When a piece has been scooped out: apparent capacity against actual

What to assume they know

• Why volumes do add even though surface areas do not — that the volumes of joined solids add, and why
• Decomposing an everyday object into the basic solids — decomposing an object into basic solids and reading part dimensions out of overall ones
• The Class IX volume formulas, in particular cylinder πr²h, cone (1/3)πr²h, sphere (4/3)πr³ and hemisphere (2/3)πr³
• That 1 cm³ of water is 1 millilitre, so a capacity in cubic centimetres is directly a capacity in millilitres
• Percentage of a quantity, and division of one volume by another to get a count
• That a solid fully under water occupies space the water previously occupied

What they should be able to do

• Compute the apparent capacity of a vessel from its outline alone
• Compute the actual capacity by subtracting the volume of any raised or intruding portion, and state the shortfall
• Express the shortfall as a fraction of the apparent capacity, and show that this fraction is independent of the value taken for π
• Find the volume of material remaining in a block after conical or hemispherical hollows have been made in it
• Use the fact that an immersed solid displaces its own volume of water to find a count, a level or a remaining volume
• Compute the water left in a container after a composite solid is stood in it
• Test a stated measurement against a computed volume and say whether the claim stands
• Explain why a hollow changes volume by subtraction while it changes surface area by addition

Where it usually goes wrong

• "Apparent and actual capacity are two different formulas." They are one computation with a term removed. Apparent is what you get if you forget the hemisphere; actual is what you get if you remember it.
• "The raised base is glass, so it should be added." It is glass, and glass is not juice. The question asks what the glass holds, so the space the glass itself occupies is subtracted from the outline.
• "Hollowing a solid must reduce its surface area as well as its volume." It reduces the volume and increases the surface area, by πr² for a hemispherical scoop. The two measurements move in opposite directions.
• "You have to know π accurately to say how much the customer loses." The fraction lost is 2r/(3h), with every π cancelled — one sixth for this glass, whatever value is used. The same cancellation makes the lead-shot count exactly 100.
• "The shots make the water rise, so add their volume to the vessel's." The vessel was already full. The shots take up room inside it and the same volume of water leaves; nothing rises.
• "A quarter of the water flows out, so the shots are a quarter of a sphere's worth." A quarter of the water, which is a quarter of the cone's capacity. Work out that volume first and only then divide by one shot.
• "The solid in question 7 might not fit." Check it: 120 + 60 = 180, exactly the cylinder's height. It fits precisely, and noticing that is part of reading the question.
• "Four small holes in a block are too small to matter." They remove about 0.28% here — but the same reasoning removes a sixth of the juice glass. Whether a correction is negligible is something you compute, not something you assume.
• "If the child's number is close, she is correct." About 346.5 against a reported 345 is close and still not equal. The question is a yes-or-no about a measurement, and the answer is no.

Questions to check understanding

• Find the apparent and actual capacity of a vessel with a raised or intruding base, and state the difference
• Express the loss in capacity as a fraction or percentage of the apparent capacity
• Find the volume of material left in a block after cylindrical, conical or hemispherical hollows are made
• Given that a stated fraction of the water in a full container spills, find the number of identical solids dropped in
• Find the volume of water remaining after a composite solid is stood in a full container
• Decide whether a reported volume for a described vessel is correct, and by how much it is out
• Convert a computed capacity in cubic centimetres to litres or millilitres
• Explain why hollowing a solid lowers its volume but raises its surface area

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 161–170.

• Example 6, the juice seller's glass (p. 168, Fig. 12.13). Inputs: the glass is a cylinder measuring 5 cm across on the inside and 10 cm tall; the base of the glass rises into the bowl as a hemisphere, which cuts down what it holds; π taken as 3.14. Read from the printed page: Fig. 12.13 draws a plain cylinder with a dome bulging upward from the bottom, its lower outline dashed where it is hidden by the glass. No dimension is printed on the figure at all; the question supplies three numbers — the 5 cm inner diameter, the 10 cm height and π as 3.14 — and the radius of 2.5 cm is derived rather than given. Verified: radius 2.5 cm; apparent capacity πr²h = 3.14 × 6.25 × 10 = 196.25 cm³; the intruding hemisphere measures (2/3)π(2.5)³ = (2/3)(3.14)(15.625) ≈ 32.71 cm³; so the glass really holds 196.25 − 32.71 = 163.54 cm³.
• The shortfall as a fraction — the point of the section, and not in the book. Verified as algebra: the ratio of the raised hemisphere to the apparent cylinder is (2/3)πr³ ÷ πr²h = 2r/(3h). Every π cancels, so the fraction does not care what value you take for it. Here 2 × 2.5 ÷ (3 × 10) = 1/6 exactly: the customer loses one sixth of the glass, and would lose one sixth whether π were 3.14 or 22/7. This is the version of the result a student can carry out of the room, and it turns an arithmetic exercise into a statement about the object.
• Where the minus sign comes from. A solid occupies the points inside its outer boundary and outside every hollow. So its volume is the outline's volume less each hollow's, for the same reason that joined volumes add: the surfaces separating the regions have no volume of their own. Write it once as a signed sum and every remaining example in the chapter is an instance of it.
• The sign flips between area and volume. The same hemispherical hollow that subtracts (2/3)πr³ from a volume adds πr² to a surface area, as Which faces vanish at the join, and why you cannot simply add establishes from the bird-bath and the scooped article. Students who have just learned that hollowing raises the area will expect it to raise the volume too. Put the two ledgers side by side.
• Exercise 12.2 q. 4, the pen stand (p. 170, Fig. 12.16). Inputs: a wooden cuboid 15 cm by 10 cm by 3.5 cm with four conical hollows drilled to hold pens, each of radius 0.5 cm and depth 1.4 cm; π taken as 22/7. Verified: the cuboid is 15 × 10 × 3.5 = 525 cm³; each cone is (1/3)(22/7)(0.25)(1.4) = 1.1/3 cm³, so four of them come to 4.4/3 ≈ 1.4667 cm³; the wood left is about 523.53 cm³. Read from the printed page: Fig. 12.16 has two panels. The left is an oblique pictorial of the block, seen from above and in front, with the four depressions drawn as shaded ellipses on its top face — not a side view and not dark holes. The right panel is a sectional elevation: four pens with V-shaped tips standing inside a plain rectangular outline of the block. It carries no measurements, but it is not decoration — it is the only place in the chapter where the depressions are drawn as cones, which is the fact the whole question turns on.
• How small the correction is, and why that matters. Verified: the four hollows remove about 0.28% of the block.
• The chapter's other hollowed objects, for the cross-link in section 8. The bird-bath, a cylinder 1.45 m high of radius 30 cm with a hemisphere hollowed into its top (Example 4, p. 166, Fig. 12.9); the cylinder 2.4 cm high and 1.4 cm across with a matching cone drilled out (Exercise 12.1 q. 8, p. 167); the cylinder 10 cm high of radius 3.5 cm with a hemisphere taken from each end (Exercise 12.1 q. 9, p. 167, Fig. 12.11). The chapter asks only for their surface areas and never for their volumes. Verified, as an added extension: the bird-bath's remaining material is π(30)²(145) − (2/3)π(30)³ = 130 500π − 18 000π = 112 500π cm³, and the scooped article of q. 9 is π(3.5)²(10) − 2 × (2/3)π(3.5)³ ≈ 122.5π − 57.17π ≈ 65.33π cm³. Offer these as extension work and say plainly that the chapter does not ask them.
• Exercise 12.2 q. 5, the lead shots (p. 170). Inputs: a vessel shaped as an inverted cone, 8 cm high, open top of radius 5 cm, filled to the brim; lead shots are spheres of radius 0.5 cm; dropping them in makes a quarter of the water spill out; the number of shots is wanted. Verified: the vessel holds (1/3)π(25)(8) = 200π/3 cm³, so the water that leaves is 50π/3 cm³; each shot has volume (4/3)π(0.125) = π/6 cm³; the count is (50π/3) ÷ (π/6) = 100. The π cancels again, so the count is exact and independent of the value used. The physical step — that the water lost equals the volume of the shots put in — is the whole question, and it is the one thing the arithmetic cannot supply.
• Exercise 12.2 q. 7, the solid standing in water (p. 170). Inputs: a cone 120 cm high, radius 60 cm, mounted on a 60 cm hemisphere; the whole thing is set upright inside a water-filled cylinder of the same 60 cm radius and 180 cm tall, low enough to touch its floor; the water remaining is wanted. Verified: the solid measures (1/3)π(3600)(120) + (2/3)π(216 000) = 144 000π + 144 000π = 288 000π cm³; the cylinder holds π(3600)(180) = 648 000π cm³; the water left is 360 000π cm³, which at π = 22/7 is about 1 131 429 cm³, near 1.13 m³. Two things worth pointing out: the solid's total height is 120 + 60 = 180 cm, exactly the cylinder's height, so it fits with nothing to spare; and the cone and the hemisphere here happen to have equal volumes, since a cone of height 2r has the same (2/3)πr³ as the hemisphere.
• Exercise 12.2 q. 8, checking a child's measurement (p. 170). Inputs: a spherical glass vessel with a cylindrical neck; the neck is 8 cm long and 2 cm across; the spherical part is 8.5 cm across; a child measures the water it holds and reports 345 cm³; π taken as 3.14, and all figures to be treated as inside measurements. Verified: neck radius 1 cm, sphere radius 4.25 cm; the volume is π(1)²(8) + (4/3)π(4.25)³ = 8π + 102.354π = 110.354π ≈ 346.51 cm³. The reported 345 cm³ is therefore about 1.5 cm³ low — close, but not right. This is the only question in the chapter that hands the student an answer and asks whether to believe it, which makes it the natural close of the topic.
• Where the syrup question sits. Exercise 12.2 q. 3, the gulab jamun at 30% of volume, is an additive computation followed by a percentage, and belongs to Why volumes do add even though surface areas do not. Mention it here only to distinguish taking a fraction of a volume from subtracting a hollow.

Figures to have open

• Fig. 12.13's glass redrawn so the hemisphere can be toggled in and out, with the juice level and the two capacity figures updating. Chapter figure (p. 168); the toggle is added here and carries section 4.
• A three-region diagram — outline, hollow, remaining solid — that can be reused for the glass, the pen stand and the bird-bath. Standard schematic; not in the book.
• The cancellation of π written out as a single line of algebra, large enough to read, with 2r/(3h) evaluated at r = 2.5 and h = 10. Standard notation panel.
• Fig. 12.16's pen stand (p. 170), redrawn with the four conical hollows shown as solid cones lifted clear of the block. The printed left panel is a small oblique pictorial and the hollows read only as shaded ellipses at that size; the right panel already shows them as cones in section, and the redraw should build on that rather than replace it.
• A full container with a composite solid being lowered into it and the displaced water spilling over a lip into a measuring vessel. Standard schematic; not in the book, and it is the picture questions 5 and 7 both depend on. The chapter draws no such figure.
• A side elevation for question 7 showing the 120 cm cone and the 60 cm hemisphere stacked against the cylinder's 180 cm wall, so the exact fit is visible.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 12 "Surface Areas and Volumes", §12.3 "Volume of a Combination of Solids", Example 6 with Fig. 12.13, p. 168
• Exercise 12.2 questions 4, 5, 7 and 8, p. 170, with Fig. 12.16 (p. 170)
• Example 4 with Fig. 12.9, p. 166, and Exercise 12.1 questions 8 and 9, p. 167, with Fig. 12.11 — the chapter's hollowed objects, used here for the area-against-volume contrast of section 8
• §12.3's opening paragraph, p. 167, for the additive rule this topic runs backwards
• The chapter's summary, §12.4, p. 170, second point