Conversion of Solids (Melting and Recasting)

The One Principle: Volume Is Conserved

When a solid is melted and recast, drawn into a wire, or reshaped in any way, its material does not change — so its volume stays the same: $$\text{Volume before} = \text{Volume after}.$$

Every problem in this section is really one equation: $$\text{Volume of the original shape} = \text{Volume of the new shape(s)}.$$

Key Point: Set up "old volume $=$ new volume" and solve for whatever is unknown (a new dimension, or the number of pieces).

Melting and Recasting into Many Pieces

If one big solid is melted and recast into $n$ identical small pieces: $$n = \dfrac{\text{Volume of the big solid}}{\text{Volume of one small piece}}.$$

For example, a sphere melted into small cones, or a big sphere into small spheres, or a cylinder into coins.

[Board Important] The $\pi$ usually cancels on both sides — leave it as $\pi$ and it disappears, saving arithmetic.

Drawing into a Wire (or a Long Cylinder)

A lump of metal drawn into a long thin wire is just a cylinder of very small radius and very large length. Again, volume is conserved: $$\text{Volume of the lump} = \pi r_{\text{wire}}^2 \times \text{length}.$$

So the wire's length $= \dfrac{\text{Volume of the lump}}{\pi r_{\text{wire}}^2}$.

Key Point: "Drawn into a wire of diameter $d$" means a cylinder of radius $d/2$ — find its length from the conserved volume.

Water Displacement and Rising Levels

When a solid is dropped into water in a container, the water it pushes up equals the solid's volume (or the submerged part): $$\text{Volume of solid} = \text{Volume of water displaced} = (\text{base area}) \times (\text{rise in level}).$$

For "how many lead shots make the water rise/overflow by a given amount," use $$\text{number} = \dfrac{\text{volume of water displaced}}{\text{volume of one shot}}.$$

Key Point: Rise in water level $\times$ base area $=$ volume of the object submerged. Same conservation idea, just with water.

Solved Examples

Example 1: Sphere melted into cones

A metallic sphere of radius 3 cm is melted and recast into small cones, each of radius 1 cm and height 3 cm. How many cones are formed?

Solution:

1. Volume of sphere $= \dfrac43\pi (3)^3 = 36\pi$ cm$^3$.
2. Volume of one cone $= \dfrac13\pi (1)^2(3) = \pi$ cm$^3$.
3. Number $= \dfrac{36\pi}{\pi} = 36$.

Final Answer: 36 cones.

Takeaway: Number $= \dfrac{V_{\text{big}}}{V_{\text{small}}}$; $\pi$ cancels.

Example 2: Sphere drawn into a wire

A copper sphere of radius 3 cm is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.

Solution:

1. Volume of sphere $= \dfrac43\pi (3)^3 = 36\pi$ cm$^3$.
2. Wire radius $= 0.1$ cm; volume $= \pi (0.1)^2 L = 0.01\pi L$.
3. $0.01\pi L = 36\pi \Rightarrow L = 3600$ cm $= 36$ m.

Final Answer: 36 m.

Takeaway: Wire is a thin cylinder; length $= V/(\pi r^2)$.

Example 3: Recasting a cylinder into a sphere

A solid metallic cylinder of radius 6 cm and height 32 cm is melted and recast into a single sphere. Find the radius of the sphere.

Solution:

1. Cylinder volume $= \pi (6)^2(32) = 1152\pi$ cm$^3$.
2. Set $\dfrac43\pi R^3 = 1152\pi \Rightarrow R^3 = \dfrac{3\times 1152}{4} = 864$.
3. $R = \sqrt[3]{864} \approx 9.52$ cm.

Final Answer: $R = \sqrt[3]{864} \approx 9.52$ cm.

Takeaway: Equate volumes and solve for the new radius.

Example 4: Lead shots in a cone of water

A conical vessel (height 8 cm, top radius 5 cm) is full of water. When lead shots (spheres of radius 0.5 cm) are dropped in, one-fourth of the water overflows. How many shots were dropped? $\left(\pi\text{ cancels}\right)$

Solution:

1. Volume of cone (water) $= \dfrac13\pi (5)^2(8) = \dfrac{200\pi}{3}$ cm$^3$.
2. Water displaced $= \dfrac14\times\dfrac{200\pi}{3} = \dfrac{50\pi}{3}$ cm$^3$.
3. One shot $= \dfrac43\pi (0.5)^3 = \dfrac{\pi}{6}$ cm$^3$.
4. Number $= \dfrac{50\pi/3}{\pi/6} = \dfrac{50}{3}\times 6 = 100$.

Final Answer: 100 shots.

Takeaway: Displaced water volume $\div$ one shot's volume $=$ number.