Surface Areas and Volumes: Conservation of Volume & Melting Class 10

In metallurgy, civil excavation, and industrial manufacturing, physical matter changes its geometric shape continuously without altering its total quantity. When a solid metallic sphere is melted down and recast into hundreds of tiny spherical ball bearings, or when earth excavated from a circular water well is spread around its rim to form a raised protective embankment, the underlying physical principle is the Law of Conservation of Volume: $\mathbf{\text{Volume of Solid Before Reshaping} = \text{Volume of Solid After Reshaping}}$

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), melting, recasting, and excavation problems carry between $4$ and $5$ marks in Section D.

In this master guide, we break down the two most famous 5-mark board exam conversion archetypes: Melting and Recasting Multiple Objects, and The Well and Embankment Excavation Problem.

What You Will Learn

• The Conservation of Volume Principle in geometry
• Finding the number ($n$) of smaller objects cast from a large metallic solid: $n = \frac{V_{\text{large}}}{V_{\text{small}}}$
• Master Problem 1: Melting Three Metallic Spheres into a Single Solid Sphere (NCERT Classic)
• Master Problem 2: The Well and Circular Embankment Problem (Excavation into an Annulus Ring)
• Master Problem 3: Melting a Metallic Cone into a Cylindrical Wire (Finding wire diameter)
• Algebraic shortcuts that eliminate unnecessary decimal multiplications

1. The Core Algebraic Rule of Conversions

    Scenario 1: One Solid ───> Cast into n Small Solids
    Volume of Large Solid = n × (Volume of 1 Small Solid)
    ===> Number of solids: n = Volume(Large) / Volume(Small)

    Scenario 2: Excavated Well ───> Earth Spread into Embankment
    Volume of Earth Dug Out (Cylinder) = Volume of Embankment (Hollow Cylinder / Annulus)

Important: <u>In melting and recasting problems, NEVER calculate numerical values of $\pi$ ($22/7$ or $3.14$) on both sides! The constant $\pi$ appears on BOTH sides of the volume equation and cancels out completely!</u>

2. Master Problem 1: Melting Three Spheres into One Large Sphere

Problem Statement:

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.

          Sphere 1 (r1=6)   +   Sphere 2 (r2=8)   +   Sphere 3 (r3=10)   ───>   Large Sphere (R)

Step-by-Step Solution:

1. Analyze Radii:
   • Radius of sphere 1: $r_1 = 6\text{ cm}$
   • Radius of sphere 2: $r_2 = 8\text{ cm}$
   • Radius of sphere 3: $r_3 = 10\text{ cm}$
   • Let the radius of the resulting large sphere be $R$.
2. Apply Conservation of Volume: $\text{Volume of Resulting Sphere} = \text{Volume}(1) + \text{Volume}(2) + \text{Volume}(3)$ $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3$
3. Factor Out Common Terms: $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi (r_1^3 + r_2^3 + r_3^3)$ Cancel $\frac{4}{3}\pi$ from both sides: $\mathbf{R^3 = r_1^3 + r_2^3 + r_3^3}$
4. Compute Cubes: $R^3 = 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = \mathbf{1728}$
5. Take Cube Root: $R = \sqrt[3]{1728} = \mathbf{12\text{ cm}}$ (Since $12 \times 12 \times 12 = 144 \times 12 = 1728$).
6. Therefore, <u>the radius of the resulting sphere is $12\text{ cm}$</u>.

3. Master Problem 2: The Well and Circular Embankment (5-Mark Classic)

Problem Statement:

A well of diameter $3\text{ m}$ is dug $14\text{ m}$ deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width $4\text{ m}$ to form an embankment. Find the height of the embankment.

                          Well (Excavated Cylinder):
                          Diameter = 3 m  ===>  Internal Radius r = 1.5 m
                          Depth h = 14 m
                          
                          Embankment (Hollow Annulus Ring):
                          Internal Radius r = 1.5 m
                          Width w = 4 m  ===>  External Radius R = 1.5 + 4 = 5.5 m
                          Height of Embankment = H

Step-by-Step Solution:

1. Dimensions of the Excavated Well:
   • Radius of well: $r = \frac{3}{2}\text{ m} = \mathbf{1.5\text{ m}}$.
   • Depth of well: $h = \mathbf{14\text{ m}}$.
2. Calculate Volume of Earth Dug Out ($V_{\text{earth}}$): $V_{\text{earth}} = \pi r^2 h = \pi \times (1.5)^2 \times 14 = \pi \times 2.25 \times 14 = \mathbf{31.5\pi\text{ m}^3}$
3. Dimensions of the Embankment (Hollow Ring):
   • Internal radius of ring: $r = 1.5\text{ m}$.
   • Width of embankment: $w = 4\text{ m}$.
   • External radius of ring: $R = r + w = 1.5 + 4 = \mathbf{5.5\text{ m}}$.
   • Let the height of the embankment be $H$.
4. Area of the Embankment Base (Circular Ring / Annulus): $\text{Base Area} = \pi (R^2 - r^2)$ $\text{Base Area} = \pi [(5.5)^2 - (1.5)^2]$ Apply $a^2 - b^2 = (a - b)(a + b)$: $\text{Base Area} = \pi (5.5 - 1.5)(5.5 + 1.5) = \pi (4)(7) = \mathbf{28\pi\text{ m}^2}$
5. Apply Conservation of Volume: $\text{Volume of Embankment} = \text{Volume of Earth Dug Out}$ $\text{Base Area} \times H = V_{\text{earth}}$ $28\pi \times H = 31.5\pi$ Cancel $\pi$ from both sides: $H = \frac{31.5}{28} = \frac{63}{56} = \frac{9}{8} = \mathbf{1.125\text{ metres}}$
6. Therefore, <u>the height of the embankment is $1.125\text{ metres}$</u>.

4. Summary and Examination Tips

Exam Tip: In the embankment problem, remember that the embankment is an open hollow ring (annulus) because the well mouth itself remains open! Its base area is strictly $\pi(R^2 - r^2)$, NOT $\pi R^2$.

Common Mistake: In calculating the external radius $R$ of the embankment, writing $R = 4\text{ m}$. The width of the embankment is $4\text{ m}$, so the external radius from the center is $R = r + \text{width} = 1.5 + 4 = 5.5\text{ m}$!