Conversion of Solid from One Shape to Another for CBSE Class 10

When an artisan melts scrap copper vessels and pours the molten liquid into a mold to cast solid cylindrical wires, or when a child rolls a lump of clay from a spherical ball into an elongated cone, what happens to the physical properties of the material? The shape changes completely; the surface area expands or contracts; but the total quantity of matter remains strictly unchanged.

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), the conversion of solids from one shape to another is governed by a single foundational physical axiom: The Conservation of Volume.

What You Will Learn

• The Law of Conservation of Volume during melting, recasting, and reshaping
• Solving for unknown dimensions: Equating Initial Volume to Final Volume
• Melting large metallic spheres to recast into cylindrical wires or smaller spheres
• Melting metallic spheres of multiple different radii to form a single large sphere
• Calculating the number of small solids formed: $n = \frac{\text{Total Volume}}{\text{Volume of One Small Solid}}$
• Step-by-step solved CBSE board examination problems and algebraic shortcuts

1. The Principle of Conservation of Volume

Whenever a solid is melted and recast into a different shape (or when water from a cylindrical container is poured into a hemispherical bowl):

The Fundamental Principle: <u>The volume of the newly formed solid(s) is ALWAYS EXACTLY EQUAL to the volume of the original solid(s) before melting!</u>

$\mathbf{\text{Volume of Solid Before Melting} = \text{Volume of Solid After Recasting}}$

2. Calculating the Number of Small Solids ($n$)

When a large solid of volume $V_{\text{large}}$ is melted to manufacture multiple identical smaller objects, each of volume $V_{\text{small}}$:

$\mathbf{n = \frac{\text{Volume of Original Large Solid}}{\text{Volume of One Small Recast Solid}}}$

• Common Examples: Number of spherical lead shots obtained from a solid cuboid, number of circular coins melted to form a solid bar, or number of cones made from a sphere.

3. High-Yield Solved Board Examination Problems

Solved Example 1: Melting Sphere to Form a Cylinder (NCERT Classic)

Problem: A metallic sphere of radius $4.2\text{ cm}$ is melted and recast into the shape of a cylinder of radius $6\text{ cm}$. Find the height of the cylinder.

Solution:

1. Analyze Dimensions:
   • Sphere radius: $R = 4.2\text{ cm} = \frac{42}{10} = \frac{21}{5}\text{ cm}$.
   • Cylinder radius: $r = 6\text{ cm}$.
   • Let height of the cylinder be $h\text{ cm}$.
2. Equate Volumes: $\text{Volume of Cylinder} = \text{Volume of Sphere}$ $\pi r^2 h = \frac{4}{3}\pi R^3$
3. Cancel $\pi$ on Both Sides: $r^2 h = \frac{4}{3} R^3$ $6^2 \times h = \frac{4}{3} \times (4.2)^3$ $36 \times h = \frac{4}{3} \times 4.2 \times 4.2 \times 4.2$
4. Solve for $h$: $36 h = 4 \times 1.4 \times 4.2 \times 4.2$ $h = \frac{4 \times 1.4 \times 4.2 \times 4.2}{36} = \frac{1.4 \times 4.2 \times 4.2}{9} = 1.4 \times 1.4 \times 1.4 = (1.4)^3 = \mathbf{2.744\text{ cm}}$
5. Therefore, <u>the height of the recast cylinder is $2.74\text{ cm}$</u>.

Solved Example 2: Three Spheres Melted into One Large Sphere

Problem: Metallic spheres of radii $6\text{ cm}$, $8\text{ cm}$, and $10\text{ cm}$, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:

1. List the Given Radii:
   • $r_1 = 6\text{ cm}, \quad r_2 = 8\text{ cm}, \quad r_3 = 10\text{ cm}$.
   • Let the radius of the resulting single large sphere be $R\text{ cm}$.
2. Apply Conservation of Volume: $\text{Volume of Large Sphere} = \text{Vol}(r_1) + \text{Vol}(r_2) + \text{Vol}(r_3)$ $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3$
3. Factor and Cancel $\frac{4}{3}\pi$: $R^3 = r_1^3 + r_2^3 + r_3^3$ $R^3 = 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728$
4. Take the Cube Root: $R = \sqrt[3]{1728} = \mathbf{12\text{ cm}}$
5. Therefore, <u>the radius of the resulting large sphere is $12\text{ cm}$</u>.

Solved Example 3: Number of Coins Melted to Form a Cuboid (NCERT Classic)

Problem: How many silver coins, $1.75\text{ cm}$ in diameter and of thickness $2\text{ mm}$, must be melted to form a cuboid of dimensions $5.5\text{ cm} \times 10\text{ cm} \times 3.5\text{ cm}$? (Use $\pi = 22/7$).

Solution:

1. Analyze Dimensions (Convert All to Centimetres!):
   • A coin is a short, wide cylinder!
   • Diameter $d = 1.75\text{ cm} \implies$ Radius $r = \frac{1.75}{2} = \frac{175}{200} = \frac{7}{8}\text{ cm}$.
   • Thickness (height) of coin: $h = 2\text{ mm} = \frac{2}{10} = \frac{1}{5}\text{ cm} = 0.2\text{ cm}$.
   • Dimensions of cuboid: $L = 5.5\text{ cm}, B = 10\text{ cm}, H = 3.5\text{ cm}$.
2. Calculate Volume of the Cuboid: $V_{\text{cuboid}} = 5.5 \times 10 \times 3.5 = 55 \times 3.5 = \mathbf{192.5\text{ cm}^3}$
3. Calculate Volume of ONE Silver Coin (Cylinder): $V_{\text{coin}} = \pi r^2 h = \frac{22}{7} \times \left(\frac{7}{8}\right)^2 \times \frac{1}{5} = \frac{22}{7} \times \frac{49}{64} \times \frac{1}{5} = \frac{11 \times 7}{32 \times 5} = \mathbf{\frac{77}{160}\text{ cm}^3}$
4. Calculate Number of Coins ($n$): $n = \frac{V_{\text{cuboid}}}{V_{\text{coin}}} = \frac{192.5}{\frac{77}{160}} = \frac{192.5 \times 160}{77} = \frac{30800}{77} = \mathbf{400}$
5. Therefore, <u>exactly $400$ silver coins must be melted to form the cuboid</u>.

4. Summary and Examination Tips

Exam Tip: In melting problems, NEVER multiply out the numerical value of $\pi$ ($3.14$ or $22/7$) on both sides! Leave $\pi$ intact as a symbol; it cancels out on both sides of the equation in the first step.

Common Mistake: Forgetting that a coin is a cylinder. Students often get confused by the word "coin"—a coin has a circular face and a finite thickness, which makes it a cylinder of height equal to its thickness!