Volume Ratio Calculation for Recast Spheres

This problem involves calculating the ratio of volumes when a sphere is melted and recast into smaller spheres.

1. Volume and Diameter Relationship

The volume ($V$) of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$, where $r$ is the radius. Since the radius is half the diameter ($r = d/2$), the volume can also be expressed in terms of diameter ($d$):

$V = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{4}{3}\pi \frac{d^3}{8} = \frac{\pi d^3}{6}$

This shows that the volume of a sphere is directly proportional to the cube of its diameter ($V \propto d^3$).

2. Volumes of the Two New Balls

Let the diameters of the two new balls be $d_1$ and $d_2$. We are given that their diameters are in the ratio $1:2$. Let $d_1 = k$ and $d_2 = 2k$ for some constant $k$.

Let $V_1$ be the volume of the smaller new ball (diameter $k$) and $V_2$ be the volume of the larger new ball (diameter $2k$).

Using the proportionality $V \propto d^3$:

• $V_1 \propto (k)^3 = k^3$
• $V_2 \propto (2k)^3 = 8k^3$

Therefore, the ratio of the volumes of the two new balls is $V_1 : V_2 = k^3 : 8k^3 = 1:8$. This means the volume of the larger new ball is 8 times the volume of the smaller new ball ($V_2 = 8V_1$).

3. Volume Conservation Principle

The original iron ball was melted to create the two new balls. Assuming no material is lost, the volume of the original ball ($V_{original}$) must be equal to the sum of the volumes of the two new balls ($V_1 + V_2$).

$V_{original} = V_1 + V_2$

Substitute $V_2 = 8V_1$ into the equation:

$V_{original} = V_1 + 8V_1 = 9V_1$

4. Final Ratio Calculation

The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{original}$).

From the previous step, we have $V_{original} = 9V_1$. Rearranging this gives:

$\frac{V_1}{V_{original}} = \frac{1}{9}$

So, the ratio is $1:9$.