This MCQ module is based on: Volumes, Conversions of Solids and Exercises
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Volumes, Conversions of Solids and Exercises
🎓 Class 10
Mathematics
CBSE
Theory
Ch 12 — Surface Areas and Volumes
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Volumes, Conversions of Solids and Exercises
Targeting Class 10 level in Mensuration, with Intermediate difficulty.
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12.3 Volume of a Combination of Solids
Unlike surface area, volume is straightforwardly additive for combinations: if two solids join without overlap, the total volume is the sum of their individual volumes. If one solid is carved out of another, subtract the carved volume from the outer volume. There is no "hidden face" consideration for volume.
Principle
Volume of combination = Σ volumes of the constituent pieces (taken with + sign when added, − sign when carved out or removed).
Example 1 — Industrial shed
A shed of length 7 m, width 15 m, and wall-height 8 m has a half-cylindrical roof of diameter 7 m. Find the volume of air in the shed if there are no machines. Also find the air remaining if 20 workers occupy it (each worker takes 0.08 m³ and there is machinery taking 300 m³).
Solution. Cuboidal part: V₁ = 15 × 7 × 8 = 840 m³. Half-cylinder roof: r = 3.5, length = 15. V₂ = (1/2)π r² × length = (1/2)(22/7)(12.25)(15) = 288.75 m³.
Total volume = 840 + 288.75 = 1128.75 m³. Occupied volume = 300 + 20(0.08) = 301.6 m³. Air remaining = 1128.75 − 301.6 = 827.15 m³.
Example 2 — Juice glass
A juice glass is cylindrical with inner diameter 5 cm, but the bottom has a raised hemispherical portion (of radius 2.5 cm) that reduces its capacity. If the height of the glass is 10 cm, find the apparent capacity and actual capacity. (Use π = 3.14.)
Solution. Apparent capacity (ignoring the bump) = πr²h = 3.14(6.25)(10) = 196.25 cm³.
Volume lost to the hemisphere = (2/3)πr³ = (2/3)(3.14)(15.625) = 32.71 cm³.
Actual capacity = 196.25 − 32.71 ≈ 163.54 cm³.
Example 3 — Wooden toy rocket
A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 4.2 cm and total height of the toy is 10.2 cm. Find the volume.
Solution. r = 4.2, h_cone = 10.2 − 4.2 = 6 cm.
\[V=V_{cone}+V_{hemi}=\tfrac13\pi r^2 h+\tfrac23\pi r^3=\tfrac{1}{3}\pi r^2(h+2r)=\tfrac13(\tfrac{22}{7})(17.64)(6+8.4).\] \[V=\tfrac{22\times 17.64\times 14.4}{21}=\tfrac{5588.35}{21}\approx 266.11\ \text{cm}^3.\]12.4 Conversion of Solid from One Shape to Another
When a solid is melted and recast, or clay is reshaped, the volume is preserved; only the shape (and hence the surface area) changes. This simple fact — conservation of volume — is the key to many practical problems.
Volume Conservation
If a solid of volume V is melted and reshaped, the new solid also has volume V. For N identical recast pieces: V(original) = N × V(piece).
Example 1 — Sphere into wire
A solid sphere of radius 4.2 cm is melted and recast into the shape of a cylindrical wire of diameter 2 cm. Find the length of the wire. (π = 22/7.)
Solution. V_sphere = (4/3)π(4.2)³ = (4/3)(22/7)(74.088) ≈ 310.464 cm³.
V_wire = πr² h = π(1)² h ⇒ h = V_sphere / π = 310.464 / π ≈ 98.784 cm ≈ 98.8 cm.
Example 2 — Cone of ice-cream reshaping
A cone of radius 5 cm and height 20 cm is filled with ice-cream; it is to be served to 10 children in cylindrical cups (with hemispherical tops) of diameter 2 cm. Find the height of each cup so the ice-cream equals the cone's volume.
Solution. V_cone = (1/3)π(25)(20) = (500/3)π. Each cup volume = πr²h + (2/3)πr³ = π(h + 2/3) with r = 1. So 10 × π(h + 2/3) = (500/3)π ⇒ h + 2/3 = 50/3 ⇒ h = 48/3 = 16 cm. (Unexpectedly large! Actual NCERT variant uses different numbers.)
Example 3 — Well dug and platform made
A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly around it in the shape of a circular ring of outer radius 6 m. Find the height of the platform.
Solution. Volume of earth = π(2²)(14) = 56π m³. Ring outer radius 6 m, inner 2 m, area = π(36 − 4) = 32π. Height of platform = 56π / 32π = 7/4 = 1.75 m.
Example 4 — Ice-cream cone with hemisphere
A container in the form of a right circular cylinder with radius 6 cm and height 15 cm is full of ice-cream. The ice-cream is distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the ice-cream cone. (Use π = 3.14.)
Solution. V_cylinder = π(36)(15) = 540π. Each distribution = (1/3)πr²(4r) + (2/3)πr³ = (4/3)πr³ + (2/3)πr³ = 2πr³. So 10(2πr³) = 540π ⇒ r³ = 27 ⇒ r = 3 cm.
Activity — Clay reshaping
Take a lump of modelling clay. Shape it into a ball, measure its diameter. Reshape it into a cylinder and measure the base radius and height. Verify that (4/3)πR³ ≈ πr²h. If not, you've probably squeezed air pockets out! Record your measurements to verify volume conservation.
For ball R = 3 cm: volume = 36π ≈ 113.1 cm³. If reshaped into cylinder of radius 2 cm: required height = 113.1/(4π) ≈ 9 cm. Your experimental values should be close, within measurement error.
Exercise 12.2
(Use π = 22/7 unless stated.)
Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii equal to 1 cm and height of cone equal to its radius. Find the volume of the solid in terms of π.
V = (1/3)π(1)²(1) + (2/3)π(1)³ = π/3 + 2π/3 = π cm³.
Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its ends by using a thin aluminium sheet. The diameter is 3 cm and length is 12 cm, and if each cone has height 2 cm, find the volume of air contained.
r = 1.5, cylinder height = 12 − 4 = 8. V = πr²h_cyl + 2·(1/3)πr²h_cone = π(2.25)(8) + (2/3)π(2.25)(2) = 18π + 3π = 21π = (22/7)(21) = 66 cm³.
Q3. A gulab-jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup is there in 45 gulab-jamuns, each shaped like a cylinder with two hemispherical ends of length 5 cm and diameter 2.8 cm.
r = 1.4. Cylinder length = 5 − 2(1.4) = 2.2. V = π(1.96)(2.2) + (4/3)π(1.4)³ = 4.312π + (4/3)(2.744)π = 4.312π + 3.659π = 7.971π ≈ 25.05 cm³ per piece. 45 pieces ≈ 1127 cm³. Syrup 30% ≈ 338 cm³.
Q4. A pen stand made of wood is in the shape of a cuboid 15 cm × 10 cm × 3.5 cm with four conical depressions (radius 0.5 cm, depth 1.4 cm) to hold pens. Find the volume of wood.
Cuboid V = 525 cm³. Four cones: 4 × (1/3)π(0.25)(1.4) = 4 × (1/3)(22/7)(0.35) = (4 × 22 × 0.35)/21 = 30.8/21 ≈ 1.47 cm³. Wood = 525 − 1.47 ≈ 523.53 cm³.
Q5. A vessel is in the form of an inverted cone, height 8 cm, radius 5 cm, full of water. Some lead spheres each of radius 0.5 cm are dropped. One-fourth of water flows out. Find the number of spheres dropped in.
V_cone = (1/3)π(25)(8) = 200π/3. Water out = 50π/3. Each sphere = (4/3)π(0.125) = π/6. Count N: N(π/6) = 50π/3 ⇒ N = 100. So 100 spheres.
Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass if iron is 8 g/cm³.
V = π(144)(220) + π(64)(60) = π(31680 + 3840) = 35520π cm³ ≈ 111575 cm³. Mass = 8 × 111575 = 892 600 g = 892.6 kg.
Q7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a cylinder of water (height 180, radius 60). Find the water displaced. (Or "left in cylinder" — standard NCERT answer is 1131428.57 cm³.)
V_cyl = π(3600)(180) = 648000π. V_solid = (1/3)π(3600)(120) + (2/3)π(216000) = 144000π + 144000π = 288000π. Water left = (648000 − 288000)π = 360000π = (360000)(22/7) = 1 131 428.57 cm³ ≈ 1.131 m³.
Q8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm diameter, and spherical part diameter 8.5 cm. By measuring water in it, a child finds the volume to be 345 cm³. Check whether she is correct taking ±above measurements. (π = 3.14.)
Sphere r = 4.25. V_sphere = (4/3)π(76.77) ≈ 321.39. Cylinder r = 1, h = 8. V_cyl = π(1)(8) = 25.12. Total ≈ 346.51 cm³. Child's answer 345 is approximately correct, slight rounding discrepancy.
12.5 Summary
Key Points
- To find the surface area of a combination of solids, add the curved/exposed surfaces only; the touching faces are hidden and must be subtracted (or simply not counted).
- To find the volume of a combination, add the volumes of the individual parts without any correction — volumes are always additive.
- On converting (melting/reshaping) one solid into another, the total volume remains unchanged even though the surface area changes.
Competency-Based Questions
Q1. Apply volume conservation: a rectangular pool 10 m × 5 m is being filled at a rate of 2 m³/min. How long to fill it to a depth of 1.5 m?
L3 ApplyVolume required = 10 × 5 × 1.5 = 75 m³. Time = 75/2 = 37.5 minutes.
Q2. Analyse: a sphere of radius r and a cylinder of radius r and height 2r have how related volumes?
L4 AnalyseV_sphere = (4/3)πr³. V_cyl = πr²(2r) = 2πr³. Ratio sphere : cyl = (4/3) : 2 = 2 : 3. Archimedes famously proved this: a sphere is two-thirds of its circumscribing cylinder.
Q3. Evaluate: if a farmer waters a field of area 1 hectare by a pipe of cross-section 2 cm × 2 cm at speed 10 km/h, how long to deposit 10 cm of water?
L5 EvaluateVolume needed = 10000 m² × 0.1 m = 1000 m³ = 10⁹ cm³. Flow rate = (2×2) cm² × (1 000 000 cm/h) = 4 × 10⁶ cm³/h. Time = 10⁹ / 4×10⁶ = 250 hours ≈ 10 days, 10 hours.
Q4. Design a problem: A drinking glass is shaped as a frustum (truncated cone) of cone. Upper radius 3 cm, lower radius 2 cm, height 10 cm. Compute its volume.
L6 CreateV_frustum = (1/3)πh(R² + Rr + r²) = (1/3)π(10)(9 + 6 + 4) = (1/3)π(10)(19) = 190π/3 ≈ 199.0 cm³.
Assertion & Reason
A: Volume of a combination of solids is simply the sum of individual volumes.
R: There is no "hidden volume" at the joint — each point of space belongs to exactly one solid or the other.
R: There is no "hidden volume" at the joint — each point of space belongs to exactly one solid or the other.
Answer: A. Unlike surfaces, volumes never overlap at a joint of non-overlapping solids.
A: Melting a sphere into a cylinder preserves mass.
R: Volume is preserved, and if density is unchanged, mass = ρV is preserved as well.
R: Volume is preserved, and if density is unchanged, mass = ρV is preserved as well.
Answer: A. Standard mass–volume reasoning.
A: If a conical cavity is scooped from a cylinder of same base and height, the remaining volume equals (2/3)πr²h.
R: V_cyl − V_cone = πr²h − (1/3)πr²h = (2/3)πr²h.
R: V_cyl − V_cone = πr²h − (1/3)πr²h = (2/3)πr²h.
Answer: A. Simple subtraction.
Closing Note
Surface areas depend on shape — they change when a solid is reshaped. Volumes depend only on how much material is present — they are invariant under shape change. This elegant contrast underlies much of real-world engineering, casting, and moulding.Frequently Asked Questions
What is the summary of Chapter 12 Surface Areas and Volumes?
Chapter 12 handles surface areas and volumes of combinations of basic solids (cylinder, cone, sphere, hemisphere, cuboid) and conversion of one solid into another by melting and recasting.
What is the volume of a cone?
Volume of a cone = (1/3) pi r squared h, which is one-third the volume of a cylinder with the same base and height.
What is the volume of a sphere?
Volume of a sphere = (4/3) pi r cubed, and of a hemisphere is half this = (2/3) pi r cubed.
What is a frustum of a cone?
A frustum is what remains of a cone when its top is cut off by a plane parallel to the base. It has two circular ends of different radii.
What is the volume of a frustum?
Volume of frustum = (1/3) pi h (R^2 + r^2 + R r), where R and r are the two radii and h is the height of the frustum.
What mistakes should I avoid in Chapter 12 problems?
Mixing radius and diameter, using slant height where vertical height is required, and forgetting unit conversion (cm to m) before final answer.
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