height 20 cm. Find its curved surface area, its total surface area and its volume.
Answer:
CSA = 2 × \( \frac{22}{7} \) × 10.5 × 20 = 1320 cm².
TSA = 2πr(h + r) = 66 × 30.5 = 2013 cm².
V = \( \frac{22}{7} \) × 110.25 × 20 = 6930 cm³.
In simple words: 2πr = 66 cm here; use it for both surface areas.

Teacher's Note:
1. The two lids add 693 cm².
2. 10.5² = 110.25.

Question 6. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 3:2. Find the ratio of their volumes and the ratio of their curved surface areas.
Answer:
Volumes: (2² × 3) : (3² × 2) = 12 : 18 = 2 : 3.
Curved surfaces: (2 × 3) : (3 × 2) = 1 : 1.
In simple words: Radius counts twice in volume but once in curved surface area.

Teacher's Note:
1. Use 2k, 3k and 3m, 2m.
2. Equal CSA, unequal volume.

Question 7. A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each of radius 2 cm and height 12 cm. Find the number of complete rods that can be made, and explain why some metal is left over.
Answer:
Cube: 12³ = 1728 cm³. Rod: \( \frac{22}{7} \) × 4 × 12 = \( \frac{1056}{7} \) ≈ 150.86 cm³.
1728 ÷ 150.86 ≈ 11.45, so 11 complete rods.
Leftover ≈ 1728 − 1659.4 = 68.6 cm³, less than one rod, because 1728 is not a whole multiple of the rod volume.
In simple words: Only whole rods count; the remainder is too small for another.

Teacher's Note:
1. Round down for complete items.
2. Leftover metal is not lost.

Question 8. A cone has height 16 cm and base radius 12 cm. Taking π = 3.14, find its curved surface area and its total surface area.
Answer:
l = √(256 + 144) = 20 cm.
CSA = 3.14 × 12 × 20 = 753.6 cm².
TSA = 3.14 × 12 × 32 = 1205.76 cm².
In simple words: Find the slant height first, then use πrl and πr(l + r).

Teacher's Note:
1. Never use h in πrl.
2. 3–4–5 scaled by 4.

Question 9. A conical tent has base radius 6 m and height 8 m. Taking π = 3.14, find the area of tarpaulin needed for its curved surface, and then the length required if the tarpaulin is 3 m wide and 20 cm extra is allowed for stitching and wastage.
Answer:
l = 10 m; CSA = 3.14 × 6 × 10 = 188.4 m².
Length = 188.4 ÷ 3 = 62.8 m; add 0.2 m → 63 m.
In simple words: Area ÷ width gives length; then add the extra.

Teacher's Note:
1. Convert 20 cm to 0.2 m.
2. No tarpaulin for the floor.

Question 10. A joker's cap is a hollow cone of base radius 7 cm and height 24 cm. Find the area of sheet needed to make 10 such caps.
Answer:
l = 25 cm; CSA = \( \frac{22}{7} \) × 7 × 25 = 550 cm².
10 caps: 5500 cm².
In simple words: Only the curved surface is made from sheet.

Teacher's Note:
1. Hollow, open base.
2. 7–24–25.

Question 11. A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of length 8 cm. Find the volume and the curved surface area of the solid formed, in terms of π.
Answer:
Cone with h = 8, r = 6, l = 10.
V = ⅓π × 36 × 8 = 96π cm³.
CSA = π × 6 × 10 = 60π cm².
In simple words: The axis side is the height, the other leg the radius, the hypotenuse the slant height.

Teacher's Note:
1. Identify r, h, l first.
2. Leave answers in π.

Question 12. A cup is in the shape of a right circular cone. It is filled with water to exactly half its depth. Find what fraction of the volume of the cup is occupied by the water, and explain why the answer is not one half.
Answer:
The water is a smaller cone with depth h/2 and radius r/2 (similar triangles).
Ratio = (½)² × (½) = \( \frac{1}{8} \).
Halving the depth also halves the radius, so every length is halved and volume shrinks by ½³.
In simple words: Half the depth of a cone holds only one eighth of it.

Teacher's Note:
1. To be half full, fill to about 0.79 of the depth.
2. Similar shapes: volume scales by the cube.

Question 13. Two solid spheres made of the same metal weigh 5920 g and 740 g. The diameter of the smaller sphere is 5 cm. Find the radius of the larger sphere.
Answer:
Mass ∝ volume ∝ r³. 5920 ÷ 740 = 8, so the radius ratio is ∛8 = 2.
Smaller radius 2.5 cm, so the larger radius = 5 cm.
In simple words: Eight times the mass means twice the radius.

Teacher's Note:
1. Same metal: mass ratio = volume ratio.
2. Halve the diameter.

Question 14. The diameter of the moon is approximately one fourth the diameter of the earth. Treating both as spheres, find the ratio of their surface areas and the ratio of their volumes.
Answer:
Radius ratio 1 : 4.
Surface areas: 1 : 16. Volumes: 1 : 64.
In simple words: Square the ratio for area, cube it for volume.

Teacher's Note:
1. Constants cancel in ratios.
2. Diameter ratio = radius ratio.

Question 15. Sixteen metal spheres, each of radius 2 cm, are packed into a rectangular box whose inside measures 16 cm by 8 cm by 8 cm. The remaining space is filled with a preservative liquid. Find the volume of the liquid, correct to the nearest cubic centimetre.
Answer:
One sphere: (4/3) × \( \frac{22}{7} \) × 8 = \( \frac{704}{21} \) cm³; 16 spheres ≈ 536.38 cm³.
Box: 16 × 8 × 8 = 1024 cm³ (4 × 2 × 2 = 16 spheres fit exactly).
Liquid = 1024 − 536.38 ≈ 488 cm³.
In simple words: Liquid = box volume − sphere volume.

Teacher's Note:
1. Check the spheres actually fit.
2. Round only at the end.

Question 16. The radius of a sphere is increased by 10%. Show that its volume increases by 33.1%.
Answer:
New volume ÷ old = (1.1)³ = 1.331.
Increase = (1.331 − 1) × 100 = 33.1%.
In simple words: 10% on length → 21% on area → 33.1% on volume.

Teacher's Note:
1. Use the multiplier 1.1.
2. Compare with the area result.

Question 17. The hemispherical dome of a building is to be painted on its curved surface. The circumference of the circular base of the dome is 35.2 m. Find the cost of painting it at ₹10 per 100 cm².
Answer:
2πr = 35.2 → r = 5.6 m.
CSA = 2πr² = 2 × \( \frac{22}{7} \) × 31.36 = 197.12 m² = 1 971 200 cm².
Cost = (1 971 200 ÷ 100) × 10 = ₹1 97 120.
In simple words: Convert m² to cm² before using a per-100 cm² rate.

Teacher's Note:
1. ₹10 per 100 cm² = ₹1000 per m².
2. A common unit trap.

Question 18. A school serves milk in cylindrical glasses of diameter 7 cm, filled to a height of 12 cm. Find how many litres of milk are needed to serve 1600 students.
Answer:
One glass: \( \frac{22}{7} \) × 3.5² × 12 = 462 cm³.
1600 × 462 = 739 200 cm³ = 739.2 litres.
In simple words: Volume per glass × number of students, then convert to litres.

Teacher's Note:
1. Radius 3.5, not 7.
2. Divide by 1000.

Question 19. A cylindrical glass of radius 4 cm holds water to a height of 16 cm. Marbles, each a sphere of radius 1 cm, are dropped in until the water rises to a height of 20 cm. Find how many marbles were dropped in.
Answer:
Water rise = 4 cm; displaced volume = π × 16 × 4 = 64π cm³.
One marble = (4/3)π cm³.
Number = 64π ÷ (4/3)π = 48.
In simple words: The marbles' volume equals the volume of water pushed up.

Teacher's Note:
1. π cancels.
2. Only the rise matters, not the original 16 cm.

Question 20. A stallholder has a cylindrical vessel of radius 12 cm holding spiced water to a depth of 40 cm. He serves it in small spherical shells of radius 2 cm, each of which he fills to about 70% of its volume. Estimate how many shells he can serve, stating the assumption you are using.
Answer:
Vessel: π × 144 × 40 = 5760π cm³.
Each shell: 0.7 × (4/3)π × 8 = \( \frac{22.4\pi}{3} \) cm³.
Number ≈ 5760 × 3 ÷ 22.4 ≈ 771 shells.
Assumption: all the water is served with none spilt, and each shell is filled to exactly 70%.
In simple words: Total volume ÷ volume per serving; the real number is a little lower.

Teacher's Note:
1. State the assumption clearly.
2. An estimate, not an exact count.