1. Basics — Building Blocks
What is a Solid?
A solid (3-D figure) occupies space. Its outer shell is its surface. We study two measurements: surface area (how much wrapping is needed) and volume (how much space it fills).
Prisms vs Pyramids
Solids formed by stacking congruent figures are prisms (e.g., cylinder). Solids tapering to a point are pyramids (e.g., cone). A sphere is neither — it is a perfectly round solid.
Key Measurements
| Symbol | Meaning | Used In |
|---|---|---|
| r | Base radius | Cone, Sphere, Hemisphere |
| h | Height (perpendicular) | Cone |
| l | Slant height | Cone |
| π | ≈ 22/7 ≈ 3.14159… | All circular shapes |
Pythagoras link: In a cone, l² = r² + h² → l = √(r² + h²). Always find the missing dimension before applying surface area or volume formulas.
Surface Area vs Volume — The Difference
Surface Area is measured in square units (cm², m²). Think: painting, tiling, wrapping.
Volume is measured in cubic units (cm³, m³). Think: filling, capacity, weight.
2. Core Concepts
2.1 Right Circular Cone
Definition
A cone formed by rotating a right-angled triangle about one of its perpendicular sides. The line from vertex to centre of base is perpendicular to the base.
- Vertex (A) — the tip/apex
- Height (h = AB) — perpendicular from vertex to base centre
- Radius (r = BC) — radius of the circular base
- Slant height (l = AC) — distance from vertex to any point on the base circle
Curved Surface Area (CSA) of a Cone
Unfold the cone → you get a sector of a circle with radius = slant height l. Cutting this sector into tiny triangles (each height = l):
CSA = πrl
where r = base radius, l = slant height
Total Surface Area (TSA) of a Cone
TSA = curved surface + circular base
TSA = πrl + πr² = πr(l + r)
2.2 Sphere
Definition
A sphere is a 3-D figure where every point on the surface is at the same distance (radius r) from the centre. It has one curved face, no flat face, no edge, no vertex.
Activity insight: The string covering a sphere of radius r exactly fills 4 circles of the same radius → Surface Area = 4 × πr².
Surface Area of Sphere = 4πr²
2.3 Hemisphere
Definition
Half a sphere. It has two faces: one curved (half the sphere's surface) and one flat circular base.
CSA of Hemisphere = 2πr²
TSA of Hemisphere = 3πr²
2.4 Volume of a Cone
Experimentally: filling a cone 3 times fills the cylinder of same base and height once:
Volume of Cone = (1/3) πr²h
2.5 Volume of a Sphere
Experimentally derived by water-displacement method:
Volume of Sphere = (4/3) πr³
2.6 Volume of a Hemisphere
Volume of Hemisphere = (2/3) πr³
3. Examples — Step by Step
Example 1 (Textbook)
Find the CSA of a right circular cone with slant height 10 cm and base radius 7 cm.
Solution
Given: r = 7 cm, l = 10 cm. Formula needed: CSA = πrl
CSA = (22/7) × 7 × 10
= 22 × 10 = 220 cm²
Example 2 (Textbook)
Height of cone = 16 cm, base radius = 12 cm. Find CSA and TSA. (π = 3.14)
Solution
Find slant height: l = √(h² + r²) = √(16² + 12²) = √(256 + 144) = √400 = 20 cm
CSA = πrl = 3.14 × 12 × 20 = 753.6 cm²
TSA = CSA + πr² = 753.6 + 3.14 × 144 = 753.6 + 452.16 = 1205.76 cm²
Example 3 (Corn Cob)
A corn cob shaped like a cone has broadest radius 2.1 cm and height 20 cm. Each 1 cm² carries 4 grains. Find number of grains.
Solution
Find slant height: l = √(r² + h²) = √(2.1² + 20²) = √(4.41 + 400) = √404.41 ≈ 20.11 cm
CSA = πrl = (22/7) × 2.1 × 20.11 = 132.73 cm² (approx)
Number of grains = 132.73 × 4 ≈ 531 grains
Example 4 (Sphere)
Find the surface area of a sphere of radius 7 cm.
Solution
SA = 4πr² = 4 × (22/7) × 7 × 7
= 4 × 22 × 7 = 616 cm²
Example 5 (Hemisphere)
Find (i) CSA and (ii) TSA of a hemisphere of radius 21 cm.
Solution
CSA = 2πr² = 2 × (22/7) × 21 × 21 = 2 × 22 × 63 = 2772 cm²
TSA = 3πr² = 3 × (22/7) × 441 = 4158 cm²
Example 6 (Motorcyclist)
A hollow sphere of diameter 7 m. Find area available to the motorcyclist.
Solution
Diameter = 7 m → r = 3.5 m
Available area = Surface area = 4πr² = 4 × (22/7) × 3.5 × 3.5 = 154 m²
Example 7 (Dome Painting)
Circumference of base of hemispherical dome = 17.6 m. Cost of painting = ₹5 per 100 cm². Find total cost.
Solution
2πr = 17.6 → r = 17.6 × 7 / (2 × 22) = 2.8 m
CSA = 2πr² = 2 × (22/7) × 2.8² = 49.28 m²
Cost per m² = ₹5 per 100 cm² = ₹500 per m²
Total cost = 500 × 49.28 = ₹24,640
Example 8 (Volume of Cone)
Height = 21 cm, slant height = 28 cm. Find volume of cone.
Solution
r = √(l² − h²) = √(784 − 441) = √343 = 7√7 cm
V = (1/3)πr²h = (1/3) × (22/7) × 7√7 × 7√7 × 21
= (1/3) × (22/7) × 343 × 21 = 7546 cm³
Example 9 (Canvas Tent)
Canvas area = 551 m², wastage = 1 m², base radius = 7 m. Find volume of conical tent.
Solution
Effective area = 550 m² = πrl → (22/7) × 7 × l = 550 → l = 25 m
h = √(l² − r²) = √(625 − 49) = √576 = 24 m
V = (1/3)πr²h = (1/3) × (22/7) × 49 × 24 = 1232 m³
Example 10–12 (Sphere & Hemisphere Volume)
Find volume of sphere r = 11.2 cm; mass of shot-putt r = 4.9 cm, density = 7.8 g/cm³; volume of hemispherical bowl r = 3.5 cm.
Solution
V(sphere) = (4/3)πr³ = (4/3)(22/7)(11.2)³ ≈ 5887.32 cm³
V(shot-putt) = (4/3)(22/7)(4.9)³ ≈ 493 cm³ → mass = 7.8 × 493 ≈ 3.85 kg
V(hemisphere) = (2/3)πr³ = (2/3)(22/7)(3.5)³ ≈ 89.8 cm³
4. Textbook Exercises — Solved
Exercise 11.1
Q1
Diameter = 10.5 cm, slant height = 10 cm. Find CSA.
Solution
r = 10.5/2 = 5.25 cm, l = 10 cm
CSA = πrl = (22/7) × 5.25 × 10 = 22 × 0.75 × 10 = 165 cm²
Q2
Slant height = 21 m, diameter = 24 m. Find TSA.
Solution
r = 12 m, l = 21 m
TSA = πr(l + r) = (22/7) × 12 × (21 + 12) = (22/7) × 12 × 33 = 1244.57 m²
Q3
CSA = 308 cm², slant height = 14 cm. Find (i) radius (ii) TSA.
Solution
πrl = 308 → (22/7) × r × 14 = 308 → 44r = 308 → r = 7 cm
TSA = πr(l + r) = (22/7) × 7 × (14 + 7) = 22 × 21 = 462 cm²
Q4
Conical tent: h = 10 m, r = 24 m. Find (i) slant height (ii) cost of canvas at ₹70/m².
Solution
l = √(h² + r²) = √(100 + 576) = √676 = 26 m
CSA = πrl = (22/7) × 24 × 26 = 1961.14 m²
Cost = 1961.14 × 70 ≈ ₹1,37,280
Q5
Tarpaulin 3 m wide, cone h = 8 m, r = 6 m, wastage = 20 cm. Find length of tarpaulin. (π = 3.14)
Solution
l = √(64 + 36) = √100 = 10 m
CSA = 3.14 × 6 × 10 = 188.4 m²
Length = 188.4/3 = 62.8 m; add wastage: 62.8 + 0.2 = 63 m
Q6
Conical tomb: slant height 25 m, diameter 14 m. Cost of whitewashing at ₹210/100 m².
Solution
r = 7 m, l = 25 m; CSA = (22/7) × 7 × 25 = 550 m²
Cost = (550/100) × 210 = ₹1155
Q7
Joker's cap: r = 7 cm, h = 24 cm. Sheet for 10 caps.
Solution
l = √(49 + 576) = √625 = 25 cm
CSA of one = (22/7) × 7 × 25 = 550 cm²
Total for 10 = 5500 cm²
Q8
50 hollow cones, diameter 40 cm, h = 1 m. Painting cost ₹12/m². (π=3.14, √1.04=1.02)
Solution
r = 20 cm = 0.2 m; l = √(1 + 0.04) = √1.04 ≈ 1.02 m
CSA per cone = 3.14 × 0.2 × 1.02 = 0.64056 m²
Total CSA = 50 × 0.64056 = 32.028 m²
Cost = 32.028 × 12 ≈ ₹384.34
Exercise 11.2
Q1 — Surface area of sphere
(i) r = 10.5 cm (ii) r = 5.6 cm (iii) r = 14 cm
Solution
(i) 4π(10.5)² = 4 × (22/7) × 110.25 = 1386 cm²
(ii) 4π(5.6)² = 4 × (22/7) × 31.36 = 394.24 cm²
(iii) 4π(14)² = 4 × (22/7) × 196 = 2464 cm²
Q4 — Balloon ratio
Radius increases 7 cm → 14 cm. Ratio of surface areas.
Solution
SA₁/SA₂ = (4πr₁²)/(4πr₂²) = (7/14)² = 1/4
Ratio = 1 : 4
Q7 — Moon vs Earth
Diameter of moon = (1/4) × diameter of earth. Ratio of surface areas.
Solution
r_moon = r_earth / 4
Ratio = (4π r_moon²)/(4π r_earth²) = (1/4)² = 1 : 16
Exercise 11.3
Q1 — Volume of cone
(i) r = 6 cm, h = 7 cm (ii) r = 3.5 cm, h = 12 cm
Solution
(i) V = (1/3)(22/7)(36)(7) = (1/3)(22)(36) = 264 cm³
(ii) V = (1/3)(22/7)(12.25)(12) = 154 cm³
Q9 — Wheat heap
Cone: diameter 10.5 m, height 3 m. Find volume and canvas area.
Solution
r = 5.25 m; l = √(5.25² + 3²) = √(27.5625 + 9) = √36.5625 ≈ 6.046 m
V = (1/3)(22/7)(5.25)²(3) = (1/3)(22/7)(27.5625)(3) ≈ 86.625 m³
Canvas = πrl = (22/7)(5.25)(6.046) ≈ 99.77 m²
Exercise 11.4
Q4 — Moon volume fraction
Diameter of moon = (1/4) earth. What fraction of earth's volume is moon's volume?
Solution
r_moon = r/4; V_moon = (4/3)π(r/4)³ = (4/3)πr³ × (1/64)
Fraction = V_moon/V_earth = 1/64 = 1/64
Q9 — 27 melted spheres
27 solid iron spheres each of radius r melted into one sphere. Find new radius r′ and ratio S:S′.
Solution
27 × (4/3)πr³ = (4/3)π(r′)³ → (r′)³ = 27r³ → r′ = 3r
S = 4πr², S′ = 4π(3r)² = 36πr²
S : S′ = 1 : 9
5. Tips & Tricks
Always Find l First
For cones, find slant height l = √(r² + h²) before applying any surface area formula.
CSA vs TSA
Open containers (tents, cups) use CSA only. Closed solids use TSA. Read problems carefully.
Which π to Use
Use π = 22/7 when r is a multiple of 7. Use π = 3.14 when decimals are involved.
Volume → Mass
Mass = Volume × Density. First find volume, then multiply by density (in same units).
Ratio Problems
For ratio of surface areas or volumes, use the ratio of r directly — no need to find actual values.
Moon / Earth Type
If diameter ratio is 1:n, then surface area ratio is 1:n² and volume ratio is 1:n³.
Unit Conversion
1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³. Convert before calculating.
Melting/Recasting
Volume is conserved when a solid is melted and recast. Set old volume = new volume.
6. Formula & Fact Sheet
Cone — CSA
πrl
l = slant height; r = base radius
Cone — TSA
πr(l + r)
= πrl + πr²
Cone — Volume
(1/3)πr²h
h = perpendicular height
Slant Height
l = √(r² + h²)
Pythagoras theorem
Sphere — SA
4πr²
Single curved face
Sphere — Volume
(4/3)πr³
r = radius
Hemisphere — CSA
2πr²
Curved face only
Hemisphere — TSA
3πr²
= 2πr² + πr²
Hemisphere — Volume
(2/3)πr³
Half of sphere volume
Quick Facts
Cone vs Cylinder: V(cone) = (1/3) V(cylinder) for same base and height.
Sphere vs Cylinder: A sphere of radius r fits exactly inside a cylinder of radius r and height 2r; their surface area ratio is 1:1 and volume ratio is 2:3.
27 small spheres: Melting 27 spheres of radius r gives one sphere of radius 3r.
π values: 22/7 ≈ 3.142857; 3.14 (for decimal r); use as instructed.
Sphere vs Cylinder: A sphere of radius r fits exactly inside a cylinder of radius r and height 2r; their surface area ratio is 1:1 and volume ratio is 2:3.
27 small spheres: Melting 27 spheres of radius r gives one sphere of radius 3r.
π values: 22/7 ≈ 3.142857; 3.14 (for decimal r); use as instructed.
7. 50 Practice Questions
Covering MCQ · 1-mark · 2-mark · 3-mark · 5-mark · Proof types. Click Show Answer to reveal.
MCQ (1 mark each)
MCQ
Q1
The curved surface area of a cone with radius 6 cm and slant height 10 cm is:
Show Answer
CSA = πrl = π × 6 × 10 = 60π cm². (B)
MCQ
Q2
The total surface area of a cone with r = 3 cm and l = 5 cm is:
Show Answer
TSA = πr(l+r) = π×3×8 = 24π cm². (A)
MCQ
Q3
Surface area of a sphere of diameter 14 cm is:
Show Answer
r = 7; SA = 4×(22/7)×49 = 616 cm². (C)
MCQ
Q4
Volume of a cone is related to the volume of a cylinder of the same base and height by:
Show Answer
(1/3)V(cyl). (B)
MCQ
Q5
The slant height of a cone with h = 8 cm and r = 6 cm is:
Show Answer
l = √(64+36) = √100 = 10 cm. (C)
MCQ
Q6
Total surface area of a hemisphere of radius r is:
Show Answer
3πr² = 2πr² (curved) + πr² (base). (C)
MCQ
Q7
If the radius of a sphere is doubled, its surface area becomes:
Show Answer
SA ∝ r² → (2r)² = 4r² → 4 times. (C)
MCQ
Q8
Volume of a hemisphere of radius 3 cm is:
Show Answer
V = (2/3)π(27) = 18π cm³. (B)
MCQ
Q9
27 spheres each of radius r are melted to form a single sphere. Its radius is:
Show Answer
27×(4/3)πr³ = (4/3)πR³ → R = 3r. (C)
MCQ
Q10
The number of faces of a sphere is:
Show Answer
A sphere has 1 curved face. (B)
1-Mark Questions
1 Mark
Q11
Write the formula for slant height of a cone.
Show Answer
l = √(r² + h²)
1 Mark
Q12
The CSA of a sphere of radius r is ______.
Show Answer
4πr²
1 Mark
Q13
What is the volume of a cone with base area A and height h?
Show Answer
(1/3)Ah
1 Mark
Q14
Name the solid formed by rotating a right triangle about one of its perpendicular sides.
Show Answer
Right Circular Cone
1 Mark
Q15
If l = 13 cm and h = 12 cm for a cone, find r.
Show Answer
r = √(l²−h²) = √(169−144) = √25 = 5 cm
1 Mark
Q16
State the relationship: volume of cone vs volume of sphere (same radius, height = 2×radius for cone).
Show Answer
V(cone, h=2r) = (1/3)πr²(2r) = (2/3)πr³ = V(hemisphere). They are equal!
1 Mark
Q17
The ratio of surface areas of two spheres with radii in ratio 2:3 is ______.
Show Answer
4 : 9
1 Mark
Q18
What does "hemi" mean in hemisphere?
Show Answer
Half
2-Mark Questions
2 Mark
Q19
Find the total surface area of a cone with radius 5 cm and slant height 13 cm.
Show Answer
TSA = πr(l+r) = (22/7)×5×18 = (22×90)/7 ≈ 282.86 cm²
2 Mark
Q20
A spherical ball has surface area 154 cm². Find its radius.
Show Answer
4πr² = 154 → r² = 154×7/(4×22) = 1078/88 = 12.25 → r = 3.5 cm
2 Mark
Q21
Find the volume of a cone whose diameter is 7 cm and height 12 cm.
Show Answer
r = 3.5; V = (1/3)(22/7)(3.5)²(12) = (1/3)(22/7)(12.25)(12) = 154 cm³
2 Mark
Q22
Find the curved surface area of a hemisphere of radius 14 cm.
Show Answer
CSA = 2πr² = 2×(22/7)×196 = 1232 cm²
2 Mark
Q23
A cone and a cylinder have the same base and height. What is the ratio of their volumes?
Show Answer
V(cone) : V(cylinder) = (1/3)πr²h : πr²h = 1 : 3
2 Mark
Q24
The diameter of earth is 4 times the diameter of moon. Find the ratio of their volumes.
Show Answer
r_earth = 4 r_moon; V ∝ r³; ratio = (r_moon)³:(4r_moon)³ = 1:64
2 Mark
Q25
If the volume of a sphere is 288π cm³, find its radius.
Show Answer
(4/3)πr³ = 288π → r³ = 216 → r = 6 cm
2 Mark
Q26
A conical vessel can hold 1.232 litres of water. Its base radius is 7 cm. Find its height.
Show Answer
V = 1232 cm³; (1/3)(22/7)(49)h = 1232 → (22×7/3)h = 1232 → h = 1232×3/154 = 24 cm
3-Mark Questions
3 Mark
Q27
A cone of height 24 cm and base radius 6 cm is made of modelling clay. It is reshaped into a sphere. Find the radius of the sphere.
Show Answer
V(cone) = V(sphere)
(1/3)π(36)(24) = (4/3)πr³
288π = (4/3)πr³
r³ = 216 → r = 6 cm
(1/3)π(36)(24) = (4/3)πr³
288π = (4/3)πr³
r³ = 216 → r = 6 cm
3 Mark
Q28
The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped in. Find the ratio of surface areas. Also find how many times more canvas is used.
Show Answer
SA₁ = 4π(49) = 196π; SA₂ = 4π(196) = 784π
Ratio = 196π : 784π = 1 : 4
4 times more canvas is used.
Ratio = 196π : 784π = 1 : 4
4 times more canvas is used.
3 Mark
Q29
A hemispherical bowl of brass has inner diameter 10.5 cm. Find the cost of tin-plating it inside at ₹16 per 100 cm².
Show Answer
r = 5.25 cm
CSA = 2πr² = 2×(22/7)×27.5625 = 173.25 cm²
Cost = (173.25/100)×16 = ₹27.72
CSA = 2πr² = 2×(22/7)×27.5625 = 173.25 cm²
Cost = (173.25/100)×16 = ₹27.72
3 Mark
Q30
A right triangle with sides 5 cm, 12 cm, 13 cm is revolved about the 12 cm side. Find the volume of the solid obtained.
Show Answer
Revolving about the 12 cm side gives a cone with r = 5 cm, h = 12 cm.
V = (1/3)πr²h = (1/3)(22/7)(25)(12) = 314.28 cm³
V = (1/3)πr²h = (1/3)(22/7)(25)(12) = 314.28 cm³
3 Mark
Q31
A capsule is in the shape of a sphere of diameter 3.5 mm. How much medicine fills it?
Show Answer
r = 1.75 mm
V = (4/3)(22/7)(1.75)³ = (4/3)(22/7)(5.359) ≈ 22.46 mm³
V = (4/3)(22/7)(1.75)³ = (4/3)(22/7)(5.359) ≈ 22.46 mm³
3 Mark
Q32
A dome of a building is a hemisphere whitewashed from inside at cost ₹4989.60. Rate = ₹20 per m². Find (i) inner surface area (ii) volume of air inside.
Show Answer
(i) Area = 4989.60/20 = 249.48 m²
2πr² = 249.48 → r² = 249.48/(2π) = 39.69 → r = 6.3 m
(ii) V = (2/3)πr³ = (2/3)(22/7)(249.47) ≈ 523.9 m³
2πr² = 249.48 → r² = 249.48/(2π) = 39.69 → r = 6.3 m
(ii) V = (2/3)πr³ = (2/3)(22/7)(249.47) ≈ 523.9 m³
3 Mark
Q33
Find the volume of a sphere whose surface area is 154 cm².
Show Answer
4πr² = 154 → r = 3.5 cm
V = (4/3)(22/7)(3.5)³ = (4/3)(22/7)(42.875) = 179.67 cm³
V = (4/3)(22/7)(3.5)³ = (4/3)(22/7)(42.875) = 179.67 cm³
3 Mark
Q34
A right circular cylinder just encloses a sphere of radius r. Find (i) SA of sphere (ii) CSA of cylinder (iii) ratio.
Show Answer
Cylinder has base radius r, height = 2r (diameter of sphere).
(i) SA(sphere) = 4πr²
(ii) CSA(cyl) = 2π(r)(2r) = 4πr²
(iii) Ratio = 4πr² : 4πr² = 1 : 1
(i) SA(sphere) = 4πr²
(ii) CSA(cyl) = 2π(r)(2r) = 4πr²
(iii) Ratio = 4πr² : 4πr² = 1 : 1
3 Mark
Q35
A conical pit of top diameter 3.5 m is 12 m deep. Find capacity in kilolitres.
Show Answer
r = 1.75 m, h = 12 m
V = (1/3)(22/7)(3.0625)(12) = (1/3)(22/7)(36.75) = 38.5 m³
= 38.5 kL
V = (1/3)(22/7)(3.0625)(12) = (1/3)(22/7)(36.75) = 38.5 m³
= 38.5 kL
5-Mark Questions
5 Mark
Q36
The volume of a right circular cone is 9856 cm³. The diameter of its base is 28 cm. Find (i) height (ii) slant height (iii) curved surface area.
Show Answer
r = 14 cm
(i) V = (1/3)πr²h → 9856 = (1/3)(22/7)(196)h → 9856 = (616/3)h → h = 9856×3/616 = 48 cm
(ii) l = √(14²+48²) = √(196+2304) = √2500 = 50 cm
(iii) CSA = πrl = (22/7)(14)(50) = 2200 cm²
(i) V = (1/3)πr²h → 9856 = (1/3)(22/7)(196)h → 9856 = (616/3)h → h = 9856×3/616 = 48 cm
(ii) l = √(14²+48²) = √(196+2304) = √2500 = 50 cm
(iii) CSA = πrl = (22/7)(14)(50) = 2200 cm²
5 Mark
Q37
A solid metallic sphere of diameter 21 cm is melted and recasted into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones.
Show Answer
V(sphere) = (4/3)(22/7)(10.5)³ = (4/3)(22/7)(1157.625) = 4851 cm³
V(one cone) = (1/3)(22/7)(1.75)²(3) = (1/3)(22/7)(3.0625)(3) = 9.625 cm³
Number = 4851/9.625 = 504 cones
V(one cone) = (1/3)(22/7)(1.75)²(3) = (1/3)(22/7)(3.0625)(3) = 9.625 cm³
Number = 4851/9.625 = 504 cones
5 Mark
Q38
A bus stop uses 50 hollow cones (base diameter 40 cm, height 1 m) of recycled cardboard. The outer side is painted at ₹12/m². Find total cost. (π=3.14, √1.04=1.02)
Show Answer
r = 0.2 m, h = 1 m
l = √(0.04+1) = √1.04 ≈ 1.02 m
CSA per cone = 3.14 × 0.2 × 1.02 = 0.64056 m²
Total area = 50 × 0.64056 = 32.028 m²
Cost = 32.028 × 12 ≈ ₹384.34
l = √(0.04+1) = √1.04 ≈ 1.02 m
CSA per cone = 3.14 × 0.2 × 1.02 = 0.64056 m²
Total area = 50 × 0.64056 = 32.028 m²
Cost = 32.028 × 12 ≈ ₹384.34
5 Mark
Q39
A hemispherical tank is made of iron sheet 1 cm thick. Inner radius = 1 m. Find the volume of iron used.
Show Answer
Outer radius = 1 m + 0.01 m = 1.01 m
V(iron) = V(outer hemisphere) − V(inner hemisphere)
= (2/3)π[(1.01)³ − (1)³]
= (2/3)(22/7)[1.030301 − 1]
= (2/3)(22/7)(0.030301)
≈ 0.06348 m³
V(iron) = V(outer hemisphere) − V(inner hemisphere)
= (2/3)π[(1.01)³ − (1)³]
= (2/3)(22/7)[1.030301 − 1]
= (2/3)(22/7)(0.030301)
≈ 0.06348 m³
5 Mark
Q40
Prove that the surface area of a sphere equals the lateral surface area of the circumscribed cylinder (cylinder that just contains the sphere).
Show Answer
Proof:
Let sphere have radius r.
The circumscribed cylinder has base radius = r and height = 2r (= diameter of sphere).
Surface area of sphere = 4πr²
Lateral (curved) surface area of cylinder = 2π × r × 2r = 4πr²
∴ Surface area of sphere = CSA of circumscribed cylinder. Proved.
Let sphere have radius r.
The circumscribed cylinder has base radius = r and height = 2r (= diameter of sphere).
Surface area of sphere = 4πr²
Lateral (curved) surface area of cylinder = 2π × r × 2r = 4πr²
∴ Surface area of sphere = CSA of circumscribed cylinder. Proved.
Mixed Practice (2–3 Mark)
2 Mark
Q41
Find the amount of water displaced by a solid spherical ball of diameter 28 cm.
Show Answer
r=14; V = (4/3)(22/7)(2744) = 11498.67 cm³
2 Mark
Q42
The mass of a metallic ball of diameter 4.2 cm is to be found. Density = 8.9 g/cm³. Find the mass.
Show Answer
r=2.1; V=(4/3)(22/7)(9.261)=38.808 cm³; mass=38.808×8.9 ≈ 345.39 g
3 Mark
Q43
A joker's cap in the form of a cone has base radius 7 cm and slant height 25 cm. Find the area of sheet required to make 10 such caps.
Show Answer
CSA = πrl = (22/7)(7)(25) = 550 cm²; for 10 = 5500 cm²
3 Mark
Q44
A conical tomb has slant height 25 m and diameter 14 m. Find cost of whitewashing curved surface at ₹210 per 100 m².
Show Answer
CSA = (22/7)(7)(25) = 550 m²; cost = (550/100)×210 = ₹1155
3 Mark
Q45
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Show Answer
r=5.25; V=(2/3)(22/7)(144.7) ≈ 303.2 cm³ ≈ 0.3032 litres
2 Mark
Q46
A cone has the same base radius and height as a cylinder. The cylinder's volume is 462 cm³. Find the cone's volume.
Show Answer
V(cone) = (1/3) V(cyl) = 462/3 = 154 cm³
2 Mark
Q47
Find the radius of a cone whose volume is 1232 cm³ and height is 24 cm.
Show Answer
(1/3)(22/7)r²(24) = 1232 → r² = 1232×3×7/(22×24) = 49 → r = 7 cm
3 Mark
Q48
A right triangle with sides 5 cm, 12 cm, 13 cm is revolved about the 5 cm side. Find volume and compare with revolution about 12 cm side.
Show Answer
About 5 cm: cone with r=12, h=5 → V = (1/3)π(144)(5) = 240π cm³
About 12 cm: cone with r=5, h=12 → V = (1/3)π(25)(12) = 100π cm³
Ratio = 240π:100π = 12:5
About 12 cm: cone with r=5, h=12 → V = (1/3)π(25)(12) = 100π cm³
Ratio = 240π:100π = 12:5
5 Mark
Q49
A heap of wheat is in the form of a cone, diameter 10.5 m, height 3 m. Find its volume and the area of canvas to protect it from rain.
Show Answer
r = 5.25 m, h = 3 m
l = √(5.25² + 3²) = √(27.5625 + 9) = √36.5625 ≈ 6.05 m
V = (1/3)(22/7)(27.5625)(3) ≈ 86.63 m³
Canvas = πrl = (22/7)(5.25)(6.05) ≈ 99.77 m²
l = √(5.25² + 3²) = √(27.5625 + 9) = √36.5625 ≈ 6.05 m
V = (1/3)(22/7)(27.5625)(3) ≈ 86.63 m³
Canvas = πrl = (22/7)(5.25)(6.05) ≈ 99.77 m²
5 Mark
Q50
Twenty seven solid iron spheres each of radius r and surface area S are melted to form a sphere with surface area S'. Find (i) radius r' of new sphere (ii) ratio S:S'.
Show Answer
(i) 27 × (4/3)πr³ = (4/3)π(r')³ → (r')³ = 27r³ → r' = 3r
(ii) S = 4πr²; S' = 4π(3r)² = 36πr²
S : S' = 4πr² : 36πr² = 1 : 9
(ii) S = 4πr²; S' = 4π(3r)² = 36πr²
S : S' = 4πr² : 36πr² = 1 : 9