To find the radius of the new sphere formed by melting three metallic spheres with radii 3 cm, 4 cm, and 5 cm, we need to apply the concept of volume conservation. The total volume of the new sphere will be equal to the sum of the volumes of the three original spheres.
The formula for the volume \( V \) of a sphere with radius \( r \) is given by:
\(V = \frac{4}{3} \pi r^3\)
Let's calculate the volumes of each sphere:
- Volume of the first sphere with radius 3 cm:
\(V_1 = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi \times 27 = 36 \pi\)
- Volume of the second sphere with radius 4 cm:
\(V_2 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi \times 64 = \frac{256}{3} \pi\)
- Volume of the third sphere with radius 5 cm:
\(V_3 = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi\)
The total volume of the new sphere is the sum of these volumes:
\(V_{\text{total}} = V_1 + V_2 + V_3 = 36 \pi + \frac{256}{3} \pi + \frac{500}{3} \pi\)
Combining the terms:
\(V_{\text{total}} = 36 \pi + \left(\frac{256 + 500}{3}\right) \pi = \left(36 + \frac{756}{3}\right) \pi = \left(36 + 252\right) \pi = 288 \pi\)
The volume of the new sphere with radius \( R \) is:
\(V_{\text{new}} = \frac{4}{3} \pi R^3\)
Since the volume is conserved, we have:
\(\frac{4}{3} \pi R^3 = 288 \pi\)
We can simplify this by canceling \(\pi\) from both sides:
\(\frac{4}{3} R^3 = 288\)
Solving for \( R^3 \):
\(R^3 = \frac{288 \times 3}{4} = 216\)
Taking the cube root of both sides, we find:
\(R = \sqrt[3]{216} = 6\)
Therefore, the radius of the new sphere is 6 cm.
Hence, the correct answer is: 6 cm
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