Exercise 10.2 — Cylinder
Surface area and volume of cylinder.
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Right Circular Cylinder — Concept and Formulas
Exercise 10.2 of Class 9 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) focuses entirely on the Right Circular Cylinder. A cylinder consists of one curved (lateral) surface and two congruent circular faces at each end. It is called a right circular cylinder when the axis joining the centres of both circular faces is perpendicular to the base — meaning the cylinder stands perfectly upright, like a can or a pipe.
This exercise contains 11 real-world problems involving water tanks, metal pipes, playground rollers, petrol storage tanks, cylindrical wells, and drums. Every single problem type appears regularly in board exams.
Right Circular Cylinder
Curved (Lateral) SA
Total Surface Area
Volume → Radius
Unit Conversions
Hollow Pipe (inner & outer)
Right Circular Cylinder
Key Formulas
Curved (Lateral) SA = 2πrhTotal SA = 2πr(h + r)Volume = πr²hUse π = 22/7 when r or h involves multiples of 7.
Use π = 3.14 for decimal values.
💡 Choosing between π = 22/7 and π = 3.14: When radius or height is a multiple of 7 (7, 14, 21, 28, 42 …), use π = 22/7 for cleaner cancellation. When given decimal values (2.1, 4.2, 0.56), use π = 3.14. Both are accepted in board exams — choose the one that simplifies the arithmetic.
Question 1 — Total Surface Area of a Closed Cylindrical Tank
Question 1
A closed cylindrical tank has height 1.4 m and base radius 56 cm. How much metal sheet is required? (Answer in m²)
Since the tank is closed, all surfaces are covered: the curved side and both circular ends. We need the Total Surface Area. Note the unit mismatch — radius is in cm, height in m — so convert first.
Given: r = 56 cm = 56/100 m = 0.56 m, h = 1.4 m
Total SA = 2πr(h + r)
= 2 × 3.14 × 0.56 × (1.4 + 0.56)
= 2 × 3.14 × 0.56 × 1.96
= 3.5168 × 1.96
≈ 6.89 m²
✅ Metal sheet required = 6.89 m²
⛔ Common error: Forgetting to convert 56 cm to 0.56 m before substituting. Always check that r and h are in the same unit before applying any formula.
Question 2 — Find Radius from Volume, then Calculate Surface Areas
Question 2
Volume of a cylinder = 308 cm³, height = 8 cm. Find its Lateral SA and Total SA.
This is a two-step reverse problem: first find r from the volume, then compute both surface areas.
Step 1 — Find r from Volume:
πr²h = 308
(22/7) × r² × 8 = 308
r² = (308 × 7) / (22 × 8) = 2156 / 176 = 12.25
r² = (7/2)² ⟹ r = 7/2 = 3.5 cm
Step 2 — Lateral SA:
= 2πrh = 2 × (22/7) × 3.5 × 8
= 2 × 22 × 0.5 × 8 = 176 cm²
Step 3 — Total SA:
= 2πr(h + r) = 2 × (22/7) × 3.5 × (8 + 3.5)
= 22 × 11.5
= 253 cm²
✅ Lateral SA = 176 cm² | Total SA = 253 cm²
📌 Trick: When using π = 22/7 with r = 3.5 = 7/2, the 7 in the numerator cancels perfectly with the 7 in the denominator. Always look for this cancellation to avoid messy decimals.
Question 3 — Cuboid Melted into a Cylinder: Find the Radius
Question 3
A cuboid (22 cm × 15 cm × 7.5 cm) is melted and recast into a cylinder of height 14 cm. Find the radius of the cylinder.
When a solid is melted and recast into another shape, its volume is conserved. Set volume of cuboid = volume of cylinder and solve for r.
Volume of cuboid = Volume of cylinder ⟹ l × b × h = πr²h₁
22 × 15 × 7.5 = (22/7) × r² × 14
2475 = (22/7) × r² × 14
r² = (22 × 15 × 7.5 × 7) / (22 × 14)
r² = (15 × 7.5 × 7) / 14 = 787.5 / 14
r² = 56.25 = (7.5)²
r = 7.5 cm
✅ Radius of the cylinder = 7.5 cm
💡 Key concept — Conservation of Volume: Melting and recasting never changes the amount of material, only its shape. This principle is used in manufacturing, casting, and Class 10 problems where spheres are melted into cones or cylinders.
Question 4 — Height of a Water Tank Given Its Volume and Diameter
Question 4
An overhead water tank (cylinder) has capacity 61.6 m³ and diameter 5.6 m. Find its height.
Note: the problem gives diameter, not radius. Always halve it first.
Given: d = 5.6 m ⟹ r = 5.6/2 = 2.8 m, V = 61.6 m³
πr²h = 61.6
(22/7) × 2.8 × 2.8 × h = 61.6
(22/7) × (28/10) × (28/10) × h = 616/10
h = (616 × 7 × 100) / (10 × 28 × 28 × 22)
h = 25/10
h = 2.5 m
✅ Height of the water tank = 2.5 m
⛔ Most common mistake: Using diameter (5.6) directly as radius. Always calculate r = d/2 = 2.8 m before substituting into formulas.
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Question 5 — Surface Areas of a Hollow Cylindrical Metal Pipe
Question 5
r = inner radius = 2 cm
R = outer radius = 2.2 cm
A metal pipe is 77 cm long. Inner diameter = 4 cm, outer diameter = 4.4 cm. Find: (i) inner CSA, (ii) outer CSA, (iii) total surface area.
Hollow Pipe
R = outer radius = 2.2 cm
This is a hollow cylinder — it has an inner surface and an outer surface, plus two annular (ring-shaped) ends.
Total SA of hollow pipe = 2πrh + 2πRh + 2π(R² − r²)
Given: h = 77 cm, r (inner) = 4/2 = 2 cm, R (outer) = 4.4/2 = 2.2 cm
(i) Inner Curved SA = 2πrh
= 2 × (22/7) × 2 × 77 ← 77/7 = 11
= 968 cm²
(ii) Outer Curved SA = 2πRh
= 2 × (22/7) × 2.2 × 77
= 1064.8 cm²
(iii) Total SA = inner CSA + outer CSA + 2 × area of ring ends
= 968 + 1064.8 + 2π(R² − r²)
= 2032.8 + 2 × 3.14 × (2.2² − 2²)
= 2032.8 + 2 × 3.14 × (4.84 − 4)
= 2032.8 + 6.28 × 0.84
= 2032.8 + 5.2752
≈ 2038.08 cm²
✅ Inner CSA = 968 cm² | Outer CSA = 1064.8 cm² | Total SA = 2038.08 cm²
📌 The two ring-shaped ends each have area = π(R² − r²). Adding two such ends gives 2π(R² − r²). This formula is unique to hollow cylinders and is different from closed solid cylinders.
Question 6 — Cost of Painting Cylindrical Pillars
Question 6
A cylindrical pillar: diameter = 56 cm, height = 35 m. There are 16 such pillars. Find the cost of painting all curved surfaces at ₹5.50 per m².
Given: d = 56 cm ⟹ r = 28 cm = 28/100 m, h = 35 m
Step 1 — Lateral SA of 1 pillar:
= 2πrh = 2 × (22/7) × (28/100) × 35
= 2 × 22 × 4 × (1/100) × 35 ← 28/7 = 4
= 2 × 22 × 4 × 0.35 = 61.6 m²
Step 2 — Cost for 1 pillar:
= 61.6 × 5.50 = ₹338.80
Step 3 — Cost for 16 pillars:
= 338.80 × 16
= ₹5420.80
✅ Total painting cost for 16 pillars = ₹5,420.80
💡 Step-by-step strategy for cost problems: (1) Convert all units to metres. (2) Find area of one shape. (3) Multiply by cost per unit area. (4) Multiply by the number of shapes. Never skip a step — each has its own mark in board exams.
Question 7 — Area of Playground Levelled by a Cylindrical Roller
Question 7
A cylindrical roller: diameter = 84 cm, length = 120 cm. It takes 500 revolutions to level the entire playground. Find the area of the playground in m².
In one complete revolution, the roller's curved surface makes full contact with the ground once. So the area covered per revolution = Lateral SA of the roller.
Area covered per revolution = Lateral SA = 2πrh
Given: d = 84 cm ⟹ r = 42 cm, h (length) = 120 cm
Area in 1 revolution:
= 2πrh = 2 × (22/7) × 42 × 120
= 2 × 22 × 6 × 120 ← 42/7 = 6
= 31,680 cm²
Area in 500 revolutions:
= 31,680 × 500 = 1,58,40,000 cm²
Convert to m² (1 m² = 10,000 cm²):
= 1,58,40,000 / 10,000
= 1584 m²
✅ Area of the playground = 1584 m²
📌 Remember: 1 m = 100 cm, so 1 m² = 100 × 100 = 10,000 cm². Dividing by 10,000 converts cm² to m². This conversion step is frequently forgotten and costs marks in exams.
Question 8 — Inner Curved Surface Area and Plastering Cost of a Well
Question 8
A circular well: inner diameter = 3.5 m, depth = 10 m. Find (i) inner curved surface area, (ii) cost of plastering at ₹40 per m².
Given: d = 3.5 m ⟹ r = 3.5/2 = 7/4 m, h = 10 m
(i) Inner Curved SA:
= 2πrh = 2 × (22/7) × (7/4) × 10
= 2 × 22 × (1/4) × 10 ← 7/7 = 1
= 2 × 22 × 2.5
= 110 m²
(ii) Cost of plastering:
= 110 × ₹40
= ₹4400
✅ Inner CSA = 110 m² | Plastering cost = ₹4400
Question 9 — Petrol Storage Tank: Total SA and Steel Sheet with Wastage
Question 9
A closed cylindrical petrol tank: diameter = 4.2 m, height = 4.5 m. (i) Find its total SA. (ii) If 1/12 of steel is wasted, find steel sheet actually used.
Given: d = 4.2 m ⟹ r = 2.1 m, h = 4.5 m
(i) Total SA:
= 2πr(h + r) = 2 × 3.14 × 2.1 × (4.5 + 2.1)
= 2 × 3.14 × 2.1 × 6.6
= 13.188 × 6.6
≈ 87 m²
🔑 Part (ii) — Understanding the Wastage Logic:
- 1/12 of steel is wasted during manufacturing.
- So only 11/12 of the total sheet purchased is actually used to form the 87 m² tank.
- If 11 parts = 87 m², then 12 parts = (87/11) × 12 = 1044/11 ≈ 95 m²
(ii) Steel actually purchased:
11 parts of steel = 87 m² (the finished tank surface)
12 parts (total bought) = (87/11) × 12 = 1044/11
≈ 95 m² (approx)
✅ Total SA = 87 m² | Steel sheet used = ≈ 95 m²
Question 10 — Water Capacity of a Cylindrical Drum
Question 10
A one-side-open cylindrical drum: inner radius = 28 cm, height = 2.1 m. How much water can it store? (Answer in litres.)
Volume of the drum = volume of water it holds. Convert height to cm first (same unit as radius), then convert cm³ to litres.
1000 cm³ = 1 litre
Given: r = 28 cm, h = 2.1 m = 210 cm
Volume = πr²h = (22/7) × 28 × 28 × 210
= 22 × 4 × 28 × 210 ← 28/7 = 4
= 22 × 4 × 5880
= 5,17,440 cm³
In litres: 5,17,440 / 1000
= 517.44 litres
✅ Water capacity of the drum = 517.44 litres
🔵 Unit conversion chain: If r is in cm and h is in m, convert h to cm first (multiply by 100), compute volume in cm³, then divide by 1000 to get litres. Never mix cm and m inside a single formula.
Question 11 — Find Height by Dividing Volume by Curved Surface Area
Question 11
CSA = 1760 cm², Volume = 12320 cm³. Find the height of the cylinder.
When both CSA and Volume are given, divide Volume by CSA — the πrh terms cancel, leaving r/2 on one side. This is an elegant trick that avoids direct quadratic solving.
Volume / CSA = πr²h / 2πrh = r/2
Step 1 — Find r:
V / CSA = 12320 / 1760 = 7
r/2 = 7 ⟹ r = 14 cm
Step 2 — Find h using CSA:
2πrh = 1760
2 × (22/7) × 14 × h = 1760
2 × 22 × 2 × h = 1760 ← 14/7 = 2
88h = 1760
h = 1760 / 88
h = 20 cm
✅ Radius = 14 cm | Height = 20 cm
🔑 Board exam shortcut: Dividing Volume by Lateral SA always gives r/2. This is a guaranteed 3–4 mark question technique. Memorise the pattern: V ÷ CSA = r/2 and you can find r instantly without any trial-and-error.
Quick Reference — All Answers at a Glance
| Q | Problem Summary | Key Formula Used | Answer |
|---|---|---|---|
| 1 | Closed tank, r=56cm, h=1.4m | TSA = 2πr(h+r) | 6.89 m² |
| 2 | V=308 cm³, h=8cm → LSA & TSA | V=πr²h → r; then 2πrh & 2πr(h+r) | LSA=176, TSA=253 cm² |
| 3 | Cuboid melted → cylinder h=14cm | Vol cuboid = Vol cylinder | r = 7.5 cm |
| 4 | Tank V=61.6m³, d=5.6m | πr²h = V → h | h = 2.5 m |
| 5 | Hollow pipe, r=2cm, R=2.2cm, h=77cm | 2πrh, 2πRh, 2πrh+2πRh+2π(R²−r²) | 968, 1064.8, 2038.08 cm² |
| 6 | 16 pillars, d=56cm, h=35m, ₹5.50/m² | LSA × rate × 16 | ₹5420.80 |
| 7 | Roller d=84cm, l=120cm, 500 rev | LSA × 500 ÷ 10000 | 1584 m² |
| 8 | Well d=3.5m, h=10m, ₹40/m² | 2πrh; cost = area × rate | 110 m²; ₹4400 |
| 9 | Petrol tank d=4.2m, h=4.5m, wastage 1/12 | TSA = 2πr(h+r); scale by 12/11 | 87 m²; ~95 m² |
| 10 | Drum r=28cm, h=2.1m → litres | V=πr²h; ÷1000 | 517.44 litres |
| 11 | CSA=1760, V=12320 → h | V/CSA = r/2 → r → h | r=14cm, h=20cm |
Common Mistakes to Avoid
- Using diameter as radius: When the problem gives diameter, always divide by 2 first. This single mistake can make every subsequent step wrong.
- Not converting units: Mix of cm and m inside a formula gives a wrong answer. Always convert so that r and h are in the same unit before substituting.
- Using wrong value of π: Use 22/7 when values are multiples of 7; use 3.14 for decimals. Using 3.14 with multiples of 7 creates messy calculations unnecessarily.
- Forgetting the hollow pipe formula: For Q5 type problems, the ring end area is π(R² − r²), not πR². Students sometimes use the full circle instead of the annular ring.
- Stopping at cm² instead of converting to m²: For Q6 and Q7 type problems where the cost rate is per m², convert area from cm² to m² by dividing by 10,000.
- Volume/CSA shortcut (Q11): Many students try to separately find r and h by guessing. The division trick V ÷ CSA = r/2 directly gives r with minimal computation — use it always.
⛔ Board Exam Alert: Questions on the cylindrical well (Q8 type) and the roller problem (Q7 type) are the most frequently repeated in Telangana SSC and AP Board 9th class exams. Practice both until the steps are automatic.
What This Exercise Prepares You For
Exercise 10.2 builds strong intuition for cylinder problems that appear throughout higher classes. In Exercise 10.3, the same strategy of finding lateral and total surface areas extends to cones and spheres — the difference is just the formula. The "melting and recasting" concept from Q3 reappears extensively in Class 10's chapter on combinations of solids.
For students preparing for CBSE, Telangana SSC, and AP Board exams, cylinder problems are guaranteed to appear in every paper. Questions 5 (hollow pipe), 7 (roller), 9 (wastage), and 11 (volume ÷ CSA trick) are the most conceptually rich and are worth 4–5 marks each when they appear as long-answer questions.
📐 Board Exam Tips (Telangana & AP):
- Always write the formula first, then substitute — this earns the formula mark even if arithmetic is wrong.
- Write units (cm, m, cm², m², litres) at every step.
- For the roller problem, state clearly: "Area per revolution = Lateral SA of roller" — this reasoning line carries marks.
- For the wastage problem, always show the proportion setup: "11 parts = 87 m² ∴ 12 parts = ?" before computing.