Calculating the Volume of a Composite Solid

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.

Understanding the Components and Given Information

The solid is composed of two parts:

1. A cone
2. A hemisphere

We are given the following dimensions:

• Height of the cone (h) = 16 cm
• Diameter of the cone = 14 cm
• Diameter of the hemisphere = 14 cm
• Value of \(\pi\) to use = 22/7

Determining the Radius

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.

Calculating the Volume of the Cone

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:

\(V_{\text{cone}} = \frac{1}{3} \times 22 \times 7 \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{1}{3} \times (22 \times 7) \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{1}{3} \times 154 \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{2464}{3} \text{ cm}^3\)

Calculating the Volume of the Hemisphere

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:

\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times (7 \times 7) \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times 49 \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2}{3} \times 1078 \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2156}{3} \text{ cm}^3\)

Calculating the Total Volume of the Solid

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:

\(4620 \div 3\)

\(4620 / 3 = 1540\)

\(V_{\text{total}} = 1540 \text{ cm}^3\)

Conclusion

The volume of the solid formed by mounting the cone on the hemisphere is 1540 cm\(^{3}\).

Revision Table: Solid Volume Calculation

Additional Information: Volumes of Geometric Shapes

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:

• Cylinder Volume: The volume of a cylinder is given by \(V = \pi r^2 h\), where \(r\) is the radius of the base and \(h\) is the height. A cone's volume is one-third of a cylinder with the same base radius and height.
• Sphere Volume: The volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius. A hemisphere is half of a sphere, so its volume is half of the sphere's volume.
• Surface Area: While this problem focuses on volume, calculating the surface area of a composite solid is often different from just adding the individual surface areas because some surfaces are hidden where the shapes join. For this solid, the total surface area would be the curved surface area of the cone plus the curved surface area of the hemisphere plus the area of the base of the hemisphere (if it's resting on a surface, but here it's joined to the cone). The base circle of the cone is hidden by the hemisphere connection.
• Other Composite Shapes: Composite solids can be formed by combining cylinders, cubes, cuboids, pyramids, etc. The approach is similar: identify the individual shapes, calculate their volumes, and sum them up.

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.