Finding the Number of Cones from a Melted Hemisphere

This problem involves the concept of volume conservation. When a solid shape is melted and recast into another shape or shapes, the total volume remains the same, assuming no material is lost in the process. Here, a solid hemisphere is melted and reshaped into several identical cones. Therefore, the volume of the hemisphere must be equal to the combined volume of all the cones.

Volume Formulas for Hemisphere and Cone

To solve this, we need the formulas for the volume of a solid hemisphere and the volume of a cone.

• The formula for the volume of a sphere with radius \(r\) is \(V_{sphere} = \frac{4}{3}\pi r^3\).
• A hemisphere is half of a sphere, so the volume of a hemisphere with radius \(r_h\) is \(V_{hemisphere} = \frac{1}{2} \times \frac{4}{3}\pi r_h^3 = \frac{2}{3}\pi r_h^3\).
• The formula for the volume of a cone with radius \(r_c\) and height \(h_c\) is \(V_{cone} = \frac{1}{3}\pi r_c^2 h_c\).

Calculating the Volume of the Hemisphere

We are given that the radius of the solid hemisphere is 8 cm.

• Radius of hemisphere, \(r_h = 8\) cm.
• Volume of hemisphere, \(V_{hemisphere} = \frac{2}{3}\pi (8)^3\)
• \(V_{hemisphere} = \frac{2}{3}\pi (512)\)
• \(V_{hemisphere} = \frac{1024}{3}\pi\) cubic cm.

Calculating the Volume of One Cone

We are given that each identical cone has a radius of 4 cm and a height of 2 cm.

• Radius of cone, \(r_c = 4\) cm.
• Height of cone, \(h_c = 2\) cm.
• Volume of one cone, \(V_{cone} = \frac{1}{3}\pi (4)^2 (2)\)
• \(V_{cone} = \frac{1}{3}\pi (16)(2)\)
• \(V_{cone} = \frac{32}{3}\pi\) cubic cm.

Finding the Number of Cones (N)

Let N be the number of identical cones formed. The total volume of N cones is \(N \times V_{cone}\). By the principle of volume conservation:

\(V_{hemisphere} = N \times V_{cone}\)

Substitute the calculated volumes:

\(\frac{1024}{3}\pi = N \times \frac{32}{3}\pi\)

To find N, we can divide the volume of the hemisphere by the volume of one cone:

\(N = \frac{\text{Volume of Hemisphere}}{\text{Volume of One Cone}}\)

\(N = \frac{\frac{1024}{3}\pi}{\frac{32}{3}\pi}\)

We can cancel out \(\pi\) from the numerator and denominator, and also the \(\frac{1}{3}\) factor:

\(N = \frac{1024}{32}\)

Now, perform the division:

\(1024 \div 32 = 32\)

So, the value of N is 32.

Summary of Calculation

\(N = \frac{V_{hemisphere}}{V_{cone}} = \frac{\frac{1024}{3}\pi}{\frac{32}{3}\pi} = \frac{1024}{32} = 32\)

The value of N, the number of identical cones, is 32.

Revision Table: Solid Shapes and Volumes

Additional Information: Volume Conservation in Mensuration

The principle of conservation of volume is fundamental in problems where solids are melted, recast, or reshaped. It states that the total volume of the material remains constant during the transformation, provided no material is added or removed.

• Melting and Recasting: As seen in this problem, the volume of the original solid (hemisphere) equals the sum of the volumes of the new solids (N cones).
• Filling a Container: If liquid from one container is poured into another, the volume of the liquid remains constant.
• Displacement: The volume of a submerged object is equal to the volume of the fluid it displaces.

Understanding the volume formulas for common 3D shapes (like spheres, hemispheres, cones, cylinders, cubes, cuboids) and the principle of volume conservation is key to solving such mensuration problems.