The total surface area of a solid hemisphere is 4158 cm 2. It is melted to form a cylinder whose height is 1 cm and radius of base 7 cm. How many cylinders will be formed?
126
The problem involves melting a solid hemisphere and recasting it into smaller cylinders. When a solid object is melted and reshaped, its volume remains constant. Therefore, the total volume of the hemisphere will be equal to the total volume of all the cylinders formed.
We are given the total surface area of the solid hemisphere and the dimensions (radius and height) of the cylinder. We need to find the number of cylinders that can be formed.
Step 1: Find the Radius of the Solid Hemisphere
The total surface area of a solid hemisphere is given by the formula: \( \text{Total Surface Area} = 3\pi R^2 \), where \(R\) is the radius of the hemisphere.
Given, Total Surface Area = 4158 cm².
So, \( 3\pi R^2 = 4158 \).
Using \( \pi = \frac{22}{7} \):
\( 3 \times \frac{22}{7} \times R^2 = 4158 \)
\( \frac{66}{7} \times R^2 = 4158 \)
\( R^2 = \frac{4158 \times 7}{66} \)
\( R^2 = \frac{63 \times 66 \times 7}{66} \)
\( R^2 = 63 \times 7 \)
\( R^2 = 441 \)
\( R = \sqrt{441} \)
\( R = 21 \) cm
The radius of the solid hemisphere is 21 cm.
Step 2: Calculate the Volume of the Solid Hemisphere
The volume of a solid hemisphere is given by the formula: \( \text{Volume} = \frac{2}{3}\pi R^3 \).
Using \( R = 21 \) cm and \( \pi = \frac{22}{7} \):
\( \text{Volume of Hemisphere} = \frac{2}{3} \times \frac{22}{7} \times (21)^3 \)
\( \text{Volume of Hemisphere} = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 \)
\( \text{Volume of Hemisphere} = \frac{2}{3} \times \frac{22}{7} \times (3 \times 7) \times 21 \times 21 \)
\( \text{Volume of Hemisphere} = 2 \times 22 \times 21 \times 21 \)
\( \text{Volume of Hemisphere} = 44 \times 441 \)
\( \text{Volume of Hemisphere} = 19404 \) cm³
Step 3: Calculate the Volume of One Cylinder
The volume of a cylinder is given by the formula: \( \text{Volume} = \pi r^2 h \), where \(r\) is the radius of the base and \(h\) is the height.
Given, radius of cylinder \(r_{cyl} = 7 \) cm and height of cylinder \(h_{cyl} = 1 \) cm.
Using \( \pi = \frac{22}{7} \):
\( \text{Volume of Cylinder} = \frac{22}{7} \times (7)^2 \times 1 \)
\( \text{Volume of Cylinder} = \frac{22}{7} \times 49 \times 1 \)
\( \text{Volume of Cylinder} = 22 \times 7 \)
\( \text{Volume of Cylinder} = 154 \) cm³
Step 4: Find the Number of Cylinders Formed
Since the volume is conserved during melting and recasting, the total volume of the hemisphere is equal to the volume of \(N\) cylinders, where \(N\) is the number of cylinders formed.
\( \text{Volume of Hemisphere} = N \times \text{Volume of One Cylinder} \)
\( 19404 = N \times 154 \)
\( N = \frac{19404}{154} \)
Let's perform the division:
\( N = 126 \)
So, 126 cylinders will be formed.
Let's summarize the key values:
| Shape | Parameter | Value | Formula |
|---|---|---|---|
| Solid Hemisphere | Total Surface Area | 4158 cm² | \(3\pi R^2\) |
| Solid Hemisphere | Radius (R) | 21 cm | Derived from TSA |
| Solid Hemisphere | Volume | 19404 cm³ | \(\frac{2}{3}\pi R^3\) |
| Cylinder | Radius (r) | 7 cm | Given |
| Cylinder | Height (h) | 1 cm | Given |
| Cylinder | Volume | 154 cm³ | \(\pi r^2 h\) |
Number of cylinders = \( \frac{\text{Volume of Hemisphere}}{\text{Volume of Cylinder}} = \frac{19404}{154} = 126 \).
Revision Table: Hemisphere and Cylinder Formulas
| Shape | Volume Formula | Surface Area Formulae (where applicable) |
|---|---|---|
| Sphere | \( \frac{4}{3}\pi R^3 \) | Total Surface Area: \( 4\pi R^2 \) |
| Solid Hemisphere | \( \frac{2}{3}\pi R^3 \) | Curved Surface Area: \( 2\pi R^2 \) Total Surface Area: \( 3\pi R^2 \) |
| Cylinder | \( \pi r^2 h \) | Curved Surface Area: \( 2\pi r h \) Total Surface Area: \( 2\pi r (r+h) \) |
Additional Information: Volume Conservation in Melting and Recasting
In problems where a solid object is melted and recast into one or more objects of different shapes, a fundamental principle is applied: the volume of the material remains constant. This is because melting and reshaping only change the form of the substance, not the amount of substance present. The density of the material is assumed to be uniform and constant throughout the process.
- Process: Melting a solid \( \rightarrow \) Forming liquid \( \rightarrow \) Recasting into new solid shapes.
- Conservation: The total volume before melting equals the total volume after recasting.
- Application: This principle allows us to equate the volume of the original shape(s) to the sum of the volumes of the new shape(s) to solve for unknown dimensions or the number of new shapes.
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