# TNPSC Group IV – Day 14 | Volume of Cubes and Cuboids
Date: Monday, 26 October 2026 | Time: 4:00 PM IST Category: TNPSC Grp IV | Subject: Aptitude & Mental Ability | Day: 14/90 Official syllabus area: Volume (SSLC standard).
## 1. Learning Goal This lesson continues yesterday's Area and Perimeter topic. We will learn the volume, lateral surface area, total surface area and capacity of cubes and cuboids. Cylinders, cones and spheres will be covered in separate lessons. Earlier percentages and interest formulas appear only in the scheduled weekly revision.
## 2. Definitions Volume: The space occupied by a three-dimensional object. Common units: cm³ and m³. Capacity: The quantity of liquid a container can hold. Common units: millilitres and litres. Surface area: The sum of the areas of the outer faces of a solid. Common units: cm² and m². Lateral Surface Area (LSA): Area of the side faces, excluding the top and bottom. Total Surface Area (TSA): Area of all outer faces. Cuboid: A solid with six rectangular faces, with length l, breadth b and height h. Cube: A solid with six square faces, each edge having the same length a.
## 3. Essential Formulas Cuboid volume V = l × b × h. Cuboid LSA = 2h(l + b). Cuboid TSA = 2(lb + bh + hl). Cube volume V = a³. Cube LSA = 4a². Cube TSA = 6a². Cube edge from its volume: a = ∛V. Cuboid height from its volume: h = V/(lb).
## 4. Unit Conversions 1 m = 100 cm; 1 m³ = 1,000,000 cm³. 1 cm³ = 1 millilitre. 1,000 cm³ = 1 litre. 1 m³ = 1,000 litres. Use square units (m²) for surface area and cubic units (m³) for volume. Convert all lengths to a common unit before multiplying.
## 5. Fully Worked Examples Examples B01–B04 are paraphrased from specified exercises in the 2019 Tamil Nadu Class 9 Mathematics textbook, Chapter 7, Exercise 7.3. Their presence and numbering in the 2026–27 edition have not yet been checked.
### B01 – Verified historical book-back: Exercise 7.3, Question 1(i), 1(ii) (2019 edition) (i) Find the volume of a cuboid measuring 12 cm by 8 cm by 6 cm. V = lbh = 12 × 8 × 6 = 576 cm³. (ii) Find the volume of a cuboid measuring 60 m by 25 m by 1.5 m. V = 60 × 25 × 1.5 = 2,250 m³. Both subparts are completed. Keep the unit appropriate to each set of measurements.
### B02 – Verified historical book-back: Exercise 7.3, Question 2 (2019 edition) Each matchbox measures 6 cm × 3.5 cm × 2.5 cm. Find the total volume of a packet containing 12 such boxes. Volume of one box = 6 × 3.5 × 2.5 = 52.5 cm³. Total volume = 12 × 52.5 = 630 cm³. Shortcut: For n identical objects, total volume = n × volume of one object.
### B03 – Verified historical book-back: Exercise 7.3, Question 6 (2019 edition) A cuboidal container has a volume of 1,440 m³, length 15 m and breadth 8 m. Find its height. From V = lbh, height h = V/(lb). h = 1440/(15 × 8) = 1440/120 = 12 m. Check: 15 × 8 × 12 = 1,440 m³.
### B04 – Verified historical book-back: Exercise 7.3, Question 7(i), 7(ii), 7(iii) (2019 edition) For a cube of edge a, volume V = a³. (i) a = 5 cm → V = 5³ = 125 cm³. (ii) a = 3.5 m → V = 3.5³ = 42.875 m³. (iii) a = 21 cm → V = 21³ = 9,261 cm³. Notice that the cubic unit changes when the length unit changes.
### M05 – Original model: Volume, LSA and TSA of a cube A cube has side 4 cm. Find its volume, lateral surface area and total surface area. V = 4³ = 64 cm³. LSA = 4 × 4² = 4 × 16 = 64 cm². TSA = 6 × 4² = 6 × 16 = 96 cm². Although 64 appears twice, 64 cm³ is a volume and 64 cm² is an area. Never interchange these units.
### M06 – Original model: Painting a room A room is 6 m long, 4 m wide and 3 m high. Find the cost of painting its four walls and ceiling at ₹25 per m². Ignore doors and windows. Area of four walls = 2h(l+b) = 2 × 3 × (6+4) = 60 m². Ceiling area = lb = 6 × 4 = 24 m². Total painting area = 60 + 24 = 84 m². Cost = 84 × ₹25 = ₹2,100. Shortcut: Four walls only → LSA. Four walls and ceiling → LSA + lb. All six outer faces → TSA.
### M07 – Original model: Water-tank capacity A rectangular water tank measures 2 m × 1.5 m × 1 m. What is its full capacity? Volume = 2 × 1.5 × 1 = 3 m³. Since 1 m³ = 1,000 litres, capacity = 3 × 1,000 = 3,000 litres.
## 6. Exam Shortcuts and Common Errors If a cube's edge doubles, its volume becomes 2³ = 8 times and its surface area becomes 2² = 4 times. If only the height of a cuboid doubles, its volume doubles. For n identical boxes, total volume = n × lbh. If a solid is melted and recast without material loss, old volume = new volume. For four walls, use 2h(l+b). For the complete outer surface, use 2(lb+bh+hl). Remember: 1,000 litres = 1 m³, NOT 1,000 m³.
## 7. Weekly Spaced Revision Spend five minutes recalling yesterday's area and perimeter formulas. For the earlier percentage lesson, remember x% of y = xy/100. For earlier interest lessons, remember SI = PRT/100 and A = P(1+r/100)^n for annual compounding. These are planned revisions, not repeated new lessons. No General Studies content is included.
## 8. Textbook and Book-Back Progress Primary reference: Tamil Nadu Samacheer Kalvi Class 9 Mathematics, Chapter 7, Mensuration (2019 edition). Volume: Section 7.5, Exercise 7.3. Surface area: Section 7.4, Exercise 7.2. The 2019 chapter contains 10 parent questions in Exercise 7.1, 7 in Exercise 7.2, 9 in Exercise 7.3 and 10 in Exercise 7.4: 36 parent questions in total. Subparts are tracked separately, not counted as additional parent questions. Exercise 7.3 fully completed today: Questions 1, 2, 6 and 7 = 4/9 parent questions. Exercise 7.3 pending: Questions 3, 4, 5, 8 and 9 = 5/9. Verified progress within the historical 2019 Chapter 7 from today's work: 4/36 parent questions completed; 32/36 pending. This count does not claim to incorporate any previous days' exercise completions. Whether the 2026–27 textbook has the same complete exercise inventory: source not yet verified. Therefore, the 2026–27 completion percentage is not asserted. TNPSC Group IV PYQs checked against an official final key today: 0. M05–M07 are original models, not past-exam questions.
## 9. Revision Dates +1 day: 27 October 2026; +3 days: 29 October; +7 days: 2 November; +14 days: 9 November. The 7 PM daily practice will include two Aptitude questions based on today's volume lesson. Additional verified book-back questions will be counted separately from the 15 headline MCQs.
## 10. ETN Push Notification Title: 📦 TNPSC Group IV | Day 14 – Volume! Message: Learn cube and cuboid volume, tank capacity, surface area and shortcuts with complete worked solutions in today's 4 PM bilingual Aptitude lesson on ETN News.
