Chapter Overview
Chapter 14 applies formulas for surface area and volume of solids. It covers cubes and cuboids, cylinders, cones, spheres and hemispheres, plus practical estimation and conservation of volume.
These worked solutions show the mathematical reasoning behind selected exercise questions and end-of-chapter problems.
Textbook-check note: Use the exact question numbers and diagrams in your copy of Ganita Manjari. Where data are read from a graph, answers are marked as estimates. This page does not reproduce every printed exercise verbatim.
Important Concepts
| Concept | Explanation |
|---|---|
| Cuboid | Volume = lwh; total surface area = 2(lw + wh + hl). |
| Cube | Volume = a³; total surface area = 6a². |
| Cylinder | Curved surface area = 2πrh; volume = πr²h. |
| Cone | Slant height l = √(r² + h²); curved surface area = πrl; volume = (1/3)πr²h. |
| Sphere | Surface area = 4πr²; volume = (4/3)πr³. |
| Hemisphere | Curved surface area = 2πr²; total surface area including base = 3πr²; volume = (2/3)πr³. |
Exercise Solutions — Chapter 14
Exercise Set 14.1 — Cubes and cuboids
Question 1
A cube has volume 64 cm³. Since a³ = 64, a = 4 cm. Total surface area = 6a² = 6×16 = 96 cm².
Question 2
A cubical box has edge 2 m = 200 cm. Small cubes have edge 20 cm, so 200 ÷ 20 = 10 fit along each edge. Total = 10³ = 1000 cubes.
Question 3
Godown volume = 40×25×10 = 10000 m³. Each box occupies 2×1.25×1 = 2.5 m³. Number of boxes = 10000 ÷ 2.5 = 4000.
Question 4
Each cube has volume 125 cm³, so its side is 5 cm. Two joined end-to-end form a cuboid 10×5×5 cm. TSA = 2(10×5 + 5×5 + 10×5) = 2(50+25+50) = 250 cm².
Question 5
A 4 cm cube has TSA 6×4² = 96 cm². Cutting it into 1 cm cubes gives 4³ = 64 small cubes, each TSA 6 cm², total TSA 384 cm². Ratio original : sum after cutting = 96:384 = 1:4.
Question 6
Let cuboid edges be l,w,h. Given lw = 6, wh = 15, hl = 10, then (lwh)² = (lw)(wh)(hl) = 6×15×10 = 900. Hence volume lwh = 30 cm³.
Question 7 — Painted cube
A 5 cm cube painted on the outside is cut into 1 cm cubes.
(i) Three painted faces: 8 corner cubes.
(ii) Two painted faces: 12 edges × (5−2) = 36 cubes.
(iii) One painted face: 6×(5−2)² = 54 cubes.
(iv) No painted face: (5−2)³ = 27 cubes. Check: 8+36+54+27 = 125 small cubes.
Question 8 — Integer cuboid with TSA 100 cm²
We need lw + wh + hl = 50. With integer edges in non-decreasing order, the two possibilities are 1×2×16 cm and 2×4×7 cm; both give 2(lw+wh+hl)=100 cm².
Exercise Set 14.2 — Cylinders
Question 1 — Compare two cylinders
Let cylinder A have radius r and height h; cylinder B has radius 2r and height h/2.
CSA(A):CSA(B) = 2πrh : 2π(2r)(h/2) = 1:1.
Volume(A):Volume(B) = πr²h : π(2r)²(h/2) = 1:2.
Question — Water displaced by marbles
A cylinder of radius 4 cm has water rise from height 16 cm to 20 cm. The rise requires extra volume π×4²×4 = 64π cm³. Each marble of radius 1 cm has volume (4/3)π cm³. Number = 64π ÷ ((4/3)π) = 48 marbles.
Exercise Set 14.3 — Cones
Question 1
Slant height 21 m and diameter 24 m give r = 12 m and l = 21 m. Total surface area = πr(l+r) = π×12×33 = 396π m² ≈ 1244.57 m² (π = 22/7).
Question 2
For a cone with r = 7 cm and l = 10 cm, curved surface area = πrl = π×7×10 = 70π = 220 cm² (π = 22/7).
Question 7 — Sphere and cone
The sphere of radius 5 cm has area 4π×25 = 100π. This is five times the cone’s curved area π×4×l, so 100π = 5×4πl ⇒ l = 5 cm. Cone height h = √(5²−4²) = 3 cm. Volume = (1/3)π×4²×3 = 16π cm³ ≈ 50.29 cm³.
Exercise Set 14.4 — Spheres
Question 1
A ball bearing has radius 0.7 cm. Surface area = 4πr² = 4×(22/7)×0.7² = 6.16 cm².
Question 2
Two solid spheres of the same metal weigh 5920 g and 740 g. Their volume ratio is 5920:740 = 8:1. Therefore radius ratio = ∛8:1 = 2:1. If the smaller diameter is 5 cm, its radius is 2.5 cm and the larger radius is 5 cm.
End-of-Chapter Exercises
Question 1 — Estimate ice-cream scoops
Container volume = 10×15×20 = 3000 cm³. Model each scoop as a sphere of radius about 2.5 cm: V ≈ (4/3)π(2.5)³ ≈ 65 cm³. 3000÷65 ≈ 46, so a practical estimate is about 40–45 scoops, allowing for packing and wastage.
Question 2 — Additional unit cubes
To grow a cube from side a to side a+1, the extra unit cubes are (a+1)³ − a³ = a³+3a²+3a+1−a³ = 3a² + 3a + 1.
Question 7 — Find a cone’s height and volume
The sphere surface area is 100π cm², five times the cone CSA; so 5π×4×l = 100π ⇒ l=5 cm. With radius 4 cm, height = 3 cm; volume = 16π cm³ ≈ 50.29 cm³.
Question 8 — Planet volumes
Volume is proportional to the cube of the radius. Hence V(Jupiter)/V(Earth) = (69900/6370)³ ≈ 1321; V(Sun)/V(Earth) = (695700/6370)³ ≈ 1.30×10⁶.
Question 12 — Change in volume
(i) Increasing cuboid length from l to l+1 changes volume by (l+1)wh−lwh = wh cubic units.
(ii) Increasing cylinder radius r to r+1 changes volume by π(r+1)²h−πr²h = 2πrh+πh.
(iii) Reducing sphere radius from r to r−1 decreases volume by (4/3)π[r³−(r−1)³] = (4/3)π(3r²−3r+1).
Question 13 — Surface area in terms of volume
For a cube, V = a³ gives a = V^(1/3). Therefore S = 6a² = 6V^(2/3).
Question 14 — Volume in terms of surface area
Since S = 6a², a = √(S/6). Hence V = a³ = (S/6)^(3/2).
Question 15 — Marbles in a glass
The required rise is 20−16 = 4 cm. Volume to displace = π×4²×4 = 64π cm³. Each marble volume = (4/3)π cm³. Number = 64π ÷ (4π/3) = 48 marbles.
One Minute Revision
- Surface area is expressed in square units; volume in cubic units.
- Convert every measurement to the same unit before substituting.
- For recast objects with no material loss, the total volume is conserved.
- For a cone, l² = r² + h².
- For rough estimates, state the assumptions and give an approximate answer.
Frequently Asked Questions
1. What is the difference between surface area and volume?
Surface area measures the outside covering of a solid; volume measures the space it occupies.
2. Does recasting change volume?
If material is neither added nor lost, the total volume stays the same.
3. What is the formula for cube total surface area?
For edge length a, TSA = 6a².