Exercise 10.4 — Sphere and Hemisphere
Surface area and volume of sphere and hemisphere.
What are a Sphere and a Hemisphere?
A sphere is a perfectly round three-dimensional solid — every point on its surface is the same distance from the centre. That distance is called the radius (r), and the full distance across the sphere through the centre is the diameter (2r). A sphere has no edges, no corners and no flat faces — only one perfectly curved surface.
A hemisphere is exactly half of a sphere, cut through its centre. It has one flat circular face (the base) and one curved surface. Both the flat base and the curved top share the same radius r. This shape appears in everyday objects such as bowls, domes, igloo-shaped structures and lampshades.
All Formulas — Sphere and Hemisphere
Memorise these six formulas. They are the only ones needed for the entire Exercise 10.4 and they appear repeatedly in board exam questions.
🔴 Sphere (radius = r)
🔵 Hemisphere (radius = r)
Questions 1 & 2 — Surface Area and Volume of a Sphere
These two questions are the most direct in the exercise. Q1 gives the radius and asks for both surface area and volume. Q2 reverses the process — you are given the surface area and must first find the radius, then compute the volume.
Tip: Convert 3.5 cm to the fraction 7/2 to keep calculations exact with π = 22/7. The 7s cancel cleanly.
Strategy: First convert the mixed fraction to an improper fraction, then use SA = 4πr² to find r, then compute volume.
Questions 3 & 4 — Finding Radius from Circumference; Leather for Balls
Key insight: The equator of a globe is a great circle, so its length is the circumference = 2πr. Use this to find r, then apply the surface area formula.
Remember: The surface area of a sphere gives the amount of material needed to cover it. For multiple identical balls, multiply by the count.
Question 5 — Ratio of Surface Areas and Volumes of Two Spheres
This is one of the most conceptually important questions in the exercise. It teaches the fundamental scaling laws: surface area scales as the square of the radius ratio, while volume scales as the cube.
Given: r₁ : r₂ = 2 : 3, so r₁/r₂ = 2/3
| Quantity | Scales as | For r₁:r₂ = 2:3 |
|---|---|---|
| Radius | r | 2 : 3 |
| Surface Area | r² (square of ratio) | 4 : 9 |
| Volume | r³ (cube of ratio) | 8 : 27 |
Questions 6, 7 & 8 — Hemisphere TSA; Balloon Ratio; Brass Bowl
Shortcut: Since surface area ∝ r², the ratio equals (r₁/r₂)². Diameters double, so radii double: ratio = (1/2)² = 1/4, giving 1 : 4. You can verify by calculating both surface areas fully.
Key step: Outer radius = inner radius + thickness = 5 + 0.25 = 5.25 cm. Then compare the curved surface areas (both are hemispherical bowls viewed from the surface-area perspective — the problem treats them as full spheres for the surface calculation).
Questions 9–12 — Density, Melting, Capacity and Bottle Problems
These final questions apply sphere and hemisphere geometry to practical scenarios including mass/density calculations, melting and recasting solids, converting cm³ to litres, and dividing liquids between containers.
Concept: Weight = Volume × Density. First find the volume of the sphere, then multiply by the density.
Principle: When a solid is melted and recast, volume is conserved. So volume of cylinder = volume of sphere.
Remember: 1 litre = 1000 cm³. Find the volume of the hemisphere in cm³, then convert.
Strategy: Number of bottles = Volume of bowl ÷ Volume of one bottle. Many terms cancel elegantly.
Quick Reference — All 12 Answers at a Glance
| Q# | Problem | Key Answer |
|---|---|---|
| Q1 | Sphere r = 3.5 cm → SA and Volume | SA = 154 cm² ; V = 179.67 cm³ |
| Q2 | SA = 1018 2/7 cm² → Volume | r = 9 cm ; V = 3052.08 cm³ |
| Q3 | Equator = 44 cm → Surface area | r = 7 cm ; SA = 616 cm² |
| Q4 | d = 21 cm ball → Leather for 5 balls | 6930 cm² |
| Q5 | Radii ratio 2:3 → SA and Volume ratios | SA = 4:9 ; V = 8:27 |
| Q6 | Hemisphere r = 10 cm → TSA | 942 cm² |
| Q7 | Balloon d: 14→28 cm → SA ratio | 1 : 4 |
| Q8 | Brass bowl, thickness 0.25 cm → SA ratio | 441 : 400 |
| Q9 | Lead ball d = 2.1 cm, density 11.34 → Weight | ≈ 55 g |
| Q10 | Cylinder d=5, h=10/3 melted → Sphere diameter | 5 cm |
| Q11 | Hemisphere bowl d = 10.5 cm → Litres | 0.303 litres |
| Q12 | Bowl (d=9) into bottles (d=3, h=3) → Count | 9 bottles |
Common Mistakes to Avoid
- Using diameter instead of radius: Q3 gives the equator length (= 2πr), Q4 gives diameter = 21 cm, Q7 gives diameters 14 and 28 cm. Always halve the diameter before substituting into any formula.
- Forgetting the flat base for hemisphere TSA: CSA of a hemisphere = 2πr² (curved surface only). TSA = 3πr² (curved + flat base). A bowl problem asking for "total surface area" needs 3πr², not 2πr².
- Wrong scaling law for ratio problems: Surface areas scale as r² (not r), and volumes scale as r³ (not r). In Q5 with ratio 2:3, the surface area ratio is 4:9 and the volume ratio is 8:27 — not 2:3 in either case.
- Not converting cm³ to litres in Q11: The question asks for litres. Divide the cm³ answer by 1000. Leaving the answer in cm³ loses the final mark.
- Choosing the wrong π value: Q6 and Q9 specifically say "use π = 3.14." All other questions use π = 22/7. Mixing them gives wrong answers.
- Volume conservation in Q10: The melted cylinder becomes a sphere of the same volume — not the same surface area, not the same radius. Always equate volumes when a solid is melted and recast.
What This Exercise Prepares You For
Exercise 10.4 lays the foundation for the most challenging part of Chapter 10 — problems involving combined or composite solids, where a hemisphere may be placed on top of a cylinder, or a cone may be inserted inside a sphere. The ability to quickly calculate volume and surface area of a sphere or hemisphere is essential for those higher-difficulty questions that appear in Class 10 board exams.
The ratio techniques from Q5 and Q7 directly connect to Similar Triangles (where sides, areas and volumes scale by the same power laws) and will reappear in Statistics when comparing proportional data. The density and weight calculation in Q9 links physics and mathematics — a common pattern in CBSE application-based questions.
For students revisiting earlier work, the volume-conservation principle in Q10 (melting and recasting) also appears in Exercise 10.3 (Cones) and will return in Class 10 Chapter 13.