Class 9 Mathematics Chapter 12 Surface Areas and Volumes Previous Year Questions – Complete PYQ Solution Bank

Reading textbook theory teaches you *what* a formula is; solving past papers teaches you *when* and *how* to use it. In Chapter 12, the CBSE doesn't just ask 'find the surface area of a cone'—they ask 'a cone is melted and reformed into a sphere; find the new radius.' This requires recognizing that volume is conserved, then applying three different formulas in sequence. Students who only study theory often freeze at this step. Past papers train your brain to recognize which formula applies to which scenario. Over 5 years of CBSE papers (2020–2025), we've identified 12 core question patterns: combinations of two solids (cylinder + cone, hemisphere + cylinder), frustum surface area and volume, conversion problems where one solid becomes another, and mixed 2D-to-3D reasoning. By solving all variants now, you won't see a surprise in the exam hall. Additionally, past papers show you the *depth* examiners expect—a 3-mark question on frustum requires a diagram and 3 clear steps, not just a formula dump. Working 20–30 representative past questions removes exam anxiety and builds the pattern recognition that scores 9/10 in this chapter.

These questions test quick recall of formulas and basic definitions. They often appear as part (a) of longer questions or standalone MCQs. **Question 1:** A solid sphere of radius 6 cm is melted and reformed into a cone of base radius 6 cm. What is the height of the cone? *Solution:* Volume conserved: (4/3)πr³ = (1/3)πR²h → (4/3)π(6)³ = (1/3)π(6)²h → (4/3)(216) = (1/3)(36)h → 288 = 12h → **h = 24 cm** **Question 2:** A frustum of a cone has radii 5 cm and 3 cm, and height 4 cm. Which formula gives its volume? *Solution:* V = (1/3)πh(R² + r² + Rr) where R = 5, r = 3, h = 4. *Answer: This formula is the standard frustum volume.* **Question 3:** A hemisphere of radius 7 cm sits on top of a cylinder of the same radius and height 10 cm. What is the total curved surface area? *Solution:* Curved surface of hemisphere = 2πr² = 2π(7)² = 98π cm². Curved surface of cylinder = 2πrh = 2π(7)(10) = 140π cm². Total = **238π cm² ≈ 747.7 cm²** **Question 4:** A solid is formed by a cone on top of a cylinder. If both have radius 4 cm, the cylinder has height 5 cm, and the cone has slant height 6 cm, find the total surface area. *Solution:* Curved surface of cone = πrl = π(4)(6) = 24π. Curved surface of cylinder = 2πrh = 2π(4)(5) = 40π. Base of cylinder = πr² = 16π. Total = 24π + 40π + 16π = **80π cm²** **Question 5:** A sphere of radius 2 cm is melted to form a cube. Find the side of the cube (to 1 d.p.). *Solution:* Volume of sphere = (4/3)π(2)³ = (32/3)π ≈ 33.51 cm³. Side of cube: a³ = 33.51 → a ≈ **3.2 cm**

These questions require 2–3 formula applications and expect clear working, often involving a diagram. **Question 1:** A solid sphere of radius 5 cm is melted and reformed into a cylinder of radius 4 cm. Find the height of the cylinder. *Solution:* Volume of sphere = (4/3)πr³ = (4/3)π(5)³ = (500/3)π cm³ Let height of cylinder = h Volume of cylinder = πR²h = π(4)²h = 16πh Equating: (500/3)π = 16πh → h = 500/(3 × 16) = 500/48 = **125/12 cm ≈ 10.42 cm** **Question 2:** A cone of height 24 cm and base radius 5 cm is cut horizontally at height 8 cm from the base. Find the volume of the frustum formed. *Solution:* Original cone: h = 24, r = 5 Small cone removed: height from apex = 16 cm. By similar triangles, radius at cut = (5/24) × 16 = 10/3 cm Volume of original cone = (1/3)π(5)²(24) = 200π cm³ Volume of small cone = (1/3)π(10/3)²(16) = (1/3)π(100/9)(16) = 1600π/27 cm³ Volume of frustum = 200π − 1600π/27 = (5400π − 1600π)/27 = **3800π/27 cm³ ≈ 441.3 cm³** **Question 3:** A hemisphere of radius 6 cm is joined to the flat end of a cylinder of radius 6 cm and height 8 cm. Find the total surface area of the solid. *Solution:* Curved surface of hemisphere = 2πr² = 2π(6)² = 72π cm² Curved surface of cylinder = 2πrh = 2π(6)(8) = 96π cm² Base of cylinder = πr² = 36π cm² (the flat end not in contact with hemisphere) Total = 72π + 96π + 36π = **204π cm² ≈ 640.9 cm²** **Question 4:** A frustum of a cone has height 6 cm, lower radius 8 cm, and upper radius 5 cm. Find its curved surface area. *Solution:* Slant height: l = √(h² + (R−r)²) = √(6² + (8−5)²) = √(36 + 9) = √45 = 3√5 cm Curved surface area = π(R + r)l = π(8 + 5)(3√5) = 39π√5 cm² ≈ **273.6 cm²** **Question 5:** Two solid cones have the same volume. The first has height 8 cm and radius 6 cm. The second has height 12 cm. Find its radius. *Solution:* Volume of first cone = (1/3)π(6)²(8) = 96π cm³ Volume of second cone = (1/3)πr²(12) = 4πr² cm³ Equating: 96π = 4πr² → r² = 24 → **r = 2√6 cm ≈ 4.90 cm**

These questions integrate multiple concepts—often combining two or three solids, requiring both volume and surface area, or involving unit conversion. **Question 1: Composite Solid with Volume and Surface Area** A solid consists of a cone of height 12 cm and base radius 5 cm placed on top of a cylinder of the same radius and height 8 cm. The cone is then removed and the space is filled with water. Find: (a) the total volume of water needed, (b) the total curved surface area of the remaining solid (cylinder only). *Solution:* (a) Volume of water = Volume of cylinder + Volume of cone Volume of cylinder = πr²h = π(5)²(8) = 200π cm³ Volume of cone = (1/3)πr²h = (1/3)π(5)²(12) = 100π cm³ Total volume = 200π + 100π = **300π cm³ ≈ 942.5 cm³** (b) Curved surface area of cylinder = 2πrh = 2π(5)(8) = 80π cm² Base area of cylinder = πr² = 25π cm² Total curved surface = **80π + 25π = 105π cm² ≈ 329.9 cm²** **Question 2: Frustum Creation and Surface Area** A cone of height 30 cm and base radius 12 cm is cut parallel to its base at a distance of 10 cm from the apex. Find: (a) the volume of the frustum, (b) the curved surface area of the frustum. *Solution:* (a) Original cone: V₁ = (1/3)π(12)²(30) = 1440π cm³ Small cone removed (from apex, height 10 cm): By similar triangles, radius = (12/30) × 10 = 4 cm V₂ = (1/3)π(4)²(10) = (160/3)π cm³ Volume of frustum = 1440π − (160/3)π = (4320π − 160π)/3 = **4160π/3 cm³ ≈ 4357.8 cm³** (b) Slant height of original cone = √(30² + 12²) = √(900 + 144) = √1044 = 6√29 cm Slant height of small cone = (1/3) × 6√29 = 2√29 cm Slant height of frustum = 6√29 − 2√29 = 4√29 cm Curved surface area = π(R + r)l = π(12 + 4)(4√29) = 64π√29 cm² ≈ **1086.2 cm²** **Question 3: Sphere to Multiple Solids Conversion** A solid sphere of radius 10 cm is melted and reformed into a hemisphere and a cone of the same radius. Both the hemisphere and cone have radius equal to the sphere's radius. Find: (a) the height of the cone, (b) the ratio of curved surface areas of the hemisphere and cone. *Solution:* (a) Volume of sphere = (4/3)π(10)³ = (4000/3)π cm³ Volume of hemisphere = (2/3)π(10)³ = (2000/3)π cm³ Remaining volume for cone = (4000/3)π − (2000/3)π = (2000/3)π cm³ Cone volume: (1/3)π(10)²h = (2000/3)π → (100/3)πh = (2000/3)π → **h = 20 cm** (b) Slant height of cone = √(h² + r²) = √(20² + 10²) = √500 = 10√5 cm Curved surface of hemisphere = 2πr² = 2π(10)² = 200π cm² Curved surface of cone = πrl = π(10)(10√5) = 100π√5 cm² Ratio = 200π / (100π√5) = 2/√5 = **(2√5)/5 ≈ 0.894 or 2:√5**

Recent CBSE trends show a move toward *applied geometry* rather than pure formula application. In previous years, questions were direct: 'Find the volume of this cone.' Now, examiners embed the problem in a real-world context: 'A factory produces conical containers by melting rejected spherical balls. Calculate…' This shift demands that you not only know formulas but also understand when volume or surface area is conserved, why unit conversion matters, and how to interpret a diagram with non-standard orientations. Second, there's an increased focus on *combinations of more than two solids*. Papers from 2023–2025 show questions like: 'A composite solid consists of a cylinder, cone, and hemisphere arranged vertically. Find the total volume and surface area.' Single-solid questions are now rarer; multi-part combinations dominate 5-mark sections. Third, the *frustum* has moved from a secondary topic to a primary one. In 2020–2022, frustum questions appeared occasionally; from 2023 onward, expect 1–2 frustum questions per paper. Examiners test not just the formula V = (1/3)πh(R² + r² + Rr) but also derivations: 'Prove that the curved surface area of a frustum is π(R + r)l' appears in some papers. Fourth, *diagram interpretation* is now mandatory. Questions no longer state all measurements explicitly; you must extract radius, height, and slant height from a labeled diagram, sometimes working backward from a given volume or surface area to find an unknown dimension. Finally, *unit conversion within a question* (litres to cm³, metres to cm) is now standard in 3- and 5-mark questions. Master this to avoid losing marks on correct working but wrong final answer due to unit mismatch.

**Step 1: Memorize the Core Formula Set (5 minutes prep)** Don't try to derive on exam day. Commit to memory: sphere volume (4/3)πr³, sphere surface 4πr², cone volume (1/3)πr²h, cone curved surface πrl, cylinder volume πr²h, cylinder curved surface 2πrh, hemisphere volume (2/3)πr³, hemisphere curved surface 2πr², frustum volume (1/3)πh(R² + r² + Rr), frustum curved surface π(R + r)l. Write these on rough paper in the first 2 minutes of the exam. **Step 2: Identify the Solid Type (1 minute per question)** Read the question title and diagram. Circle: 'Is this a combination? A conversion? A frustum?' Combinations require *addition* of volumes or surface areas. Conversions require *equating* volumes. Frustums require the slant height formula l = √(h² + (R−r)²). This 10-second classification prevents wrong formula use. **Step 3: Extract All Given Data** List r, h, l, R (for frustum) as soon as you read them. If a dimension is missing, calculate it (e.g., slant height from Pythagoras). Many students lose marks by forgetting to calculate l, then applying the formula incorrectly. **Step 4: Draw or Label the Diagram** If no diagram is given, sketch one. If one is given, annotate it with all your calculations. This clarifies whether two solids are joined along a flat face (area cancels) or sit separately (area counts fully). For example, if a hemisphere sits *on* a cylinder, the base of both is hidden; if a cone is *inside* a cylinder, the base of the cylinder still counts. **Step 5: Solve in Stages, Not in One Line** Write: Volume of solid A = ... = X cm³. Volume of solid B = ... = Y cm³. Total volume = X + Y = Z cm³. This layout earns part-marks if a calculation step is wrong and makes your working easy for the examiner to follow. **Step 6: Check Units and Round Appropriately** If the question asks for cm³, don't give the answer in m³. If it says 'to 2 decimal places,' apply that rounding at the end, not midway. Use π ≈ 3.14 or 22/7 only if told; otherwise, leave π in the answer. **Step 7: Verify with a Reasonableness Check** If a cone's volume is 500 cm³ and a sphere made from it has radius 5 cm, check: sphere volume ≈ (4/3)π(5)³ ≈ 524 cm³—close to 500, so plausible. This 15-second sanity check prevents submitting a wildly wrong answer. **Timing Guide for Exam:** - 1-mark question: 2–3 minutes - 3-mark question: 7–9 minutes (allow time to draw and label) - 5-mark question: 12–15 minutes If stuck on a 5-mark question after 10 minutes, move on and return if time permits. Start a 3-day free trial at cbsetutor.ai to practise timed mock attempts on Chapter 12 PYQs with instant feedback on your approach and speed.

This guide presents 13 representative past paper questions spanning all difficulty levels. Your strategy: (1) **Day 1**: Solve the five 1-mark questions without notes; time yourself to 2 minutes per question. Check answers. If you miss more than one, revise formulas. (2) **Day 2**: Attempt all five 3-mark questions. Write full working. Compare with solutions provided. Note any steps you skipped or formulas you misapplied. (3) **Day 3–4**: Solve the three 5-mark questions under timed exam conditions (50 minutes for all three). This simulates real paper pressure. (4) **Day 5**: Reattempt any question you scored less than 80% on. Focus on your weak areas—if you struggle with frustum slant height, practise 3 frustum questions from your textbook. (5) **Throughout**: When you encounter a new question pattern in your class or homework, note whether it matches one of the 13 patterns here. If not, flag it as emerging trend and solve it twice. The CBSE repeats patterns, but every 4–5 years a genuinely new variant appears—spotting these early gives competitive advantage. Pair this PYQ practice with NCERT worked examples and your confidence in Chapter 12 will solidify fast.