Conversion of Solids

When a solid object is melted, moulded, or reshaped into a different solid, its volume is conserved — provided no material is added or lost. This principle, called the conversion of solids, is one of the most practically important ideas in mensuration. It explains how a metal rod can be drawn into a wire, how a sphere of clay becomes a cylinder, and how engineers calculate the dimensions of a cast object from raw material. The mathematical foundation is straightforward: Volume of original solid = Volume of new solid. Because volume is a measure of the space occupied by matter, reshaping cannot change it. What does change is the shape — and therefore the surface area, height, radius, or other dimensions. Students use the standard volume formulae for spheres, cones, cylinders, cuboids, and hemispheres, then set them equal and solve for the unknown dimension. This concept sits at the heart of composite-solid problems. You may need to find how many small spheres can be cast from a large one, what length of wire is produced from a cylindrical block of copper, or what height a cone of ice cream must have to hold the same volume as a hemisphere. Mastering conversion of solids requires fluency with volume formulae and careful algebraic manipulation — skills that are tested heavily in GCSE, IGCSE, and equivalent examinations worldwide.

Key takeaways
  • Identify the original solid and write its volume formula
  • Identify the new solid and write its volume formula
  • Set the two volumes equal
  • Solve algebraically for the unknown
  • Interpret and state the answer with units

Identify the original solid and write its volume formula

Determine the shape of the object before reshaping. Write the correct volume formula: sphere V = (4/3)πr³, cylinder V = πr²h, cone V = (1/3)πr²h, cuboid V = lwh, hemisphere V = (2/3)πr³. Substitute all known dimensions and calculate the numerical volume, keeping π in exact form where possible to avoid rounding errors.

Identify the new solid and write its volume formula

Determine the target shape. Write its volume formula with the unknown dimension represented by a variable (e.g., h for height, r for radius, n for number of pieces). If multiple identical solids are formed, multiply the single-piece volume by n.

Set the two volumes equal

Write the equation: V_original = V_new (or n × V_one piece = V_original if counting pieces). This single equation embodies the conservation of volume. Do not add or subtract anything unless waste material is explicitly mentioned in the problem.

Solve algebraically for the unknown

Isolate the unknown variable. Cancel common factors (π cancels in many problems), then perform arithmetic carefully. Always check dimensional consistency — if working in cm, the volume is in cm³ and lengths in cm.

Interpret and state the answer with units

Give the answer to the precision requested. For a number of objects, round down (you cannot cast a fraction of a sphere). For a length or radius, round to the specified decimal places. State the unit explicitly — cm, m, mm — and verify the answer makes physical sense.

The Key Equation

Voriginal=Vnew⟹43πR3=n⋅43πr3⟹n=R3r3V_{\text{original}} = V_{\text{new}} \quad \Longrightarrow \quad \frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3 \quad \Longrightarrow \quad n = \frac{R^3}{r^3}
  • V_{\text{original}} — Volume of original solid (cm³ (or m³, mm³)): Total volume of material before reshaping; this quantity is conserved throughout.
  • R — Radius of original sphere (cm): The radius of the large sphere being melted or subdivided.
  • r — Radius of each new sphere (cm): The radius of each of the n smaller spheres formed from the original.
  • n — Number of new solids (dimensionless (count)): The number of identical smaller objects that can be formed; must be a positive integer in practice.

Worked Example

Problem. A solid metallic sphere of radius 9 cm is melted and recast into small solid cones, each of base radius 3 cm and height 4 cm. Find the number of cones formed. (Take π to be exact and cancel.)

n=43π(9)313π(3)2(4)=972π12π=81n = \dfrac{\tfrac{4}{3}\pi(9)^3}{\tfrac{1}{3}\pi(3)^2(4)} = \dfrac{972\pi}{12\pi} = 81

Answer. 81 cones

Figure
Diagram
Conversion of Solids: Volume is ConservedOriginal SphereR = 9 cmV = 972π cm³melt & recastr=3,h=4r=3,h=4r=3,h=4…81 Small ConesEach: V = 12π cm³Total = 81×12π = 972π cm³V_original = V_new ✓
Origins · 1 minArchimedes of Syracuse and how this was discovered
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Archimedes of Syracuse
-287–-212 · Greek (Sicilian)

Archimedes derived the exact formulae for the volume and surface area of a sphere — results he considered his greatest achievement. He proved that the volume of a sphere is exactly two-thirds the volume of the circumscribed cylinder, a relationship directly used in conversion-of-solids problems whenever spheres and cylinders are compared. His method of exhaustion (an early form of integration) underpins all modern volume calculations. Archimedes was so proud of his sphere-cylinder result that he reportedly requested a diagram of a sphere inscribed in a cylinder be engraved on his tombstone. The Roman statesman Cicero later claimed to have found the neglected tomb and identified it by this very carving.

  • Volume of a Sphere — Converting a sphere into other shapes requires V = (4/3)πr³ as the starting volume.
  • Volume of a Cone — Many conversion problems recast spherical or cylindrical solids into cones; V = (1/3)πr²h is essential.
  • Volume of a Cylinder — Wire-drawing and rod problems use V = πr²h to find length when a solid is recast as a long thin cylinder.
  • Composite Solids — Conversion problems often involve shapes made from combined solids; understanding composite volumes is a prerequisite.
  • Surface Area of Solids — Although volume is conserved during reshaping, surface area changes; related exam questions often ask for both.