Melting and Reshaping Solids

In the previous chapter, "Volume of Combined Solids", we learned how to calculate the total space occupied by objects that are stuck together—like a cone sitting on top of a hemisphere. But what happens if we take that entire solid, throw it into a furnace, melt it down into a liquid, and pour that liquid into a brand new mould?

This is the art of melting and reshaping solids. It is one of the most predictable, high-scoring, and logical topics in geometry. Once you understand the single golden rule that governs this process, every single problem—whether it asks you about melting wax, forging iron, or casting silver coins—becomes a simple balancing act of equations.

Let us build this concept from the ground up.

The Golden Rule: Conservation of Volume

Imagine you have a ball of play-dough. If you roll it between your hands, you can turn it into a long snake (a cylinder). If you squash it on the table, you can shape it into a box (a cuboid).

During this entire process, the shape of the play-dough changed drastically. The surface area—the amount of skin you can touch—changed completely. But did the amount of play-dough change? No. You didn't add any extra dough from the box, and you didn't throw any away. The total amount of material remained exactly the same.

In mathematics, we call this the Conservation of Volume.

This is the engine that drives every single problem in this chapter. The exam will try to distract you with complex shapes, changing radii, and different heights. Ignore the noise. Just remember that the liquid metal poured out of the first shape must perfectly fill the second shape.

The Three Scenarios of Reshaping

When we melt and recast objects, we generally face one of three scenarios:

1. 1One-to-One: Melting one large object to make one different large object (e.g., melting a sphere to make a cylinder).
2. 2One-to-Many: Melting one large object to make many smaller identical objects (e.g., melting a large block of lead to make hundreds of small spherical bullets).
3. 3Many-to-One: Melting several smaller objects to combine them into one massive object (e.g., melting three small iron spheres to cast one giant iron sphere).

We will look at how to set up the math for each of these, but first, we need to talk about how to calculate smartly.

The Golden Rule of Calculation: Never Multiply Early

Before we solve our first problem, we need to establish a rule that will save you minutes of precious time and protect you from calculation errors.

When you set up an equation where Volume A = Volume B, do not calculate the final volume of the first shape before setting up the equation.

Why? Because both sides of your equation will often contain π, fractions, and common multiples. If you multiply everything out on the left side, you create a massive, ugly number. Then, you have to divide that massive number by the numbers on the right side.

Instead, leave everything as a multiplication string. Let the numbers sit there. When you bring the numbers from the right side over to the left side, you will find that almost everything cancels out beautifully. π will cancel with π. A 3 in the denominator will cancel with a 9 in the numerator.

Scenario 1: The One-to-One Transformation

This is the simplest type of problem. You have one object, you melt it, and you make one new object.

The Strategy:

1. 1Write the formula for the volume of the old shape.
2. 2Write the formula for the volume of the new shape.
3. 3Put an equals sign between them.
4. 4Plug in the numbers you know, and solve for the one number you don't know.

Walkthrough: Sphere to Cylinder

The Problem: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a solid cylinder of radius 6 cm. Find the height of the cylinder.

Step 1: Identify the shapes and the rule. Old shape = Sphere. New shape = Cylinder. Rule: Volume of Sphere = Volume of Cylinder.

Step 2: Write the formulas. 4⁄3 π r_sphere³ = π r_cylinder² h

Step 3: Plug in the values. Notice that we have π on both sides. We can immediately cancel it out! This is why we never multiply early. 4⁄3 × (4.2)³ = (6)² × h

Let's expand this so it's easy to see: 4⁄3 × 4.2 × 4.2 × 4.2 = 36 × h

Step 4: Isolate the unknown (h) and solve. Move the 36 to the other side (it goes to the denominator): h = 4 × 4.2 × 4.2 × 4.2⁄3 × 36

Now, let's cancel. Cancel 4 and 36: The 36 becomes a 9. h = 4.2 × 4.2 × 4.2⁄3 × 9

Cancel 3 with 4.2: 4.2 ÷ 3 = 1.4. h = 1.4 × 4.2 × 4.2⁄9

To make it even easier, let's write the decimals as fractions: h = 14⁄10 × 42⁄10 × 42⁄10 × 1⁄9

Cancel 9 (which is 3 × 3) with the two 42s (since 42 ÷ 3 = 14): h = 14⁄10 × 14⁄10 × 14⁄10

h = 14 × 14 × 14⁄1000 h = 2744⁄1000 h = 2.744 cm.

The height of the new cylinder is 2.744 cm. By cancelling fractions instead of multiplying 4.2 × 4.2 × 4.2 right away, we avoided a massive decimal headache.

Scenario 2: One Big Shape into Many Small Shapes

This is the most frequently tested scenario. You are given a large block of metal (usually a cuboid or a large sphere) and told it is melted down to make hundreds of small, identical objects (like bullets, marbles, or coins). You are asked to find the number of small objects produced.

Think of this like having a large jug of water (the big volume) and pouring it into several small glasses (the small volumes). How do you find the number of glasses you can fill? You divide the total amount of water by the capacity of one glass.

Beware the Hidden Shapes: Coins and Wires

Before we solve a "many objects" problem, we must address a classic trap. The exam will often ask you to melt something into "coins" or "wires".

If you read "coin", your brain might immediately think "circle". But a circle is a flat, 2D shape. It has no volume. A real coin has a radius, but it also has a small thickness. A coin is a very short cylinder. The thickness of the coin is the height (h) of the cylinder.

Similarly, if you read "wire", your brain might think of a 1D line. But a wire has a circular cross-section and a length. A wire is a very long, thin cylinder. The length of the wire is the height (h) of the cylinder, and its thickness is the diameter.

Walkthrough: The Silver Coins Problem

The Problem: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a solid cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Step 1: Check your units! This is the first trap. The cuboid is in cm. The coin's diameter is in cm. But the coin's thickness is in mm. We must convert everything to the same unit. Let's use centimeters. Thickness of coin (h) = 2 mm = 0.2 cm.

Step 2: Identify the shapes and dimensions. Large Shape = Cuboid. Dimensions: l = 5.5, b = 10, h = 3.5.

Small Shape = Coin (Cylinder). Diameter = 1.75 cm. Radius (r) = 1.75⁄2 cm. (Leave it as a fraction for now!) Height (h) = 0.2 cm.

Step 3: Set up the formula. n = Volume of Cuboid⁄Volume of ONE Coin n = l × b × h⁄π r² h

Step 4: Plug in the numbers and use fraction magic. Let's calculate the volume of the cuboid first, as it's easy: V_cuboid = 5.5 × 10 × 3.5 = 55 × 3.5 = 192.5 cm³. To make cancellation easier later, let's write 192.5 as a fraction: 1925⁄10, which simplifies to 385⁄2.

Now, let's write out the volume of one coin. We will convert the decimals to fractions because fractions are much easier to cancel. r = 1.75⁄2 = 175⁄200 = 7⁄8 cm. (Dividing top and bottom by 25). h = 0.2 = 2⁄10 = 1⁄5 cm.

V_coin = 22⁄7 × (7⁄8)² × 1⁄5 V_coin = 22⁄7 × 7⁄8 × 7⁄8 × 1⁄5

Cancel one 7 from the top and bottom: V_coin = 22 × 7⁄8 × 8 × 5 V_coin = 154⁄320 = 77⁄160 cm³.

Step 5: Divide to find n. n = V_cuboid⁄V_coin n = 385/2⁄77/160

When dividing by a fraction, we flip it and multiply: n = 385⁄2 × 160⁄77

Cancel 2 and 160: n = 385 × 80⁄77

Does 77 go into 385? Let's check the last digit. 7 × 5 = 35. Let's try 77 × 5. 70 × 5 = 350. 7 × 5 = 35. 350 + 35 = 385. Yes! It goes in exactly 5 times. n = 5 × 80 n = 400

You need exactly 400 coins to form that cuboid. Notice how converting decimals to fractions (1.75/2 → 7/8) turned a nightmare calculation into a smooth, satisfying series of cancellations.

Scenario 3: Many-to-One (Combining Solids)

Sometimes, instead of breaking a large object apart, we melt several smaller objects and pool their liquid metal together to cast one giant object.

The logic is exactly the same, just in reverse: Sum of Volumes of Small Solids = Volume of the New Large Solid

Walkthrough: Three Spheres into One

The Problem: Three solid metallic spheres of radii 6 cm, 8 cm, and 10 cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Step 1: Set up the equation. Let the radius of the new giant sphere be R. Volume of Sphere 1 + Volume of Sphere 2 + Volume of Sphere 3 = Volume of Giant Sphere 4⁄3π(r₁)³ + 4⁄3π(r₂)³ + 4⁄3π(r₃)³ = 4⁄3π(R)³

Step 2: Factor out the common terms. Notice that every single term has a 4⁄3π in it. We can factor that out on the left side: 4⁄3π × [ (r₁)³ + (r₂)³ + (r₃)³ ] = 4⁄3π × (R)³

Because 4⁄3π is multiplying both sides, we can divide both sides by it. It completely vanishes! (r₁)³ + (r₂)³ + (r₃)³ = R³

Step 3: Plug in the numbers. 6³ + 8³ + 10³ = R³ 216 + 512 + 1000 = R³ 1728 = R³

Step 4: Solve for R. We need to find the cube root of 1728. If you don't have it memorized, you can estimate. 10³ = 1000. 20³ = 8000. So the answer is between 10 and 20. Look at the last digit: 8. What number, when cubed, ends in 8? 2 × 2 × 2 = 8. So let's test 12. 12 × 12 = 144. 144 × 12 = 1728. Perfect.

R = 12 cm. The radius of the new, combined sphere is 12 cm.

The Universal Step-by-Step Method

No matter how strange the shapes are, or how wordy the problem is, you can solve any melting and recasting problem by following this exact sequence:

One More Trap: The "Drawn into a Wire" Problem

We mentioned earlier that a wire is a cylinder. Let's look at exactly how this is tested, because it often involves a massive difference in units.

The Problem: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Note: "Drawn into" is just metallurgical language for stretching and reshaping. The volume is still conserved.

Step 1: Units and Shapes. Old shape: Rod (Cylinder). Diameter = 1 cm → Radius = 0.5 cm. Length (h) = 8 cm. New shape: Wire (Cylinder). Length (h) = 18 m. Trap spotted! Convert 18 m to cm. 18 × 100 = 1800 cm. Let the radius of the wire be r.

Step 2: Set up the equation. Volume of Rod = Volume of Wire π × (R_rod)² × H_rod = π × (r_wire)² × H_wire

Step 3: Cancel π and plug in numbers. (0.5)² × 8 = r² × 1800 0.25 × 8 = r² × 1800 2 = r² × 1800

Step 4: Solve for r. r² = 2⁄1800 r² = 1⁄900

Take the square root of both sides: r = 1⁄30 cm.

Step 5: Read the final question carefully. The question asked for the thickness of the wire. The thickness of a wire is its diameter, not its radius! Diameter = 2 × r = 2 × 1⁄30 = 1⁄15 cm.

The thickness of the wire is 1⁄15 cm (which is about 0.67 mm).

Quick Revision

Before you walk into the exam, make sure these core truths are locked in your memory:

• The Golden Rule: When a solid is melted and recast, its Volume remains exactly the same.
• Surface Area Changes: Never assume surface area is conserved. A flattened shape has much more surface area than a spherical one, even if they have the same volume.
• The "n" Formula: To find how many small objects can be made from a big one, use n = Volume of Big Object⁄Volume of One Small Object.
• Hidden Cylinders: A "coin" is a cylinder where the thickness is the height. A "wire" is a cylinder where the length is the height.
• Thickness = Diameter: If a question asks for the "thickness" of a wire or a rod, it is asking for the diameter, which is 2 × radius.
• Unit Check: Always scan the problem for mixed units (cm mixed with mm or m). Convert everything to a single unit before writing your equation.
• Don't Multiply Early: Write out your equations fully and cancel π and other common factors across the equals sign. Convert decimals to fractions to make cancellation easier.