Sphere and Hemisphere: Surface Area, Volume, Ratios and Recasting
Sphere and Hemisphere: Surface Area, Volume, Ratios and Recasting
A sphere has one surface; a solid hemisphere has a curved surface plus a flat circular base. NTPC asks surface area and volume, the radius from one of them, ratios between two spheres or between a sphere and a hemisphere, and melting one solid into others.
1. Formula Box
| Solid | Curved surface area | Total surface area | Volume |
|---|---|---|---|
| Sphere (r) | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere (r) | 2πr² | 3πr² (solid) | (2/3)πr³ |
| Spherical shell (R, r) | (4/3)π(R³ − r³) |
Radius ratio k : 1 gives surface area ratio k² : 1 and volume ratio k³ : 1
Melting and recasting: volume is conserved (number of small solids = big volume ÷ one small volume)
✗ Total surface area of a solid hemisphere = 2πr² | ✓ Add the flat base: 3πr²
✗ Radii 2 : 3, so volumes 4 : 9 | ✓ Volumes go by the cube: 8 : 27
हिंदी नोट: गोले का पृष्ठीय क्षेत्रफल 4πr² और आयतन (4/3)πr³ होता है। ठोस अर्धगोले का कुल पृष्ठीय क्षेत्रफल 3πr² (वक्र 2πr² + आधार πr²) होता है। पिघलाकर दोबारा ढालने पर आयतन वही रहता है।
Exam Pointer: Verified NTPC patterns: sphere surface area (January 2017 CBT-2), solid hemisphere total surface area (June 2022 CBT-2), sphere versus hemisphere surface comparison (June 2022 CBT-2) and a wire melted into a sphere (June 2022 CBT-2).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Sphere: surface area, volume and radius
[PYQ: NTPC CBT-2 17-Jan-2017 Shift-1]
EXAM LEVEL
Q. Find the surface area and volume of a sphere of radius 21 cm.
SA = 4 × 22/7 × 441 = 5,544 cm². Volume = 4/3 × 22/7 × 9,261 = 38,808 cm³.
Answer: 5,544 cm²; 38,808 cm³
EXAMATLAS LEVEL
Q. The volume of a sphere is 4,851 cm³. Find its surface area.
(4/3)(22/7)r³ = 4,851 gives r³ = 4,851 × 21/88 = 1,157.625 = 10.5³, so r = 10.5 cm. SA = 4 × 22/7 × 110.25 = 1,386 cm².
Answer: 1,386 cm²
Pattern 2: Hemisphere: curved and total surface area, volume
[PYQ: NTPC CBT-2 12-Jun-2022 Shift-2]
EXAM LEVEL
Q. Find the curved surface area, total surface area and volume of a solid hemisphere of radius 7 cm.
CSA = 2 × 22/7 × 49 = 308 cm². TSA = 3 × 154 = 462 cm². Volume = 2/3 × 22/7 × 343 = 718 2/3 cm³.
Answer: 308 cm², 462 cm², 718 2/3 cm³
EXAMATLAS LEVEL
Q. The total surface area of a solid hemisphere is 1,848 cm². Find its volume.
3πr² = 1,848 gives πr² = 616 and r = 14 cm. Volume = 2/3 × 22/7 × 2,744 = 5,749 1/3 cm³.
Answer: 5,749 1/3 cm³
Pattern 3: Ratios of spheres and hemispheres
[PYQ: NTPC CBT-2 16-Jun-2022 Shift-1]
EXAM LEVEL
Q. The radii of two spheres are in the ratio 2 : 3. Find the ratios of their surface areas and volumes.
Surface areas 4 : 9, volumes 8 : 27.
Answer: 4 : 9 and 8 : 27
EXAMATLAS LEVEL
Q. A sphere and a solid hemisphere have the same radius. Find the ratio of their total surface areas, and by what percent the sphere's surface exceeds the hemisphere's.
4πr² : 3πr² = 4 : 3. The sphere's surface exceeds by 1/3 = 33 1/3%.
Answer: 4 : 3; 33 1/3%
Pattern 4: Melting and recasting
[PYQ: NTPC CBT-2 14-Jun-2022 Shift-2]
EXAM LEVEL
Q. A metal sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the cylinder's height.
(4/3)π × 216 = π × 16 × h, so 288 = 16h and h = 18 cm.
Answer: 18 cm
EXAMATLAS LEVEL
Q. A metal sphere of radius 9 cm is melted and drawn into a wire of radius 0.3 cm. Find the length of the wire in metres.
(4/3)π × 729 = π × 0.09 × L, so 972 = 0.09L and L = 10,800 cm = 108 m.
Answer: 108 m
60-Second Revision
- Sphere 4πr², (4/3)πr³; hemisphere CSA 2πr², TSA 3πr², V (2/3)πr³.
- Ratios: area by k², volume by k³.
- Recasting: equate volumes; π cancels when both solids use it.