Conversion between solids (melting and recasting)

A goldsmith melts a lump of gold shaped like a sphere and draws it into a long thin wire. A foundry melts down old metal cylinders and casts them into new cones and cuboids. In every such process the shape of the material changes completely, but nothing is added and nothing is lost — the amount of material, measured as volume, stays exactly the same before and after. This single fact, that melting and recasting conserves volume, is the whole engine behind a large family of Class 10 problems that ask you to find an unknown radius, height, or number of pieces once a solid is reshaped. This article takes that one idea — volume before equals volume after — and shows how to turn it into an equation, solve it cleanly, and avoid the handful of traps (mixed units, forgotten fractions, confusing volume with surface area) that cost the most marks in this topic.

The core insight

Melting and recasting a solid changes its shape but never its volume, so setting $V_{\text{original}} = V_{\text{new}}$ (or $V_{\text{original}} = n \times V_{\text{one new piece}}$ for many identical pieces) turns any conversion-of-solids problem into a single solvable equation.

The core idea

When a solid is melted and reshaped — or a liquid is poured from one container into another — the quantity of material is conserved even though its shape is not. Measuring "quantity of material" as volume gives the governing equation of this entire topic:

V_{\text{original solid}} = V_{\text{new solid}}

where: $V_{\text{original solid}}$ is the volume of the solid before melting, and $V_{\text{new solid}}$ is the volume of the solid it is recast into — the two must be equal because no material is created or destroyed in the process. If the same lump of material is recast into $n$ identical smaller solids instead of one new solid, the equation gains a factor of $n$ :

V_{\text{original solid}} = n \times V_{\text{one new solid}}

where: $n$ is the number of identical new pieces produced, and $V_{\text{one new solid}}$ is the volume of just one of them. This version is what you use for problems like "how many spherical shots can be cast from a melted cone." Every problem in this topic reduces to the same three-part plan: write the volume of the original solid using the standard formula for its shape, write the volume of the new solid (or one new piece) leaving the unknown as a variable, and then equate the two sides and solve. The formulae you need are the same ones from the surface-areas-and-volumes chapter — cuboid $V = lbh$ , cylinder $V = \pi r^2 h$ , cone $V = \tfrac13\pi r^2 h$ , sphere $V = \tfrac43\pi r^3$ , hemisphere $V = \tfrac23\pi r^3$ — nothing new is introduced here beyond applying them to a "before and after" pair of shapes. The picture below shows why this works: the material inside the original solid is simply repacked into the new shape, so the two enclosed volumes are the same lump of matter viewed at two different moments.

Figure

A sphere of metal on the left transforming through an arrow into a long thin cylindrical wire on the right, both labelled with the same volume

Why it works

The reasoning behind the core equation is simple conservation of matter, but it is worth stating precisely because exam answers are marked on exactly this justification. Melting turns a solid into a liquid without removing any of its mass; recasting turns that liquid back into a solid of a new shape without adding or removing any of it either. Since the material is assumed to be of uniform density throughout the process, "same mass" and "same volume" mean the same thing here — nothing evaporates, nothing is trimmed away, and the mould is filled completely. That is why the single equation $V_{\text{original}} = V_{\text{new}}$ (or its $n$ -piece version) fully captures the physics of the situation: it is not a coincidence or an approximation, it is a direct restatement of "the same stuff, repacked." The same logic extends to liquids without any change in the argument. Water poured from a full cylindrical bucket onto the ground and heaped into a cone, or transferred between two differently shaped tanks, obeys exactly the same "volume in = volume out" principle, provided none of it spills or evaporates along the way. This is why textbook problems freely mix solid-melting and liquid-pouring scenarios under the same heading — the underlying equation is identical either way.

Worked example

A solid metallic sphere of radius 6 cm is melted and recast into a cylindrical wire of radius 0.2 cm. Find the length of the wire.

Step 1. Find the volume of the sphere, the original solid. Using $V = \tfrac43\pi r^3$ with $r = 6$ cm: $V_{\text{sphere}} = \tfrac43\pi(6)^3 = \tfrac43\pi \times 216 = 288\pi \text{ cm}^3$ .

Step 2. Write the volume of the cylindrical wire, the new solid, leaving its length $h$ as the unknown. Using $V = \pi r^2 h$ with $r = 0.2$ cm: $V_{\text{cylinder}} = \pi (0.2)^2 h = 0.04\pi h \text{ cm}^3$ .

Step 3. Equate the two volumes, since melting and recasting loses no material: $288\pi = 0.04\pi h$ .

Step 4. Solve for $h$ : $h = \dfrac{288}{0.04} = 7200$ cm, which converts to $h = 72$ m. The wire is 72 metres long — a striking demonstration of how a fairly compact sphere can be drawn into a very long, very thin wire while its volume stays fixed.

Competitive stretch

A cylindrical bucket 32 cm in diameter and 45 cm high is full of sand. The sand is emptied onto the ground and forms a conical heap of height 24 cm. Find the radius of the heap.

Step 1. Volume of sand in the bucket: radius $= \tfrac{32}{2} = 16$ cm, height $= 45$ cm, so $V_{\text{cylinder}} = \pi(16)^2(45) = 11520\pi \text{ cm}^3$ .

Step 2. Volume of the conical heap, with unknown radius $R$ and given height $24$ cm: $V_{\text{cone}} = \tfrac13\pi R^2(24) = 8\pi R^2 \text{ cm}^3$ .

Step 3. Equate and solve: $11520\pi = 8\pi R^2 \Rightarrow R^2 = 1440 \Rightarrow R = \sqrt{1440} \approx 37.9$ cm. The conical sand heap has a base radius of about 37.9 cm.

Common traps

Equating surface areas instead of volumes. Melting and recasting conserves the amount of material, not the surface area — the new shape almost always has a different surface area from the original.
Mismatched units before equating. Convert every length to the same unit before writing the equation (the wire example needs a final cm-to-m conversion); mixing cm and m mid-calculation is the single commonest slip.
Dropping the fractional factors. Forgetting the $\tfrac13$ in the cone formula or the $\tfrac43$ in the sphere formula throws off the whole equation, even when every other step is correct.
Assuming a dimension carries over unchanged. Unless a problem explicitly says a radius or height stays the same, the new solid generally has its own dimensions — solving for one of them is usually the whole point of the question.

Ask the tutor: If you have a melting-and-recasting problem and are not sure which shape's volume formula to use or whether you need the $n$ -piece version, describe the two shapes to the AI tutor and it will help you set up the right equation.

Why is it volume that stays the same and not surface area

Volume measures how much material fills the solid, and melting followed by recasting neither adds nor removes any material, so the volume before must equal the volume after. Surface area, in contrast, depends entirely on how the material is arranged into a shape — a sphere and a very long thin wire made from the same lump of metal enclose the same volume but have wildly different surface areas, because the wire's surface is spread over a much larger, thinner region. Always equate volumes in these problems, never surface areas.

Does this idea work if the original solid is hollow, or if some material is lost while melting

The basic principle assumes all of the original solid's material ends up in the new solid, so for a hollow solid you must first find the volume of material actually present, which usually means subtracting an inner volume from an outer one, before equating it to the new solid's volume. If a question explicitly says some material is wasted, lost, or does not fully fill the new mould, you would equate the new solid's volume to only the fraction of the original volume that survives the process — but unless a CBSE problem says so, assume no material is lost.

How do I know whether to equate a single volume or use the n-piece version with n

Look at whether the question describes one new object or many identical new objects being made from the same original material. If it is one recast solid — a sphere melted into one cylinder, one bucket of sand poured into one conical heap — use $V_{\text{original}} = V_{\text{new}}$ . If the material is divided into several identical smaller pieces, such as casting a number of identical spherical shots or coins from one melted block, use $V_{\text{original}} = n \times V_{\text{one new solid}}$ , where $n$ is exactly what you are usually asked to find.