Cylinder and Cone: Surface Area, Volume, Hollow Pipes and Ratios
Cylinder and Cone: Surface Area, Volume, Hollow Pipes and Ratios
The cylinder and the cone share the same base, and a cone holds exactly one-third of the cylinder with the same base and height. NTPC asks curved and total surface area, radius or height from volume, cylinders compared by ratio or with equal volumes, water flowing through pipes, and cone volume from its slant height.
1. Formula Box
| Solid | Curved surface area | Total surface area | Volume |
|---|---|---|---|
| Cylinder (r, h) | 2πrh | 2πr(h + r) | πr²h |
| Hollow cylinder (R, r, h) | 2π(R + r)h (inner + outer) | adds 2π(R² − r²) for the rims | π(R² − r²)h |
| Cone (r, h, slant l) | πrl | πr(l + r) | (1/3)πr²h |
Cone slant height l = √(r² + h²)
Water through a pipe: volume per second = πr² × speed of flow
✗ Cone volume = πr²h | ✓ A cone is one-third of the matching cylinder: (1/3)πr²h
✗ Radius cut by 20% at the same volume, so height rises 20% | ✓ Height factor = 1/0.8² = 1.5625, a 56.25% rise
हिंदी नोट: बेलन का आयतन πr²h और वक्र पृष्ठ 2πrh होता है। उसी आधार और ऊँचाई वाले शंकु का आयतन बेलन का एक-तिहाई, (1/3)πr²h होता है। शंकु की तिर्यक ऊँचाई l = √(r² + h²)।
Exam Pointer: Verified NTPC patterns: cylinder with a relation between curved and total surface area (June 2025 Graduate CBT-1), cylinders with equal volume and changed radius (June 2022 CBT-2), and cone volume from slant height or a partly filled conical vessel (June 2022 CBT-2, June 2025 Graduate CBT-1). Hollow cylinders and pipe-flow questions are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Cylinder: surface areas and volume
[PYQ: NTPC Graduate CBT-1 17-Jun-2025 Shift-3]
EXAM LEVEL
Q. Find the curved surface area, total surface area and volume of a cylinder with radius 7 cm and height 10 cm.
CSA = 2 × 22/7 × 7 × 10 = 440 cm². TSA = 2 × 22/7 × 7 × 17 = 748 cm². Volume = 22/7 × 49 × 10 = 1,540 cm³.
Answer: 440 cm², 748 cm², 1,540 cm³
EXAMATLAS LEVEL
Q. The curved surface area of a cylinder is two-thirds of its total surface area, and its volume is 2,156 cm³. Find its radius and height.
2πrh = (2/3) × 2πr(h + r) gives 3h = 2h + 2r, so h = 2r. Volume = πr² × 2r = 2πr³ = 2,156, so r³ = 343 and r = 7 cm, h = 14 cm.
Answer: r = 7 cm, h = 14 cm
Pattern 2: Comparing cylinders: ratios and equal volumes
[PYQ: NTPC CBT-2 15-Jun-2022 Shift-1]
EXAM LEVEL
Q. Two cylinders have radii in the ratio 2 : 3 and heights in the ratio 5 : 4. Find the ratio of their volumes.
Volume ∝ r²h: (4 × 5) : (9 × 4) = 20 : 36 = 5 : 9.
Answer: 5 : 9
EXAMATLAS LEVEL
Q. The radius of a cylinder is reduced by 20% while its volume stays the same. By what percent does its height increase?
r² falls to 0.64 of itself, so h must be multiplied by 1/0.64 = 1.5625, an increase of 56.25%.
Answer: 56.25%
Pattern 3: Hollow cylinders and water flowing through a pipe
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A hollow metal pipe has outer radius 8 cm, inner radius 6 cm and length 14 cm. Find the volume of metal.
π(R² − r²)h = 22/7 × (64 − 36) × 14 = 22/7 × 28 × 14 = 1,232 cm³.
Answer: 1,232 cm³
EXAMATLAS LEVEL
Q. Water flows at 5 m/s through a pipe of internal diameter 14 cm into a tank that holds 46.2 m³. How long does it take to fill the tank?
Radius = 0.07 m. Flow per second = 22/7 × 0.0049 × 5 = 0.077 m³. Time = 46.2/0.077 = 600 seconds = 10 minutes.
Answer: 10 minutes
Pattern 4: Cone: slant height, surface area and volume
[PYQ: NTPC CBT-2 12-Jun-2022 Shift-2 | NTPC Graduate CBT-1 5-Jun-2025 Shift-3]
EXAM LEVEL
Q. A cone has radius 5 cm and height 12 cm. Find its slant height, curved surface area and volume.
l = √(25 + 144) = 13 cm. CSA = π × 5 × 13 = 65π cm². Volume = (1/3)π × 25 × 12 = 100π cm³.
Answer: 13 cm; 65π cm²; 100π cm³
EXAMATLAS LEVEL
Q. The base area of a cone is 154 cm² and its slant height is 25 cm. Find its volume.
πr² = 154 gives r = 7 cm. h = √(625 − 49) = 24 cm. Volume = (1/3) × 154 × 24 = 1,232 cm³.
Answer: 1,232 cm³
60-Second Revision
- Cylinder: CSA 2πrh, TSA 2πr(h + r), V πr²h.
- Cone: l = √(r² + h²), CSA πrl, TSA πr(l + r), V (1/3)πr²h.
- Same volume, radius factor k: height factor 1/k².
- Hollow: π(R² − r²)h; pipe flow per second = πr² × speed.