This problem involves the concept of volume conservation. When a solid object is melted and recast into different shapes, the total volume of the material remains constant. Here, a solid cylinder is melted and transformed into solid spherical balls. Therefore, the volume of the original cylinder is equal to the sum of the volumes of all the spherical balls formed.

Calculating the Volume of the Cylinder

The formula for the volume of a cylinder is given by:

\(\text{Volume of cylinder} = \pi r_{\text{cylinder}}^2 h_{\text{cylinder}}\)

Where:

• \(r_{\text{cylinder}}\) is the radius of the base of the cylinder.
• \(h_{\text{cylinder}}\) is the height of the cylinder.

Given in the problem:

• Radius of the cylinder, \(r_{\text{cylinder}}\) = 8 cm
• Height of the cylinder, \(h_{\text{cylinder}}\) = 36 cm

Substituting these values into the formula:

\(\text{Volume of cylinder} = \pi \times (8 \text{ cm})^2 \times (36 \text{ cm})\)

\(\text{Volume of cylinder} = \pi \times (64 \text{ cm}^2) \times (36 \text{ cm})\)

\(\text{Volume of cylinder} = 2304\pi \text{ cm}^3\)

Calculating the Volume of One Spherical Ball

The formula for the volume of a sphere is given by:

\(\text{Volume of sphere} = \frac{4}{3} \pi r_{\text{sphere}}^3\)

Where:

• \(r_{\text{sphere}}\) is the radius of the sphere.

Given in the problem:

• Diameter of each spherical ball = 1.5 cm

The radius is half of the diameter:

\(r_{\text{sphere}} = \frac{\text{Diameter}}{2} = \frac{1.5 \text{ cm}}{2} = 0.75 \text{ cm}\)

We can write 0.75 as a fraction: \(0.75 = \frac{75}{100} = \frac{3}{4}\)

So, \(r_{\text{sphere}} = \frac{3}{4} \text{ cm}\)

Substituting this radius into the formula for the volume of a sphere:

\(\text{Volume of one sphere} = \frac{4}{3} \pi \times \left(\frac{3}{4} \text{ cm}\right)^3\)

\(\text{Volume of one sphere} = \frac{4}{3} \pi \times \left(\frac{3^3}{4^3}\right) \text{ cm}^3\)

\(\text{Volume of one sphere} = \frac{4}{3} \pi \times \left(\frac{27}{64}\right) \text{ cm}^3\)

We can simplify this expression:

\(\text{Volume of one sphere} = \frac{4}{3} \times \frac{27}{64} \pi \text{ cm}^3\)

\(\text{Volume of one sphere} = \frac{1 \times 9}{1 \times 16} \pi \text{ cm}^3 \quad (\text{since } 4 \text{ goes into } 64 \text{ sixteen times and } 3 \text{ goes into } 27 \text{ nine times})\)

\(\text{Volume of one sphere} = \frac{9}{16} \pi \text{ cm}^3\)

Finding the Number of Spherical Balls

The total volume of the material from the cylinder is used to make the spherical balls. If \(n\) is the number of spherical balls made, then:

\(\text{Volume of cylinder} = n \times \text{Volume of one sphere}\)

We want to find \(n\), so we can rearrange the formula:

\(n = \frac{\text{Volume of cylinder}}{\text{Volume of one sphere}}\)

Substituting the calculated volumes:

\(n = \frac{2304\pi \text{ cm}^3}{\frac{9}{16} \pi \text{ cm}^3}\)

The \(\pi\) and \(\text{cm}^3\) units cancel out:

\(n = \frac{2304}{\frac{9}{16}}\)

Dividing by a fraction is the same as multiplying by its reciprocal:

\(n = 2304 \times \frac{16}{9}\)

Now, we can perform the multiplication. We can simplify by dividing 2304 by 9 first:

\(2304 \div 9 = 256\)

So, \(n = 256 \times 16\)

Calculating the final product:

\(256 \times 16 = 4096\)

Therefore, 4,096 solid spherical balls can be made.

Revision Table: Volume Calculation Key Points

Additional Information: Understanding Volume Conservation

Volume conservation is a fundamental principle in physics and geometry when dealing with phase changes (like melting) or reshaping of materials without loss. When a solid material is melted, it changes its state but not the amount of substance. If this molten material is then poured into moulds or reshaped into new forms, the total amount of space it occupies (its volume) remains the same, assuming no material is added or lost in the process. This principle allows us to solve problems like the one above by equating the volume of the initial shape(s) to the total volume of the final shape(s).

For example, if you melt a block of ice and refreeze it into ice cubes, the total volume of all the ice cubes will be equal to the volume of the original block of ice (ignoring minor changes due to temperature or trapped air).

In quantitative aptitude problems, this principle is often applied to various 3D shapes like cylinders, cones, spheres, cubes, and cuboids. The key is to correctly calculate the volume of the initial shape and the volume of the final individual shape, and then divide the total initial volume by the volume of one final shape to find the number of objects made.