Three solid spheres have their radii `r_(1),r_(2)`and`r_(3)` .The spheres are melted to form a solid sphere of bigger radius of the new sphere is:

Options

A. `(r_(1)+r_(2)+r_(3))`
B. `(r_(1)^(2)+r_(2)^(2)+r_(3)^(2))^((1)/(2))`
C. `(r_(1)^(3)+r_(2)^(3)+r_(3)^(3))^((1)/(3))`
D. `(r_(1)^(4)+r_(2)^(4)+r_(3)^(4))^((1)/(4))`

Answer

To find the radius of the new sphere formed by melting three solid spheres with radii \( r_1, r_2, \) and \( r_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Volume of Each Sphere**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the volumes of the three spheres are:
- Volume of sphere 1: \( V_1 = \frac{4}{3} \pi r_1^3 \)
- Volume of sphere 2: \( V_2 = \frac{4}{3} \pi r_2^3 \)
- Volume of sphere 3: \( V_3 = \frac{4}{3} \pi r_3^3 \)

2. **Total Volume of the New Sphere**:
When the three spheres are melted, their volumes add up to form a new sphere. Thus, the total volume \( V_{total} \) is:
\[ V_{total} = V_1 + V_2 + V_3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3 \]
This simplifies to:
\[ V_{total} = \frac{4}{3} \pi (r_1^3 + r_2^3 + r_3^3) \]

3. **Volume of the New Sphere**:
Let the radius of the new sphere be \( R \). The volume of the new sphere can also be expressed as:
\[ V_{new} = \frac{4}{3} \pi R^3 \]

4. **Setting the Volumes Equal**:
Since the total volume of the melted spheres equals the volume of the new sphere, we set the two volume equations equal to each other:
\[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (r_1^3 + r_2^3 + r_3^3) \]

5. **Canceling Common Terms**:
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = r_1^3 + r_2^3 + r_3^3 \]

6. **Finding the Radius of the New Sphere**:
To find \( R \), we take the cube root of both sides:
\[ R = \sqrt[3]{r_1^3 + r_2^3 + r_3^3} \]

### Final Answer:
The radius of the new sphere is:
\[ R = \sqrt[3]{r_1^3 + r_2^3 + r_3^3} \]