Understanding Volume Conservation in Melting Spheres

When solid objects are melted and reformed into new shapes, their total volume remains constant. This principle is called conservation of volume. In this problem, 64 small solid iron spheres are melted and combined to form a single large solid sphere. This means the total volume of the 64 small spheres is equal to the volume of the one large sphere.

Volume of a Sphere

The formula for the volume of a sphere with radius \(s\) is given by:

\(V = \frac{4}{3}\pi s^3\)

Calculating Volumes of Small and Large Spheres

Let the radius of each small sphere be \(r\). The volume of one small sphere is:

\(V_{small} = \frac{4}{3}\pi r^3\)

There are 64 such small spheres. So, the total volume of the 64 small spheres is:

\(V_{total\_small} = 64 \times V_{small} = 64 \times \frac{4}{3}\pi r^3\)

Let the radius of the large sphere be \(R\). The volume of the large sphere is:

\(V_{large} = \frac{4}{3}\pi R^3\)

Applying Conservation of Volume

According to the principle of conservation of volume, the total volume of the small spheres is equal to the volume of the large sphere formed by melting them:

\(V_{large} = V_{total\_small}\)

Substitute the volume formulas:

\(\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3\)

Solving for the Ratio R : r

We need to find the ratio of the radius of the large sphere (\(R\)) to the radius of a small sphere (\(r\)), which is \(R : r\) or \(\frac{R}{r}\). Let's simplify the equation:

Divide both sides of the equation by \(\frac{4}{3}\pi\):

\(\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi} = \frac{64 \times \frac{4}{3}\pi r^3}{\frac{4}{3}\pi}\)

\(R^3 = 64 r^3\)

To find \(R\) in terms of \(r\), take the cube root of both sides of the equation:

\(\sqrt[3]{R^3} = \sqrt[3]{64 r^3}\)

\(R = \sqrt[3]{64} \times \sqrt[3]{r^3}\)

The cube root of 64 is 4 (since \(4 \times 4 \times 4 = 64\)), and the cube root of \(r^3\) is \(r\). So:

\(R = 4r\)

Now, find the ratio \(R : r\):

\(\frac{R}{r} = 4\)

This can be written as the ratio \(R : r = 4 : 1\).

Summary of Steps

1. Calculate the volume of one small sphere: \(V_{small} = \frac{4}{3}\pi r^3\).
2. Calculate the total volume of 64 small spheres: \(V_{total\_small} = 64 \times \frac{4}{3}\pi r^3\).
3. Calculate the volume of the large sphere: \(V_{large} = \frac{4}{3}\pi R^3\).
4. Equate the volumes using conservation principle: \(V_{large} = V_{total\_small}\).
5. Solve for \(R\) in terms of \(r\): \(\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3 \implies R^3 = 64r^3 \implies R = 4r\).
6. Determine the ratio \(R : r\): \(\frac{R}{r} = 4\), which is \(4 : 1\).

The ratio of the radius of the large sphere to the radius of a small sphere is \(4 : 1\).

Revision Table: Key Concepts

Additional Information: Properties of Spheres and Melting

A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a perfectly round ball. All points on the surface of a sphere are equidistant from its center.

Melting is a phase transition from a solid to a liquid. When iron is melted, its state changes, but the total amount of iron substance, and therefore its total volume (assuming constant density and no material loss), is conserved. This is a fundamental concept in physics and chemistry when dealing with states of matter and transformations.

Understanding how volume scales with radius for a sphere (\(V \propto s^3\)) is crucial. If the radius is multiplied by a factor, say \(k\), the volume is multiplied by \(k^3\). In this case, since the volume increased by a factor of 64, the radius must have increased by a factor of \(\sqrt[3]{64} = 4\).