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

Very Short Answer Questions

Question 1. The volume of a cube is 216 cm³. Find the length of its side and its total surface area.
Answer:
a = ∛216 = 6 cm.
TSA = 6 × 36 = 216 cm². (Same number, different units: cm³ and cm².)

Teacher's Note:
1. Only a cube of side 6 gives equal numbers.
2. Always write units.

Question 2. Find the total surface area of a cuboid measuring 12 cm by 8 cm by 5 cm.
Answer:
TSA = 2(96 + 40 + 60) = 2 × 196 = 392 cm².

Teacher's Note:
1. Three different face areas, each twice.
2. Add before doubling.

Question 3. The areas of the three faces of a cuboid that meet at one corner are 6 cm², 15 cm² and 10 cm². Find the volume of the cuboid.
Answer:
(lw)(wh)(hl) = (lwh)², so (lwh)² = 6 × 15 × 10 = 900.
Volume = √900 = 30 cm³ (the cuboid 2 × 3 × 5).

Teacher's Note:
1. Multiply the three areas.
2. Take the square root.

Question 4. How many cubes of side 20 cm can be packed tightly into a cubical box of side 2 m?
Answer:
2 m = 200 cm; 200 ÷ 20 = 10 along each edge.
Number = 10 × 10 × 10 = 1000.

Teacher's Note:
1. Convert to the same unit first.
2. Exact fit, so volume division also works.

Question 5. Find the curved surface area of a cylinder of base radius 7 cm and height 15 cm.
Answer:
CSA = 2πrh = 2 × \( \frac{22}{7} \) × 7 × 15 = 660 cm².

Teacher's Note:
1. Cancel 7.
2. Curved surface only.

Question 6. Find the volume of a cylinder of base radius 7 cm and height 15 cm, and state how many litres it holds.
Answer:
V = πr²h = \( \frac{22}{7} \) × 49 × 15 = 2310 cm³.
2310 ÷ 1000 = 2.31 litres.

Teacher's Note:
1. 1000 cm³ = 1 litre.
2. Base area 154 × 15.

Question 7. A closed cylindrical tin has base diameter 14 cm and height 10 cm. Find its total surface area.
Answer:
r = 7 cm. TSA = 2πr(h + r) = 44 × 17 = 748 cm².

Teacher's Note:
1. Halve the diameter.
2. Closed: both ends counted.

Question 8. A cone has base radius 7 cm and height 24 cm. Find its slant height and its curved surface area.
Answer:
l = √(24² + 7²) = √625 = 25 cm.
CSA = πrl = \( \frac{22}{7} \) × 7 × 25 = 550 cm².

Teacher's Note:
1. 7–24–25 triple.
2. Use l, not h, for CSA.

Question 9. A cone has base radius 7 cm and slant height 25 cm. Find its total surface area, and state which two areas you have added together.
Answer:
TSA = πr(l + r) = 22 × 32 = 704 cm².
It is the curved surface (550 cm²) plus the circular base (154 cm²).

Teacher's Note:
1. Name the two parts.
2. Check 550 + 154 = 704.

Question 10. Find the volume of a cone whose base radius is 7 cm and whose height is 24 cm.
Answer:
V = ⅓πr²h = ⅓ × \( \frac{22}{7} \) × 49 × 24 = 1232 cm³.

Teacher's Note:
1. Divide 24 by 3 early.
2. One third of the cylinder.

Question 11. Find the surface area of a sphere of radius 21 cm.
Answer:
4πr² = 4 × \( \frac{22}{7} \) × 441 = 4 × 1386 = 5544 cm².

Teacher's Note:
1. πr² = 1386 for r = 21.
2. Multiply by 4.

Question 12. Find the volume of a sphere of radius 21 cm, and express your answer in litres.
Answer:
V = (4/3) × \( \frac{22}{7} \) × 9261 = 38 808 cm³.
= 38.808 litres.

Teacher's Note:
1. 21³ = 9261.
2. Divide by 1000 for litres.

Question 13. Find the curved surface area and the total surface area of a hemisphere of radius 21 cm.
Answer:
CSA = 2πr² = 2 × 1386 = 2772 cm².
TSA = 3πr² = 3 × 1386 = 4158 cm².

Teacher's Note:
1. TSA adds the flat face πr².
2. Check: 2772 + 1386 = 4158.

Question 14. Find the volume of a hemisphere of radius 10.5 cm, and hence the volume of the whole sphere from which it was cut.
Answer:
V = (2/3)πr³ = (2/3) × \( \frac{22}{7} \) × 1157.625 = 2425.5 cm³.
Whole sphere = 2 × 2425.5 = 4851 cm³.

Teacher's Note:
1. 10.5³ = 1157.625.
2. Double for the sphere.

Question 15. A cone and a cylinder stand on the same base and have the same height. Write the ratio of the volume of the cone to the volume of the cylinder, and state how this was confirmed experimentally in this chapter.
Answer:
Cone : cylinder = 1 : 3.
A paper cone of the same base and height was filled with salt and emptied into the cylinder; it took exactly three conefuls to fill it.

Teacher's Note:
1. Do the experiment if possible.
2. Clay version: one cylinder makes three cones.

Question 16. The edge of a solid cube measures r cm. The largest possible right circular cylinder is cut out of it. Write down the radius and the height of that cylinder, and hence its volume in terms of r.
Answer:
Radius = r/2, height = r.
V = π(r/2)² × r = \( \frac{\pi r^3}{4} \) cm³, about 78.5% of the cube (so about 21.5% is cut away).

Teacher's Note:
1. The circle fits inside a square face.
2. π/4 ≈ 0.785 is the fraction kept.

Question 17. The radii of two spheres are in the ratio 2:3. Find the ratio of their surface areas.
Answer:
Surface area ∝ r², so the ratio is 2² : 3² = 4 : 9.

Teacher's Note:
1. 4π cancels.
2. Square the ratio.

Question 18. The radii of two spheres are in the ratio 2:3. Find the ratio of their volumes, and explain why this ratio is not the same as the ratio of their surface areas.
Answer:
Volume ∝ r³, so the ratio is 2³ : 3³ = 8 : 27.
Area is measured in square units and scales with the square; volume scales with the cube. So they differ from 4 : 9.

Teacher's Note:
1. Square for area, cube for volume.
2. Link to scaling.

Question 19. The volume of a cube is 512 cm³. Find the length of one edge and the total length of all twelve edges.
Answer:
a = ∛512 = 8 cm.
Total edge length = 12 × 8 = 96 cm.

Teacher's Note:
1. 8³ = 512.
2. A cube has 12 equal edges.

Question 20. A solid cube of side 6 cm is melted and recast into small cubes of side 2 cm. How many small cubes are obtained?
Answer:
Number = 6³ ÷ 2³ = 216 ÷ 8 = 27.
(3 along each edge: 3 × 3 × 3.)

Teacher's Note:
1. Volume is conserved in recasting.
2. Check with the arrangement.

Question 21. The curved surface area of a cylinder is 1760 cm² and its base radius is 14 cm. Find its height.
Answer:
2 × \( \frac{22}{7} \) × 14 × h = 1760 → 88h = 1760 → h = 20 cm.

Teacher's Note:
1. 2πr = 88 here.
2. Divide to find h.

Question 22. The volume of a cone is 1570 cm³ and its height is 15 cm. Taking π = 3.14, find the radius of its base.
Answer:
⅓ × 3.14 × r² × 15 = 1570 → 15.7r² = 1570 → r² = 100 → r = 10 cm.

Teacher's Note:
1. Simplify ⅓ × 15 = 5 first.
2. Square root at the end.

Question 23. The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of its base.
Answer:
\( \frac{22}{7} \) × r × 14 = 308 → 44r = 308 → r = 7 cm.

Teacher's Note:
1. CSA uses slant height.
2. 22/7 × 14 = 44.

Question 24. The surface area of a sphere is 616 cm². Find its radius.
Answer:
4 × \( \frac{22}{7} \) × r² = 616 → r² = 49 → r = 7 cm.

Teacher's Note:
1. 88/7 × r² = 616.
2. Take the square root.

Question 25. A cylindrical water tank has base radius 7 m and height 5 m. Find its capacity in cubic metres and in litres.
Answer:
V = \( \frac{22}{7} \) × 49 × 5 = 770 m³.
= 770 × 1000 = 770 000 litres.

Teacher's Note:
1. 1 m³ = 1000 litres.
2. Base area 154 m².

Question 26. A sphere fits exactly inside the smallest cylinder that can contain it. Write the ratio of the volume of the sphere to the volume of that cylinder.
Answer:
Cylinder: radius r, height 2r, volume 2πr³. Sphere: (4/3)πr³.
Ratio = 2 : 3.

Teacher's Note:
1. Archimedes' tomb result.
2. Height of cylinder = diameter.

Question 27. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated through 360° about the side of length 3 cm. Name the solid formed, find its volume in terms of π, and state whether turning the same triangle about its 4 cm side would give the same volume.
Answer:
A cone with height 3 cm and radius 4 cm: V = ⅓π × 16 × 3 = 16π cm³.
About the 4 cm side: radius 3, height 4, V = ⅓π × 9 × 4 = 12π cm³. So not the same.

Teacher's Note:
1. The radius is squared, so it matters more.
2. Axis side = height.

Question 28. The radius of a sphere is increased by 10%. Find the percentage increase in its surface area.
Answer:
New area ÷ old = (1.1)² = 1.21.
Increase = 21%.

Teacher's Note:
1. Not 10%: area scales with the square.
2. Use a multiplier.

Question 29. A solid cube has edge a. Find the volume of the largest sphere that can be carved from it, as a fraction of the volume of the cube.
Answer:
Radius a/2: V = (4/3)π(a/2)³ = \( \frac{\pi a^3}{6} \).
Fraction = \( \frac{\pi}{6} \) ≈ 0.524 (about 52%).

Teacher's Note:
1. Independent of the cube's size.
2. About 48% is cut away.

Question 30. A solid block is cut into several smaller pieces. Explain why the total volume of the pieces is the same as the volume of the block, while their total surface area is greater.
Answer:
Volume is unchanged: cutting only rearranges the material; nothing is added or removed.
Surface area increases: each cut exposes two new faces that were inside before.
Example: a 4 cm cube (96 cm²) cut into 64 unit cubes gives 384 cm².

Teacher's Note:
1. Why powdered sugar dissolves faster.
2. Use the 4 cm cube example.