Three solid lead spheres of radius 6 cm, 8 cm and 10 cm are melted together and recast as a solid sphere. What is the percentage diminution of the surface area as compared to the sum of the surface areas of the three spheres ?
28%
Understanding the Geometry Problem: Melting Spheres
The question asks us to find the percentage decrease in the total surface area when three solid lead spheres are melted together and reformed into a single solid sphere. The key principle here is that when materials are melted and recast, the total volume remains constant. However, the surface area usually changes.
We are given the radii of the three initial spheres:
- Radius of sphere 1: \(r_1 = 6\) cm
- Radius of sphere 2: \(r_2 = 8\) cm
- Radius of sphere 3: \(r_3 = 10\) cm
We need to find the radius of the new, larger sphere and then compare the total surface area before and after melting.
Calculating the Total Volume Before Melting
The volume of a sphere is given by the formula \(V = \frac{4}{3} \pi r^3\).
Let's calculate the volume of each of the three initial spheres:
- Volume of sphere 1: \(V_1 = \frac{4}{3} \pi r_1^3 = \frac{4}{3} \pi (6 \text{ cm})^3 = \frac{4}{3} \pi (216) \text{ cm}^3\)
- Volume of sphere 2: \(V_2 = \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi (8 \text{ cm})^3 = \frac{4}{3} \pi (512) \text{ cm}^3\)
- Volume of sphere 3: \(V_3 = \frac{4}{3} \pi r_3^3 = \frac{4}{3} \pi (10 \text{ cm})^3 = \frac{4}{3} \pi (1000) \text{ cm}^3\)
The total volume of the three spheres before melting is the sum of their individual volumes:
Total Volume (\(V_{total}\)) \(= V_1 + V_2 + V_3\)
\(V_{total} = \frac{4}{3} \pi (216) + \frac{4}{3} \pi (512) + \frac{4}{3} \pi (1000)\)
\(V_{total} = \frac{4}{3} \pi (216 + 512 + 1000)\)
\(V_{total} = \frac{4}{3} \pi (1728) \text{ cm}^3\)
Finding the Radius of the New Solid Sphere
When the three spheres are melted and recast into a single sphere, the total volume remains the same. Let \(R\) be the radius of the new, larger sphere. Its volume will be \(V_{new} = \frac{4}{3} \pi R^3\).
Since \(V_{new} = V_{total}\):
\(\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (1728)\)
We can cancel \(\frac{4}{3} \pi\) from both sides:
\(R^3 = 1728\)
To find \(R\), we need to calculate the cube root of 1728.
\(R = \sqrt[3]{1728}\)
We know that \(10^3 = 1000\) and \(12^3 = 1728\).
So, the radius of the new sphere is \(R = 12\) cm.
Calculating Surface Areas
The surface area of a sphere is given by the formula \(SA = 4 \pi r^2\).
First, let's find the total surface area of the three initial spheres:
- Surface area of sphere 1: \(SA_1 = 4 \pi r_1^2 = 4 \pi (6 \text{ cm})^2 = 4 \pi (36) = 144 \pi \text{ cm}^2\)
- Surface area of sphere 2: \(SA_2 = 4 \pi r_2^2 = 4 \pi (8 \text{ cm})^2 = 4 \pi (64) = 256 \pi \text{ cm}^2\)
- Surface area of sphere 3: \(SA_3 = 4 \pi r_3^2 = 4 \pi (10 \text{ cm})^2 = 4 \pi (100) = 400 \pi \text{ cm}^2\)
The sum of the surface areas of the three spheres (\(SA_{initial}\)) is:
\(SA_{initial} = SA_1 + SA_2 + SA_3 = 144 \pi + 256 \pi + 400 \pi = 800 \pi \text{ cm}^2\)
Now, let's find the surface area of the new, larger sphere (\(SA_{new}\)) with radius \(R=12\) cm:
\(SA_{new} = 4 \pi R^2 = 4 \pi (12 \text{ cm})^2 = 4 \pi (144) = 576 \pi \text{ cm}^2\)
Calculating Percentage Diminution of Surface Area
Diminution means decrease. The decrease in surface area is the difference between the initial total surface area and the new surface area.
Diminution in Surface Area \(= SA_{initial} - SA_{new}\)
Diminution \(= 800 \pi - 576 \pi = 224 \pi \text{ cm}^2\)
The percentage diminution is calculated as:
Percentage Diminution \(= \frac{\text{Diminution in Surface Area}}{\text{Initial Total Surface Area}} \times 100\%\)
Percentage Diminution \(= \frac{224 \pi}{800 \pi} \times 100\%\)
We can cancel \(\pi\) from the numerator and denominator:
Percentage Diminution \(= \frac{224}{800} \times 100\%\)
Percentage Diminution \(= \frac{224}{8} \%\)
Percentage Diminution \(= 28\%\)
So, the percentage diminution of the surface area is 28%.
Summary of Results
Here is a quick summary of the calculations:
| Sphere | Radius (cm) | Volume (\(\frac{4}{3} \pi r^3\)) (\(\text{cm}^3\)) | Surface Area (\(4 \pi r^2\)) (\(\text{cm}^2\)) |
|---|---|---|---|
| Sphere 1 | 6 | \(\frac{4}{3} \pi (216)\) | \(144 \pi\) |
| Sphere 2 | 8 | \(\frac{4}{3} \pi (512)\) | \(256 \pi\) |
| Sphere 3 | 10 | \(\frac{4}{3} \pi (1000)\) | \(400 \pi\) |
| Total Initial | - | \(\frac{4}{3} \pi (1728)\) | \(800 \pi\) |
| New Sphere | 12 (since \(12^3=1728\)) | \(\frac{4}{3} \pi (1728)\) | \(576 \pi\) |
Initial total surface area \(= 800 \pi \text{ cm}^2\)
New surface area \(= 576 \pi \text{ cm}^2\)
Diminution \(= 800 \pi - 576 \pi = 224 \pi \text{ cm}^2\)
Percentage Diminution \(= \frac{224 \pi}{800 \pi} \times 100\% = 28\%\)
Revision Table: Key Concepts for Melting & Recasting
| Concept | Description | Impact on Melting/Recasting |
|---|---|---|
| Volume | The amount of space a solid occupies. Formula for sphere: \(\frac{4}{3} \pi r^3\). | Conserved: Total volume before equals total volume after. This is used to find the new dimension (radius). |
| Surface Area | The total area of the outer surface of a solid. Formula for sphere: \(4 \pi r^2\). | Usually Not Conserved: The total surface area changes because the shape/size changes. We calculate the percentage change. |
| Radius (r or R) | The distance from the center to the surface of a sphere. | A fundamental dimension needed for volume and surface area calculations. Finding the new radius is a key step. |
| Percentage Diminution | The relative decrease expressed as a percentage. | \((\text{Initial Value} - \text{Final Value}) / \text{Initial Value} \times 100\%\). Used here for surface area. |
Additional Information: Sphere Properties and Volume Conservation
Understanding the properties of spheres is crucial for solving this type of geometry problem. A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a perfectly round ball.
- All points on the surface of a sphere are equidistant from its center. This distance is the radius (\(r\)).
- The formulas for the volume (\(\frac{4}{3} \pi r^3\)) and surface area (\(4 \pi r^2\)) are standard and should be memorized for geometry and mensuration problems.
The principle of volume conservation is fundamental in problems involving melting and recasting solids. It assumes no loss of material during the process. This means the sum of the volumes of the original objects equals the volume of the new object formed.
In this specific problem, three spheres combine their volumes to form one larger sphere. This is why we first calculated the total volume and then used it to find the radius of the resulting sphere. The change in surface area happens because a single larger sphere has a smaller surface-area-to-volume ratio than multiple smaller spheres with the same total volume. Imagine combining small bubbles into one large bubble; the total surface area decreases.
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