MathsICSE Class 10 · Area and Volume of Solids – Cylinder, Cone and Sphere

Melting and Recasting: Converting One Solid into Another (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

When a solid is melted and recast, or liquid is poured from one vessel into another, the volume stays the same. Write volume before = volume after and solve; $\pi$ usually cancels. To count small solids, divide the big volume by one small volume. Surface area is not conserved.

The shape changes but the material does not, so equate the two volumes, cancel π, and solve; surface area is not conserved.

Illustration drawn for QuestionBuddy.

Understand Melting and Recasting of Solids

Melting and recasting questions appeared in both the 2024 and 2025 ICSE papers (2024 Q8(iii), 2025 Q9(i)). They use only the volume formulas of the CISCE syllabus: cylinder, cone, sphere and hemisphere (frustum excluded).

The one principle. The amount of material does not change, so

$volume of the original solid(s) = volume of the new solid(s)$ .

The volume formulas.

Cylinder $π r^{2} h$ ; cone $\frac{1}{3} π r^{2} h$ ; sphere $\frac{4}{3} π r^{3}$ ; hemisphere $\frac{2}{3} π r^{3}$ .
Hollow sphere $\frac{4}{3} π (R^{3} - r^{3})$ ; hollow cylinder $π (R^{2} - r^{2}) h$ .

Standard types.

1. One into one: sphere into cylinder, cone into sphere. Equate and solve for the unknown dimension.
2. One into many: number $= \frac{big volume}{one small volume}$ . Sphere into balls: $n = (\frac{R}{r})^{3}$ .
3. Many into one: add the volumes first. Spheres of radii $a$ , $b$ , $c$ make one of radius $3 a^{3} + b^{3} + c^{3}$ .
4. Wire: a wire is a long thin cylinder; use the length in the same unit as the radius.
5. Liquids: pouring, or a solid dropped into water, gives the rise in level from volume ÷ base area.
6. Digging: earth dug out of a well or trench is spread as an embankment or a platform with the same volume.

**Why $π$ cancels.** Every formula has $π$ , so cancel it on both sides before substituting. This keeps the arithmetic exact; use $π = \frac{22}{7}$ only for an area or volume asked as a number.

Surface area changes. Breaking one sphere into many small ones increases the total surface area; combining decreases it. Only volume is conserved.

Rounding. Keep exact values and round only the final answer to what is asked: in 2024 Q8(iii) the height came to $\frac{2 8 ^{3}}{2 1 ^{2}} \approx 49.78$ cm, to be given as 50 cm.

Melting and Recasting of Solids: formulas to remember

Principle: $V_{before} = V_{after}$; Material is neither lost nor gained.
Number of small solids: $n = \frac{V _{big}}{V _{small}}$; For spheres, $n = (R / r)^{3}$ .
Spheres combined: $R^{3} = a^{3} + b^{3} + c^{3}$; Three spheres melted into one.
Rise in water level: $h = \frac{V _{solid}}{π R ^{2}}$; $R$ = radius of the cylindrical vessel.
Hollow sphere: $V = \frac{4}{3} π (R^{3} - r^{3})$; Outer minus inner.

How to solve Melting and Recasting of Solids questions

Write one equation: volume before = volume after.

1Identify the original solid(s) and the new solid(s); convert diameters to radii and all lengths to one unit.
2Write the volume formula for each side, multiplying by the number of pieces where needed.
3Cancel $π$ and common factors, then solve for the unknown.
4For 'how many', divide and check the answer is a whole number.
5Round only at the end and add units; answer any follow-up (surface area, cost) separately.

Melting and Recasting of Solids in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

One into many

Suppose a candle-maker melts a cylindrical block of wax of radius 3 cm and height 8 cm to make conical candles of radius 1.5 cm and height 4 cm.

How many candles can be made?

Try it first, then show the solutionSolution

Block: $π \times 9 \times 8 = 72 π$ cm³. One candle: $\frac{1}{3} π \times 2.25 \times 4 = 3 π$ cm³.
$n = \frac{72 π}{3 π} = 24$ .

Answer: 24 candles

Memory hook: Big volume ÷ small volume.

Drawing a wire

Imagine a goldsmith melting a small gold ball of radius 0.6 cm and drawing it into a wire 0.1 cm thick (diameter).

How long is the wire?

Try it first, then show the solutionSolution

Ball: $\frac{4}{3} π (0.6)^{3} = 0.288 π$ cm³. Wire radius 0.05 cm.
$π (0.05)^{2} L = 0.288 π$ , so $L = \frac{0.288}{0.0025} = 115.2$ cm.

Answer: 115.2 cm

Memory hook: A wire is just a very long cylinder.

Tricks to remember Melting and Recasting of Solids

Short memory aids. Say each one aloud once, then use it on a question the same day.

Volume is the currency
Melting changes the shape, never the amount: spend the same volume.
Use it when: Setting up every equation.
Cancel π first
π appears on both sides; strike it out before any numbers.
Use it when: Exact, quick arithmetic.
Cube for spheres
Sphere counts and radius ratios go by cubes: radius × 2 means 8 times the volume.
Use it when: MCQs on ratios and numbers of balls.

Melting and Recasting of Solids: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Counting balls · 1 mark · MCQBoard level

A solid sphere of radius 4 cm is melted and recast into small spheres of radius 1 cm. The number of small spheres is:
(a) 4
(b) 16
(c) 32
(d) 64

Try it first, then show the solutionSolution

$n = \frac{\frac{4}{3} π ( 4 ) ^{3}}{\frac{4}{3} π ( 1 ) ^{3}} = 4^{3} = 64$ .

Answer: (d)

Example 2: Sphere into a cylinder · 3 marks · Short answerBoard level

A solid metal sphere of radius 6 cm is melted and recast into a solid cylinder of radius 8 cm. Find the height of the cylinder.

Try it first, then show the solutionSolution

$π \times 8^{2} \times h = \frac{4}{3} π \times 6^{3}$ .
$64 h = \frac{4}{3} \times 216 = 288$ , so $h = 4.5$ cm.

Answer: 4.5 cm

Example 3: Cylinder into cones · 3 marks · Short answerHarder board level

A solid cylinder of radius 6 cm and height 15 cm is melted and recast into solid cones of radius 3 cm and height 4 cm. How many cones are made?

Try it first, then show the solutionSolution

Cylinder: $π \times 36 \times 15 = 540 π$ cm³.
One cone: $\frac{1}{3} π \times 9 \times 4 = 12 π$ cm³.
$n = \frac{540 π}{12 π} = 45$ .

Answer: 45 cones

Example 4: Rod into wire · 4 marks · Long answerHarder board level

A copper rod of diameter 2 cm and length 12 cm is drawn into a wire of uniform thickness and length 48 m. Find the radius and the diameter of the wire, in mm.

Try it first, then show the solutionSolution

Rod: $r = 1$ cm, $h = 12$ cm. Wire length $= 48$ m $= 4800$ cm.
$π r^{2} \times 4800 = π \times 1^{2} \times 12$ , so $r^{2} = \frac{12}{4800} = \frac{1}{400}$ .
$r = \frac{1}{20} = 0.05$ cm $= 0.5$ mm; diameter $= 1$ mm.

Answer: Radius 0.5 mm, diameter 1 mm

Example 5: Three spheres into one · 4 marks · Long answerHOTS / competency

Three solid metal spheres of radii 3 cm, 4 cm and 5 cm are melted and recast into a single solid sphere.
(a) Find its radius.
(b) By what percentage does the total surface area decrease?

Try it first, then show the solutionSolution

(a) $\frac{4}{3} π R^{3} = \frac{4}{3} π (27 + 64 + 125)$ , so $R^{3} = 216$ and $R = 6$ cm.
(b) Before: $4 π (9 + 16 + 25) = 200 π$ cm². After: $4 π \times 36 = 144 π$ cm².
Decrease $= 56 π$ , i.e. $\frac{56}{200} \times 100 = 28%$ .

Answer: (a) 6 cm (b) 28%

Example 6: Case study: digging a well · 4 marks · Case-basedHOTS / competency

A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly all round it to form an embankment 4 m wide (a ring around the well).
(a) Find the volume of earth dug out.
(b) Find the area of the embankment.
(c) Find the height of the embankment. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

(a) $V = \frac{22}{7} \times 1. 5^{2} \times 14 = 99$ m³.
(b) Outer radius $= 1.5 + 4 = 5.5$ m. Area $= \frac{22}{7} (5. 5^{2} - 1. 5^{2}) = \frac{22}{7} \times 28 = 88$ m².
(c) Height $= \frac{99}{88} = 1.125$ m.

Answer: (a) 99 m³ (b) 88 m² (c) 1.125 m

Where students lose marks in Melting and Recasting of Solids

The slips examiners see most often, and what to write instead.

Mistake: Ignoring a condition in the question
Only the given fraction of the vessel is transferred.
Wrong working
Using the full sphere when the oil fills only $\frac{3}{4}$ of it.
Right working
$\frac{3}{4} \times \frac{4}{3} π (28)^{3} = π (21)^{2} h$ .
Do this: Underline every condition (fractions, 'hollow', 'half') before writing the equation.
Usually costs: Reported in 2024 Q8(iii).
Mistake: Diameters taken as radii
Volume formulas use radii; a wrong radius is cubed or squared, so the error grows.
Wrong working
Hollow sphere of external diameter 10 cm: $\frac{4}{3} π (1 0^{3} - 6^{3})$ .
Right working
$\frac{4}{3} π (5^{3} - 3^{3})$ .
Do this: Halve every diameter as the first line of working.
Usually costs: Reported in 2025 Q9(i).
Mistake: Not rounding as instructed
The final answer must be in the form asked.
Wrong working
Height $= 49.78$ cm when the question says 'to the nearest cm'.
Right working
Height $\approx 50$ cm.
Do this: Reread the last line of the question before writing the answer.
Usually costs: Reported in 2024 Q8(iii): candidates did not round to 50 cm.

Exam tips

2024 Q8(iii) (4 marks): $\frac{3}{4}$ of a sphere of radius 28 cm into a cylinder of radius 21 cm: $h = \frac{2 8 ^{3}}{2 1 ^{2}} \approx 50$ cm. The examiners recommend equating volumes directly instead of computing each separately.
2025 Q9(i) (3 marks): hollow sphere (diameters 10 cm and 6 cm) into a cone of height 8 cm: $r^{2} = 49$ , $r = 7$ cm.
2025 Q1(xii) (1 mark): volumes $27 : 64$ give radii $3 : 4$ ; the report notes candidates who used the area ratio $9 : 16$ .
Keep units consistent: convert metres of wire to centimetres before equating.

Practice questions on Melting and Recasting of Solids

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The volumes of two spheres are in the ratio $125 : 216$ . The ratio of their radii is:
(a) $5 : 6$
(b) $25 : 36$
(c) $125 : 216$
(d) $6 : 5$
Hint
Take cube roots.
Answer
(a)
Q2 · 2 marks · Short answerBoard level
A solid cone of radius 4 cm and height 16 cm is melted and recast into a solid sphere. Find the radius of the sphere.
Hint
$\frac{4}{3} r^{3} = \frac{1}{3} \times 16 \times 16$ .
Answer
$r^{3} = 64$ , $r = 4$ cm.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a solid sphere is melted and recast into several smaller spheres, the total surface area remains the same.
Reason (R): In melting and recasting, the volume of material remains the same.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Try one sphere of radius 2 into eight of radius 1.
Answer
(b): the total surface area increases.
Q4 · 3 marks · Short answerHarder board level
How many cylindrical coins of diameter 1.75 cm and thickness 2 mm can be made from a cuboid of metal measuring 5.5 cm × 10 cm × 3.5 cm? ( $π = \frac{22}{7}$ )
Hint
One coin: $\frac{22}{7} \times 0.87 5^{2} \times 0.2$ .
Answer
Coin $= 0.48125$ cm³; cuboid $= 192.5$ cm³; $n = 400$ .
Q5 · 3 marks · Short answerHarder board level
A hemispherical bowl of internal radius 9 cm is full of water. The water is poured into a cylindrical vessel of radius 6 cm. Find the height of water in the vessel.
Hint
$π \times 36 \times h = \frac{2}{3} π \times 729$ .
Answer
$h = \frac{486}{36} = 13.5$ cm.
Q6 · 3 marks · Short answerHarder board level
A metal pipe 21 cm long has external radius 5 cm and internal radius 4 cm. It is melted and recast into a solid cylinder of the same length. Find the radius of the solid cylinder.
Hint
$π (25 - 16) \times 21 = π R^{2} \times 21$ .
Answer
$R^{2} = 9$ , $R = 3$ cm.
Q7 · 4 marks · Long answerHOTS / competency
A solid sphere of radius $r$ is melted and recast into a solid cone with the same radius $r$ .
(a) Show that the height of the cone is $4 r$ .
(b) If $r = 3.5$ cm, find the height and the slant height of the cone, to 2 decimal places.
Hint
$\frac{1}{3} π r^{2} h = \frac{4}{3} π r^{3}$ .
Answer
(a) $r^{2} h = 4 r^{3}$ , so $h = 4 r$ . (b) $h = 14$ cm; $l = 3. 5^{2} + 1 4^{2} = 208.25 \approx 14.43$ cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is conserved in melting and recasting?Show answer
Volume.
2.Sphere of radius 3 cm into balls of radius 1 cm. How many?Show answer
27
3.Volume of a hemisphere?Show answer
$\frac{2}{3} π r^{3}$
4.Radii ratio for volumes $1 : 8$ ?Show answer
$1 : 2$
5.Does total surface area stay the same?Show answer
No; it usually changes.

Melting and Recasting of Solids: frequently asked questions

What is the rule for melting and recasting questions?

The volume of the material stays the same. Write the volume of the original solid equal to the total volume of the new solid or solids, and solve.

How do I find the number of small solids made?

Divide the volume of the big solid by the volume of one small solid. For spheres this is the cube of the ratio of the radii.

Should I substitute π = 22/7 in recasting questions?

Usually not needed: π appears on both sides and cancels. Substitute only when the question asks for an actual area or volume.