Question

Metallic spheres of radii 6 cm, 8 cm, and 10 cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Hint:

To solve this problem, remember that the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \) and use the conservation of volume when combining multiple spheres into one.

Answer 1

The volume of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] The total volume of the three spheres is the sum of their individual volumes. So, \[ V_{\text{total}} = V_1 + V_2 + V_3 \] Where the radii of the three spheres are \( r_1 = 6 \, \text{cm}, r_2 = 8 \, \text{cm}, r_3 = 10 \, \text{cm} \). Thus, the volumes are: \[ V_1 = \frac{4}{3} \pi (6)^3, \quad V_2 = \frac{4}{3} \pi (8)^3, \quad V_3 = \frac{4}{3} \pi (10)^3 \] \[ V_1 = \frac{4}{3} \pi \times 216 = 288 \pi, \quad V_2 = \frac{4}{3} \pi \times 512 = 682.67 \pi, \quad V_3 = \frac{4}{3} \pi \times 1000 = 1333.33 \pi \] Total volume: \[ V_{\text{total}} = 288 \pi + 682.67 \pi + 1333.33 \pi = 2304 \pi \] Now, let the radius of the resulting sphere be \( R \). The volume of the resulting sphere is: \[ V_{\text{new}} = \frac{4}{3} \pi R^3 \] Since the total volume of the material is conserved: \[ \frac{4}{3} \pi R^3 = 2304 \pi \] Canceling \( \pi \) from both sides: \[ \frac{4}{3} R^3 = 2304 \] Multiplying both sides by 3: \[ 4 R^3 = 6912 \] Now, divide by 4: \[ R^3 = 1728 \] Taking the cube root of both sides: \[ R = \sqrt[3]{1728} = 12 \, \text{cm} \] Thus, the radius of the resulting sphere is \( \mathbf{12 \, \text{cm}} \). \hrule