Exercise 10.4 — Conversion of Solids

Conversion of solids from one shape to another.

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Exercise 10.4 — Conversion of Solids From One Shape to Another

Exercise 10.4 from Chapter 10, Mensuration, of Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) deals with what happens when a solid is melted and recast, dug out and reshaped, or poured into a different shape entirely. A metal sphere becomes a cylinder, three spheres merge into one, earth from a well becomes a platform or an embankment, ice cream is scooped from a tub into cones, coins are melted into a bar, lead shots are dropped into a cone of water, and a sphere is recast into hundreds of tiny cones.

Eight problems, eight different stories — but every single one rests on the same unshakeable fact: melting, digging, or pouring changes the shape of a solid, but never its volume. Once that idea clicks, this entire exercise becomes a matter of setting up one equation and solving for whatever is unknown.

Volume of Sphere Volume of Cylinder & Cone Volume of Cuboid Conservation of Volume

💡 Foundation fact: Whenever a solid is melted, dug out, or poured into a new shape, volume before = volume after. Nothing about that idea changes from question to question — only the shapes on either side of the equals sign change. Unless stated otherwise, this exercise uses π = 22/7.

What Does "Conversion of Solids" Mean?

Think about melting a block of wax and pouring it into a different mould — the wax doesn't disappear or multiply, it just takes on a new shape. The same logic applies to every problem in this exercise. Four everyday situations capture the whole idea:

Water in a cylinder

= water in a cuboid

An iron bar

= sum of cylindrical pipes

Volume of a cylinder

= volume of a cone

Many small spheres

= one big sphere

Two Groups Every Question Falls Into

Every problem in this exercise asks you to find one of exactly two things:

Group	What You Solve For	Questions
Find a missing length	Set Volume before = Volume after, then solve for the one unknown height or radius.	Q1, Q2, Q3, Q4
Find how many pieces	Set Volume of the big solid = n × Volume of one small piece, then solve for n.	Q5, Q6, Q7, Q8

💡 A pattern worth noticing — when does π cancel? Every volume formula for a sphere, cylinder, cone, or hemisphere contains π. So whenever both shapes in a question are "round" (Q1, Q2, Q4, Q5, Q7, Q8), the π on both sides of the equation cancels out completely — you never need its numeric value at all! The only two exceptions are Q3 and Q6, where the final shape is a cuboid. A cuboid's volume formula (l × b × h) has no π in it, so in those two questions alone you actually need to use π = 22/7 as a number.

Volume Formulas You Need for This Exercise

Every question in this exercise is built entirely from these five formulas — recognise the shapes on each side of the problem, and the rest is algebra.

Solid	Volume Formula
Cuboid (length l, breadth b, height h)	`l × b × h`
Cylinder (radius r, height h)	`πr²h`
Cone (radius r, height h)	`⅓πr²h`
Sphere (radius r)	`4/3 πr³`
Hemisphere (radius r)	`2/3 πr³`

Question 1 — A Metallic Sphere Recast Into a Cylinder

A metallic sphere of radius 4.2 cm is melted down and recast into a cylinder of radius 6 cm. We need the height of that cylinder. Since the metal is simply reshaped, the cylinder's volume must equal the sphere's original volume.

Sphere melted into a cylinder

Solution

Find the height of the cylinder

Volume of sphere = Volume of cylinder 4/3 πR³ = πr²h ⟹ h = 4R³ ÷ 3r² (π cancels — both shapes are round) h = (4 × 4.2 × 4.2 × 4.2) ÷ (3 × 6 × 6) h = 296.352 ÷ 108 = 2.744 cm

✅ Answer: The height of the cylinder is 2.744 cm.

Question 2 — Three Spheres Melted Into One Big Sphere

Three metallic spheres of radii 6 cm, 8 cm, and 10 cm are melted together and recast into a single solid sphere. We need the radius of the resulting sphere.

Three spheres merged into one

Solution

Find the radius of the resulting sphere

Volume of big sphere = sum of volumes of the small spheres 4/3 πR³ = 4/3 πr₁³ + 4/3 πr₂³ + 4/3 πr₃³ ⟹ R³ = r₁³ + r₂³ + r₃³ (π and 4/3 cancel on both sides) R³ = 6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728 R = ∛1728 = 12 cm

✅ Answer: The radius of the resulting sphere is 12 cm.

⚠️ Classic mistake: Don't add the radii directly (6 + 8 + 10 = 24, then guess R is related to that). Volumes add up, not radii — so it's the cubes of the radii that must be added: R³ = 6³ + 8³ + 10³, not R = 6 + 8 + 10.

Question 3 — Earth From a Well Spread Into a Rectangular Platform

A well 20 m deep and 7 m in diameter is dug, and all the earth removed is spread out evenly to form a rectangular platform measuring 22 m by 14 m. We need the height of that platform. The well is shaped like a cylinder; the platform is a cuboid.

Well dug, earth spread into a platform

Solution

Find the height of the platform

Radius of well, r = 7 ÷ 2 = 3.5 m; depth, H = 20 m Volume of earth dug out = πr²H = (22/7) × 3.5 × 3.5 × 20 = 770 m³ Volume of platform = volume of earth dug out l × b × h = 770 22 × 14 × h = 770 h = 770 ÷ 308 = 2.5 m

✅ Answer: The height of the platform is 2.5 m.

📌 Why π doesn't cancel here: The well is round, but the platform is a cuboid — and a cuboid's volume formula has no π in it at all. That's why this is one of only two questions in this exercise where π = 22/7 actually has to be evaluated as a number.

Question 4 — Earth From a Well Spread Into a Circular Embankment

A well 14 m in diameter is dug 15 m deep. This time, the earth taken out is spread evenly all around the mouth of the well to form a circular embankment (a ring-shaped raised platform) of width 7 m. We need the height of that embankment.

Top view — well surrounded by the embankment

Solution

Find the height of the embankment

Radius of well, r = 14 ÷ 2 = 7 m; depth, H = 15 m Inner radius of embankment, r₁ = 7 m Outer radius of embankment, r₂ = r₁ + width = 7 + 7 = 14 m Volume of well = Volume of embankment πr²H = π(r₂² − r₁²)h (π cancels — both are round shapes) 7 × 7 × 15 = (14² − 7²) × h 735 = (196 − 49) × h = 147h h = 735 ÷ 147 = 5 m

✅ Answer: The height of the embankment is 5 m.

💡 Spot the difference from Question 3: Here the earth forms a ring around the well, not a solid block — so its cross-section is the area between two circles (r₂² − r₁²), not a simple rectangle. Because the embankment is still a round shape, π cancels here, unlike in Question 3.

Question 5 — Ice Cream From a Cylinder Filled Into Cone-Shaped Cups

A cylindrical container 12 cm in diameter and 15 cm tall is full of ice cream. It needs to be scooped into cone-shaped cups, each 6 cm in diameter and 12 cm tall, topped with a hemispherical scoop of the same radius. We need to know how many such cups can be filled.

Cylinder tub of ice cream

One ice-cream cone, × n

Solution

Find the number of cones that can be filled

Cylinder (R = 6, H = 15): Volume = πR²H = π × 6 × 6 × 15 = 540π Cone (r = 3, h = 12): Volume = ⅓πr²h = ⅓ × π × 3 × 3 × 12 = 36π Hemisphere (r = 3): Volume = ⅔πr³ = ⅔ × π × 3 × 3 × 3 = 18π Volume of one ice-cream cone = 36π + 18π = 54π Number of cones, n = 540π ÷ 54π = 10 (π cancels neatly)

✅ Answer: 10 ice-cream cones can be filled.

Question 6 — Silver Coins Melted Into a Cuboid

Silver coins 1.75 cm in diameter and 2 mm thick are melted down to form a cuboid measuring 5.5 cm by 10 cm by 3.5 cm. We need to find how many coins are required.

Coins melted into a cuboid

Solution

Find the number of coins needed

Radius of one coin, r = 1.75 ÷ 2 = 0.875 cm Thickness, h = 2 mm = 0.2 cm Volume of one coin = πr²h = (22/7) × 0.875 × 0.875 × 0.2 ≈ 0.481 cm³ Volume of cuboid = 5.5 × 10 × 3.5 = 192.5 cm³ Number of coins, n = 192.5 ÷ 0.481 ≈ 400

✅ Answer: 400 silver coins are needed.

💡 Mind the units: The coin's thickness is given in millimetres while every other measurement in the problem is in centimetres — convert 2 mm to 0.2 cm before using it in the volume formula.

Question 7 — Lead Shots Dropped Into a Conical Vessel

A vessel shaped like an inverted cone (height 8 cm, top radius 5 cm) is filled with water right up to the rim. Spherical lead shots, each of radius 0.5 cm, are dropped in one by one, and exactly a quarter of the water spills out. We need to find how many lead shots were dropped in.

Inverted cone — lead shots cause overflow

Solution

Find the number of lead shots dropped in

Volume of cone = ⅓πR²h = ⅓ × π × 5 × 5 × 8 = 200π/3 Water that flows out = ¼ of the vessel's volume = ¼ × 200π/3 = 50π/3 Volume of one lead shot (r = 0.5) = 4/3 πr³ = 4/3 × π × 0.5 × 0.5 × 0.5 = π/6 Volume of water out = n × volume of one shot 50π/3 = n × π/6 n = (50π/3) × (6/π) = 100 (π cancels neatly)

✅ Answer: 100 lead shots were dropped into the vessel.

📌 Don't lose the "¼" along the way: The question gives the fraction of water that overflows, not the full vessel's volume — that ¼ has to multiply the cone's total volume before you set it equal to n × (volume of one shot).

Question 8 — A Sphere Recast Into Many Smaller Cones

A solid metallic sphere 28 cm in diameter is melted and recast into a number of small cones. Each cone has a diameter of 4⅔ cm (that is, 14/3 cm) and a height of 3 cm. We need to find how many such cones are formed.

Sphere recast into many small cones

Solution

Find the number of cones formed

Radius of sphere, R = 28 ÷ 2 = 14 cm Volume of sphere = 4/3 πR³ = 4/3 × π × 14³ = 10976π/3 Diameter of each cone = 4⅔ = 14/3 cm ⟹ radius, r = 7/3 cm ≈ 2.33 cm; height, h = 3 cm Volume of one cone = ⅓πr²h = ⅓ × π × (7/3)² × 3 = 49π/9 Number of cones, n = (10976π/3) ÷ (49π/9) n = (10976/3) × (9/49) = 672 (π cancels neatly)

✅ Answer: 672 smaller cones are formed.

Common Mistakes to Avoid

Adding radii instead of cubes of radii: When several spheres merge into one (Question 2), it's R³ = r₁³ + r₂³ + r₃³ that holds true — never R = r₁ + r₂ + r₃. Volumes add; lengths don't.
Mixing up diameter and radius: Almost every question states a diameter, but every volume formula needs a radius — always halve the diameter before substituting it into πr².
Forgetting a unit conversion: Question 6's coin thickness is in millimetres while everything else is in centimetres, and Question 8's cone diameter is a mixed fraction (4⅔ cm) that's easy to mis-convert.
Using the whole volume instead of a fraction of it: In Question 7, it's only ¼ of the water that overflows — make sure that fraction is applied to the cone's volume before equating it to n lead shots.
Assuming π always cancels: It does whenever both shapes are round (sphere, cylinder, cone, hemisphere) — but it doesn't in Questions 3 and 6, because a cuboid's volume formula has no π in it at all.
Accepting a non-whole-number answer for "n": The number of coins, cones, or lead shots must come out as a whole number. If your calculation gives something like 399.6, go back and check your unit conversions rather than rounding.

Quick Reference — All Answers at a Glance

Question	Conversion	Answer
Q1	Sphere → Cylinder	h ≈ 2.744 cm
Q2	3 Spheres → 1 Sphere	R = 12 cm
Q3	Well → Rectangular platform	h = 2.5 m
Q4	Well → Circular embankment	h = 5 m
Q5	Cylinder → Ice-cream cones	10 cones
Q6	Coins → Cuboid	400 coins
Q7	Cone (water) → Lead shots	100 shots
Q8	Sphere → Smaller cones	672 cones

What This Exercise Prepares You For

Exercise 10.4 is the natural conclusion of Chapter 10 — it takes the same volume formulas built up in Exercise 10.3 — Volume of Combination of Solids and applies them in reverse: instead of adding or subtracting volumes of joined pieces, you set two volumes equal and solve for an unknown length or count. If the surface-area side of combined solids needs a refresher first, Exercise 10.2 — Surface Area of Combination of Solids covers that groundwork.

The "melt and recast" idea here also lays the foundation for later mensuration problems involving rates of flow, such as how long a pipe takes to fill a tank — those questions use exactly the same volume-conservation reasoning, just with time added into the equation. A firm grip on this exercise is well worth it heading into the Introduction to Mensuration review before your CBSE, Telangana, or Andhra Pradesh board exam.

📐 Board Exam Tip: "Melt and recast" or "dig and reshape" questions are a near-guaranteed 4 or 5-mark question in this chapter. Before reaching for a formula, write down which shape you're starting with and which shape you're ending with — that one habit makes choosing the right two formulas almost automatic.