Mensuration — Advanced Exam Problems

🎯 Learning Objective

Solve recasting, percentage-change and water-displacement problems with the volume-invariant idea.

💡 Concept

• Recasting (melt a sphere into a wire, a cube into cubes): VOLUME is the invariant — surface area is not
• Percentage change in area/volume: multiply each dimension's factor — r² takes its factor twice
• Submersion problems: displaced water volume = the solid's volume, spread over the vessel's cross-section
• Counting pieces after melting: big volume ÷ small volume, valid only when nothing is wasted
• π almost always cancels — keep it as a symbol till the last line

🧮 Key Formulas

Recast: V(original) = V(new shape)

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Volume MF = product of each dimension's MF (squared dimensions count twice)

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Water rise h = V(solid)/(πR²) in a vessel of radius R

✏️ Easy Example

Q. A cube of side 4 cm is melted and recast into small cubes of side 1 cm. How many small cubes are formed, and what is the ratio of their total surface area to the big cube's?

1. Melting preserves VOLUME — count = big volume ÷ small volume = 4³/1³ = 64
2. Big cube TSA = 6 × 4² = 96 cm²
3. All small cubes together: TSA = 64 × 6 × 1² = 384 cm²
4. Ratio = 384 : 96 = 4 : 1 — surface area grows when material is cut smaller

Answer: 64 cubes; TSA ratio 4 : 1

🇮🇳 Real-Life Example

A goldsmith melts an old bangle into a chain — the gold stays the same, only the shape changes. His maths is example one.

📝 Exam-Level Example

Q. A solid copper sphere of radius 3 cm is melted and drawn into a wire of diameter 0.4 cm. Find the length of the wire.

1. Melting keeps volume constant — equate the sphere's volume to the wire's (a long thin cylinder)
2. Sphere: V = (4/3)πr³ = (4/3)π × 27 = 36π cm³
3. Wire radius = diameter/2 = 0.2 cm, so wire volume = π × (0.2)² × L = 0.04πL
4. Equate: 0.04πL = 36π — π cancels from both sides, the usual cleanup
5. L = 36/0.04 = 900 cm = 9 m

Answer: 900 cm (9 metres)

📝 Exam-Level Example

Q. The radius of a cylinder is increased by 20% and its height is decreased by 20%. Find the percentage change in its volume.

1. V = πr²h — r appears squared and h once, so r's change acts twice and h's change once
2. New volume factor = (1.2) × (1.2) × (0.8) — one multiplying factor per dimension occurrence
3. (1.2)² = 1.44; then 1.44 × 0.8 = 1.152
4. Factor 1.152 means the new volume is 115.2% of the original
5. Change = 115.2 − 100 = +15.2% increase

Answer: 15.2% increase

📝 Exam-Level Example

Q. A solid metal sphere of radius 6 cm is completely submerged in a cylindrical vessel of radius 12 cm containing water. Find the rise in the water level.

1. A submerged solid displaces water equal to ITS OWN volume — Archimedes in exam form
2. Sphere volume = (4/3)π × 6³ = (4/3)π × 216 = 288π cm³
3. The displaced water forms a flat cylindrical layer of the VESSEL's radius 12: volume = π × 12² × h = 144πh
4. Equate the two: 144πh = 288π → π cancels
5. h = 288/144 = 2 cm rise

Answer: 2 cm

🪄 Memory Trick

Dimension changes: give every dimension its MF and multiply — squared dimensions get the factor twice, cubed thrice. 20% up on r → 1.2 × 1.2.

⚠️ Common Mistakes

• ❌ Equating surface areas in melting questions — only VOLUME survives recasting
• ❌ Applying r's 20% only once although volume carries r² (it acts twice)
• ❌ Using the sphere's radius instead of the vessel's radius for the water-rise layer

🏆 Exam Tips

• ✅ Write 'volume before = volume after' as the first line of every recast question
• ✅ Keep π symbolic — it cancels in almost every such equation
• ✅ Anchor checks: radius doubles → area ×4, volume ×8

📌 Summary

• Melting or recasting → volumes equal, π cancels
• % changes → chain multiplying factors per dimension
• Submerged solid → its volume spreads as a πR²h layer
• Cutting smaller raises total surface area — volume alone is safe