Spherical Aluminium Ball Melting Calculation

This problem involves understanding the principle of conservation of volume. When a material like aluminium is melted and recast, its total volume remains the same, assuming no material is lost in the process. We need to figure out how many smaller spherical balls can be formed from a larger one, given the relationship between their radii.

Understanding Volume Conservation

The core idea here is that the volume of the original aluminium ball is equal to the sum of the volumes of all the smaller balls made from it. This is a direct application of the law of conservation of mass (and volume, assuming density doesn't change significantly).

Sphere Volume Formula

The volume ($V$) of a sphere is given by the formula:

$$V = \frac{4}{3}\pi r^3$$ where $r$ is the radius of the sphere.

Step-by-Step Calculation

1. Define Variables: Let the radius of the original spherical aluminium ball be $R$. Let the volume of the original ball be $V_{original}$. Let the radius of each smaller ball be $r$. Let the volume of each smaller ball be $V_{small}$. Let $N$ be the number of smaller balls that can be made.
2. Original Ball Volume: Using the formula, the volume of the original ball is:

   $$V_{original} = \frac{4}{3}\pi R^3$$

3. Smaller Ball Radius: The question states that the radius of the smaller balls is 1/3rd of the original radius. So,

   $$r = \frac{1}{3}R$$

4. Smaller Ball Volume: Now, we calculate the volume of one smaller ball using its radius $r$:

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

   Substitute $r = \frac{1}{3}R$ into the formula:

   $$V_{small} = \frac{4}{3}\pi \left(\frac{1}{3}R\right)^3$$

   Simplify the expression:

   $$V_{small} = \frac{4}{3}\pi \left(\frac{R^3}{3^3}\right)$$

   $$V_{small} = \frac{4}{3}\pi \left(\frac{R^3}{27}\right)$$

   $$V_{small} = \frac{1}{27} \left(\frac{4}{3}\pi R^3\right)$$

   Notice that $\left(\frac{4}{3}\pi R^3\right)$ is the volume of the original ball, $V_{original}$. Therefore,

   $$V_{small} = \frac{1}{27} V_{original}$$

5. Calculate the Number of Balls ($N$): Since the total volume must be conserved, the volume of the original ball equals the total volume of the smaller balls:

   $$V_{original} = N \times V_{small}$$

   Substitute the relationship $V_{small} = \frac{1}{27} V_{original}$:

   $$V_{original} = N \times \left(\frac{1}{27} V_{original}\right)$$

   To find $N$, divide both sides by $V_{original}$ (assuming $V_{original}$ is not zero) and multiply by 27:

   $$1 = N \times \frac{1}{27}$$

   $$N = 27$$

Conclusion

By melting the original spherical aluminium ball and recasting it into smaller balls, each with 1/3rd the radius of the original, we can make exactly 27 such smaller balls. This is because the volume scales with the cube of the radius, and $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$.