Calculating New Ball Diameter from Melted Balls

The problem involves finding the diameter of a single large metallic ball formed by melting 8 smaller metallic balls, conserving the total volume.

Volume of Smaller Balls

Each small ball has a radius ($r$) of 1 cm.

The formula for the volume ($V$) of a sphere is $V = \frac{4}{3}\pi r^3$.

Volume of one small ball:

$ V_{\text{small}} = \frac{4}{3}\pi (1 \text{ cm})^3 = \frac{4}{3}\pi \text{ cm}^3 $

There are 8 such balls. The total volume ($V_{\text{total}}$) is:

$ V_{\text{total}} = 8 \times V_{\text{small}} = 8 \times \frac{4}{3}\pi \text{ cm}^3 = \frac{32}{3}\pi \text{ cm}^3 $

Volume of the New Ball

Let the radius of the new, larger ball be $R$. Its volume is:

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

Equating Volumes and Solving for Radius

Since the material is conserved, the volume of the new ball equals the total volume of the small balls:

$ V_{\text{new}} = V_{\text{total}} $ $ \frac{4}{3}\pi R^3 = \frac{32}{3}\pi \text{ cm}^3 $

To find $R$, we can cancel out $\frac{4}{3}\pi$ from both sides:

$ R^3 = \frac{32}{3}\pi \times \frac{3}{4\pi} \text{ cm}^3 $ $ R^3 = 8 \text{ cm}^3 $

Taking the cube root of both sides:

$ R = \sqrt[3]{8 \text{ cm}^3} = 2 \text{ cm} $

Calculating the New Ball's Diameter

The diameter ($D$) of a sphere is twice its radius ($D = 2R$).

Therefore, the diameter of the new ball is:

$ D = 2 \times R = 2 \times (2 \text{ cm}) = 4 \text{ cm} $

Conclusion

The diameter of the new ball is 4 cm.