CBSE • Class 10 • Mathematics • Chapter 12 (Surface Areas and Volumes) • Exercise 12.2

Exercise 12.2: Surface Areas and Volumes — NCERT Solutions

Volume of a combination of solids and conversion-of-shape problems.

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What this exercise covers

Volume of combinationsFrustum of a coneRecasting problems

Step-by-step solutions — Exercise 12.2

6 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 12.2 Q1 • 2 marks

A solid is in the shape of a cone standing on a hemisphere, both having radius 1 cm1\ \text{cm} and the height of the cone equal to its radius. Find the volume of the solid in terms of π\pi.
Hint (Socratic — try this first)▾
Can you just add the volume of the cone to the volume of the hemisphere?
Step-by-step solution▾

Radius r=1 cmr = 1\ \text{cm}, cone height h=1 cmh = 1\ \text{cm}.

Volume of solid = Volume of cone + Volume of hemisphere: =13πr2h+23πr3= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 =13π(1)2(1)+23π(1)3= \frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3 =13π+23π=π cm3= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi\ \text{cm}^3

Volume of the solid =π cm3= \pi\ \text{cm}^3.

Common mistake:
Using the sphere volume 43πr3\tfrac{4}{3}\pi r^3 instead of the hemisphere volume 23πr3\tfrac{2}{3}\pi r^3.
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Exercise 12.2 Q2 • 3 marks

A vessel is in the form of an inverted cone. Its height is 8 cm8\ \text{cm} and radius of its top is 5 cm5\ \text{cm}. It is filled with water up to the brim. When lead shots, each of radius 0.5 cm0.5\ \text{cm}, are dropped in, one-fourth of the water flows out. Find the number of lead shots dropped. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
The overflowed water volume equals the total volume of all the spherical shots — how do you relate the two?
Step-by-step solution▾

Cone: r=5 cmr = 5\ \text{cm}, h=8 cmh = 8\ \text{cm}.

Volume of water in cone: V=13πr2h=13π(25)(8)=2003π cm3V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (25)(8) = \frac{200}{3}\pi\ \text{cm}^3

Water that flows out =14V=14×2003π=503π cm3= \dfrac{1}{4}V = \dfrac{1}{4}\times \dfrac{200}{3}\pi = \dfrac{50}{3}\pi\ \text{cm}^3.

Volume of one lead shot (sphere), rs=0.5 cmr_s = 0.5\ \text{cm}: =43πrs3=43π(0.5)3=43π×0.125=0.53π cm3= \frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{0.5}{3}\pi\ \text{cm}^3

Number of shots: n=water outone shot=503π0.53π=500.5=100n = \frac{\text{water out}}{\text{one shot}} = \frac{\tfrac{50}{3}\pi}{\tfrac{0.5}{3}\pi} = \frac{50}{0.5} = 100

So 100100 lead shots were dropped.

Common mistake:
Equating the shots to the full cone volume instead of only one-fourth of it (the water that overflowed).
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Exercise 12.2 Q3 • 3 marks

A metallic sphere of radius 6 cm6\ \text{cm} is melted and recast into a solid cylinder of radius 4 cm4\ \text{cm}. Find the height of the cylinder.
Hint (Socratic — try this first)▾
When one shape is melted and recast into another, which quantity stays exactly the same?
Step-by-step solution▾

When recast, volume is conserved.

Volume of sphere == Volume of cylinder: 43πR3=πr2h\frac{4}{3}\pi R^3 = \pi r^2 h

Here R=6 cmR = 6\ \text{cm}, r=4 cmr = 4\ \text{cm}. 43π(6)3=π(4)2h\frac{4}{3}\pi (6)^3 = \pi (4)^2 h 43×216=16h\frac{4}{3}\times 216 = 16 h 288=16h288 = 16 h h=18 cmh = 18\ \text{cm}

Height of the cylinder =18 cm= 18\ \text{cm}.

Common mistake:
Equating surface areas instead of volumes, since it is the material (volume) that is conserved on melting.
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Exercise 12.2 Q4 • 3 marks

A container shaped like a right circular cylinder having diameter 12 cm12\ \text{cm} and height 15 cm15\ \text{cm} is full of ice cream. This ice cream is to be distributed to children in cones of height 12 cm12\ \text{cm} and diameter 6 cm6\ \text{cm}, each having a hemispherical top. Find the number of such cones that can be filled.
Hint (Socratic — try this first)▾
Each serving is a cone PLUS a hemisphere — what total volume must you divide the cylinder's volume by?
Step-by-step solution▾

Cylinder: radius R=6 cmR = 6\ \text{cm}, height H=15 cmH = 15\ \text{cm}. Vcyl=πR2H=π(36)(15)=540π cm3V_{cyl} = \pi R^2 H = \pi (36)(15) = 540\pi\ \text{cm}^3

Each ice-cream serving: cone radius r=3 cmr = 3\ \text{cm}, cone height h=12 cmh = 12\ \text{cm}, hemisphere radius 3 cm3\ \text{cm}.

Volume of cone =13πr2h=13π(9)(12)=36π= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (9)(12) = 36\pi

Volume of hemisphere =23πr3=23π(27)=18π= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (27) = 18\pi

Total per serving =36π+18π=54π cm3= 36\pi + 18\pi = 54\pi\ \text{cm}^3.

Number of cones: n=540π54π=10n = \frac{540\pi}{54\pi} = 10

So 1010 cones can be filled.

Common mistake:
Forgetting to add the hemispherical top's volume to each cone, thus overestimating the number of servings.
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Exercise 12.2 Q5 • 3 marks

A solid iron pole consists of a cylinder of height 220 cm220\ \text{cm} and base diameter 24 cm24\ \text{cm}, surmounted by another cylinder of height 60 cm60\ \text{cm} and radius 8 cm8\ \text{cm}. Find the total volume of the pole. (Take π=3.14\pi = 3.14)
Hint (Socratic — try this first)▾
The pole is simply two stacked cylinders — can their volumes be added directly?
Step-by-step solution▾

Lower cylinder: radius r1=12 cmr_1 = 12\ \text{cm}, height h1=220 cmh_1 = 220\ \text{cm}. V1=πr12h1=3.14×144×220=99475.2 cm3V_1 = \pi r_1^2 h_1 = 3.14 \times 144 \times 220 = 99475.2\ \text{cm}^3

Upper cylinder: radius r2=8 cmr_2 = 8\ \text{cm}, height h2=60 cmh_2 = 60\ \text{cm}. V2=πr22h2=3.14×64×60=12057.6 cm3V_2 = \pi r_2^2 h_2 = 3.14 \times 64 \times 60 = 12057.6\ \text{cm}^3

Total volume: V=V1+V2=99475.2+12057.6=111532.8 cm3V = V_1 + V_2 = 99475.2 + 12057.6 = 111532.8\ \text{cm}^3

Total volume of the pole =111532.8 cm3= 111532.8\ \text{cm}^3.

Common mistake:
Using the given diameter of the lower cylinder (24 cm) directly as its radius instead of halving it to get 12 cm.
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Exercise 12.2 Q6 • 3 marks

A well of diameter 3 m3\ \text{m} is dug 14 m14\ \text{m} deep. The earth taken out has been spread evenly all around it in the shape of a circular ring of width 4 m4\ \text{m} to form an embankment. Find the height of the embankment. (Take π=227\pi = \tfrac{22}{7})
Hint (Socratic — try this first)▾
The dug-out earth (a cylinder) is reshaped into a hollow ring — how do you find the ring's base area between two circles?
Step-by-step solution▾

Well: radius r=1.5 mr = 1.5\ \text{m}, depth =14 m= 14\ \text{m}.

Volume of earth dug out: V=πr2h=227×(1.5)2×14=227×2.25×14=99 m3V = \pi r^2 h = \frac{22}{7}\times (1.5)^2 \times 14 = \frac{22}{7}\times 2.25 \times 14 = 99\ \text{m}^3

Embankment is a hollow ring: inner radius =1.5 m= 1.5\ \text{m}, width =4 m= 4\ \text{m}, so outer radius R=1.5+4=5.5 mR = 1.5 + 4 = 5.5\ \text{m}.

Base area of ring: =π(R2−r2)=227(5.52−1.52)=227(30.25−2.25)=227×28=88 m2= \pi(R^2 - r^2) = \frac{22}{7}(5.5^2 - 1.5^2) = \frac{22}{7}(30.25 - 2.25) = \frac{22}{7}\times 28 = 88\ \text{m}^2

Let height of embankment =H= H. Since volume is conserved: 88×H=9988 \times H = 99 H=9988=1.125 mH = \frac{99}{88} = 1.125\ \text{m}

Height of the embankment =1.125 m= 1.125\ \text{m}.

Common mistake:
Using the outer radius alone (a full cylinder πR2\pi R^2) instead of the ring area π(R2−r2)\pi(R^2 - r^2), ignoring the hollow well in the middle.
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How to approach Exercise 12.2

  1. Re-read the chapter summary first. Open Surface Areas and Volumes and refresh the key concepts: Cube, Cuboid, Sphere, Hemisphere.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Surface Areas and Volumes

  1. Exercise 12.1 — Surface area of a combination of solids — cone on a hemisphere, cuboid with a cylinder, and similar.
  2. Exercise 12.2 — Volume of a combination of solids and conversion-of-shape problems.

← Exercise 12.1

Surface area of a combination of solids — cone on a hemisphere, cuboid with a cylinder, and similar.

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