A cone of height 16 cm and diameter 14 cm is mounted on a hemisphere of same diameter. What is the volume of the solid thus formed? (take π = 22/7)
1540 cm 3
The problem asks us to find the total volume of a solid formed by mounting a cone on top of a hemisphere. Both the cone and the hemisphere share the same diameter.
The solid is composed of two parts:
We are given the following dimensions:
The diameter for both the cone and the hemisphere is 14 cm. The radius (r) is half of the diameter.
Radius \(r = \frac{\text{Diameter}}{2} = \frac{14 \text{ cm}}{2} = 7 \text{ cm}\)
So, the radius for both the cone and the hemisphere is 7 cm.
The formula for the volume of a cone is \(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\).
Substitute the values: \(r = 7 \text{ cm}\), \(h = 16 \text{ cm}\), and \(\pi = \frac{22}{7}\).
\(V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times (7 \text{ cm})^2 \times 16 \text{ cm}\)
\(V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times (49 \text{ cm}^2) \times 16 \text{ cm}\)
We can cancel out one 7 from the numerator (49) with the 7 in the denominator:
\(V_{\text{cone}} = \frac{1}{3} \times 22 \times 7 \times 16 \text{ cm}^3\)
\(V_{\text{cone}} = \frac{1}{3} \times (22 \times 7) \times 16 \text{ cm}^3\)
\(V_{\text{cone}} = \frac{1}{3} \times 154 \times 16 \text{ cm}^3\)
\(V_{\text{cone}} = \frac{2464}{3} \text{ cm}^3\)
The formula for the volume of a hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3} \pi r^3\).
Substitute the values: \(r = 7 \text{ cm}\) and \(\pi = \frac{22}{7}\).
\(V_{\text{hemisphere}} = \frac{2}{3} \times \frac{22}{7} \times (7 \text{ cm})^3\)
\(V_{\text{hemisphere}} = \frac{2}{3} \times \frac{22}{7} \times (7 \times 7 \times 7) \text{ cm}^3\)
We can cancel out one 7 from the numerator with the 7 in the denominator:
\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times (7 \times 7) \text{ cm}^3\)
\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times 49 \text{ cm}^3\)
\(V_{\text{hemisphere}} = \frac{2}{3} \times 1078 \text{ cm}^3\)
\(V_{\text{hemisphere}} = \frac{2156}{3} \text{ cm}^3\)
The total volume of the solid is the sum of the volume of the cone and the volume of the hemisphere because the cone is mounted on the hemisphere.
\(V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}}\)
\(V_{\text{total}} = \frac{2464}{3} \text{ cm}^3 + \frac{2156}{3} \text{ cm}^3\)
Since the denominators are the same, we can add the numerators directly:
\(V_{\text{total}} = \frac{2464 + 2156}{3} \text{ cm}^3\)
\(V_{\text{total}} = \frac{4620}{3} \text{ cm}^3\)
Now, perform the division:
\(4620 \div 3\)
\(4620 / 3 = 1540\)
\(V_{\text{total}} = 1540 \text{ cm}^3\)
The volume of the solid formed by mounting the cone on the hemisphere is 1540 cm\(^{3}\).
| Component | Shape | Formula | Dimensions | Calculated Volume |
|---|---|---|---|---|
| Bottom Part | Hemisphere | \(\frac{2}{3} \pi r^3\) | \(r=7 \text{ cm}\) | \(\frac{2156}{3} \text{ cm}^3\) |
| Top Part | Cone | \(\frac{1}{3} \pi r^2 h\) | \(r=7 \text{ cm}, h=16 \text{ cm}\) | \(\frac{2464}{3} \text{ cm}^3\) |
| Composite Solid | Cone on Hemisphere | \(V_{\text{cone}} + V_{\text{hemisphere}}\) | Combined | \(\frac{2464}{3} + \frac{2156}{3} = \frac{4620}{3} = 1540 \text{ cm}^3\) |
Calculating the volume of composite solids like the one in this problem involves understanding the formulas for basic geometric shapes and adding the volumes of the individual parts. Here are some related concepts:
Understanding these basic formulas and the concept of breaking down complex shapes into simpler ones is key to solving problems involving volumes and surface areas of composite solids.
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