Exercise 12.2 answers: Surface Areas and Volumes

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1. Radius of both the cone and hemisphere is 1 cm; the cone's height also equals 1 cm.
2. Since they join along a flat circle, which takes up no room, the volumes simply add — nothing is shared and nothing is lost.
3. Volume = (2/3)πr³ (hemisphere) + (1/3)πr²h (cone) = (2/3)π(1)³ + (1/3)π(1)²(1) = 2π/3 + π/3 = π.
4. Volume of the solid = π cm³ — leave it exactly as π, since the question asks for the answer in terms of π.

Answerπ cm³

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1. Diameter 3 cm gives radius 1.5 cm, shared by the cylinder and both cones.
2. The two cones take up 2 + 2 = 4 cm of the 12 cm length, so the cylinder is 12 − 4 = 8 cm long.
3. The sheet is thin, so the air inside is the volume of the whole model: cylinder + two cones. The pieces meet only along flat circles, so their volumes simply add.
4. Cylinder: πr²h = π × 1.5² × 8 = 18π cm³. One cone: (1/3)πr²h = (1/3) × π × 1.5² × 2 = 1.5π cm³, so the two cones make 3π cm³.
5. Total = 18π + 3π = 21π cm³. Taking π = 22/7, as the exercise says: 21 × 22/7 = 66 cm³.

Answer66 cm³

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1. Diameter 2.8 cm gives radius 1.4 cm, shared by the cylinder and both hemispherical ends.
2. The two ends take up 1.4 + 1.4 = 2.8 cm of the 5 cm length, so the cylindrical part is 5 − 2.8 = 2.2 cm long.
3. The two hemispheres together make one sphere of radius 1.4 cm. Volume of one gulab jamun = πr²h + (4/3)πr³ = (22/7) × 1.4² × 2.2 + (4/3) × (22/7) × 1.4³ = 13.552 + 11.4987 = 25.0507 cm³ (to 4 decimal places).
4. Volume of 45 gulab jamuns = 45 × 25.0507 ≈ 1127.28 cm³.
5. Syrup is about 30% of this: 0.3 × 1127.28 ≈ 338.18 cm³, which is about 338 cm³.

Answer42273/125 cm³ = 338.184 cm³, so about 338 cm³ of syrup

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1. Volume of the solid cuboid = 15 × 10 × 3.5 = 525 cm³.
2. Each conical depression has radius 0.5 cm and depth 1.4 cm, so its volume is (1/3)πr²h = (1/3)×(22/7)×0.5²×1.4.
3. One cone's volume ≈ 0.3667 cm³, so four of them together ≈ 1.4667 cm³.
4. Volume of wood = 525 − 1.4667 ≈ 523.53 cm³.

Answer7853/15 cm³, which is about 523.53 cm³

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1. Volume of the cone-shaped vessel = (1/3)πr²h = (1/3)π×5²×8 = 200π/3 cm³.
2. One-fourth of this water spills out: (1/4)×200π/3 = 50π/3 cm³ — and this equals the total volume of the lead shots dropped in.
3. Volume of one lead shot (radius 0.5 cm) = (4/3)πr³ = (4/3)π×0.5³ = π/6 cm³.
4. Number of shots = (50π/3) ÷ (π/6) = (50/3)×6 = 100.

Answer100 lead shots

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1. The first cylinder has base diameter 24 cm, so radius 12 cm, and height 220 cm.
2. The second cylinder has radius 8 cm and height 60 cm.
3. Since the two cylinders meet on a flat circle, their volumes simply add: π×12²×220 + π×8²×60 = 3.14×31680 + 3.14×3840 = 99475.2 + 12057.6 = 111532.8 cm³.
4. Mass = volume × 8 g/cm³ = 111532.8 × 8 = 892262.4 g = 892.26 kg (approximately).

Answer892.2624 kg, which is about 892.26 kg

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1. The cone (height 120 cm, radius 60 cm) stands on the hemisphere (radius 60 cm), so the solid's total height is 120 + 60 = 180 cm — exactly the cylinder's height, so it fits with nothing to spare.
2. Volume of the solid = (1/3)πr²h + (2/3)πr³ = (1/3)π×60²×120 + (2/3)π×60³ = 144000π + 144000π = 288000π cm³.
3. Volume of the cylinder = πr²h = π×60²×180 = 648000π cm³.
4. Water left = 648000π − 288000π = 360000π = 360000×22/7 = 7920000/7 ≈ 1131428.57 cm³ (about 1.13 m³).

Answer7920000/7 cm³, about 1,131,428.57 cm³, which is about 1.131 m³

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1. The cylindrical neck has diameter 2 cm, so radius 1 cm, and length 8 cm.
2. The spherical part has diameter 8.5 cm, so radius 4.25 cm.
3. Since the neck and sphere join along a flat circle, their volumes add: (4/3)πR³ + πr²h = (4/3)×3.14×4.25³ + 3.14×1²×8.
4. This works out to about 321.39 + 25.12 = 346.51 cm³, not 345 cm³ — so the child's measurement is not correct (it is about 1.51 cm³ too low).

AnswerThe actual volume is 831629/2400 cm³, about 346.51 cm³, so the child's figure of 345 cm³ is not correct

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