A solid metallic right circular cone of height 30cm and radius of the base 12 cm is melted and two solid spheres formed from it. If the volume of one of the sphere is 8 times that of the other find the radius of the smaller sphere.

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript:

### Step 1: Calculate the volume of the cone
The volume \( V \) of a right circular cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius of the base and \( h \) is the height. Here, the radius \( r = 12 \) cm and the height \( h = 30 \) cm.

Substituting the values:
\[ V = \frac{1}{3} \pi (12)^2 (30) \]
Calculating this:
\[ V = \frac{1}{3} \pi (144) (30) = \frac{1}{3} \pi (4320) = 1440 \pi \text{ cm}^3 \]

### Step 2: Set up the relationship between the volumes of the spheres
Let the radius of the smaller sphere be \( r \) and the radius of the larger sphere be \( R \). According to the problem, the volume of the larger sphere is 8 times that of the smaller sphere. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, we can write:
\[ \frac{4}{3} \pi R^3 = 8 \left( \frac{4}{3} \pi r^3 \right) \]
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 8r^3 \]
Taking the cube root of both sides, we find:
\[ R = 2r \]

### Step 3: Express the total volume of the spheres in terms of \( r \)
The total volume of the two spheres is:
\[ V_{\text{total}} = \frac{4}{3} \pi R^3 + \frac{4}{3} \pi r^3 \]
Substituting \( R = 2r \):
\[ V_{\text{total}} = \frac{4}{3} \pi (2r)^3 + \frac{4}{3} \pi r^3 \]
Calculating \( (2r)^3 \):
\[ (2r)^3 = 8r^3 \]
Thus,
\[ V_{\text{total}} = \frac{4}{3} \pi (8r^3) + \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8r^3 + r^3) = \frac{4}{3} \pi (9r^3) \]
This simplifies to:
\[ V_{\text{total}} = 12 \pi r^3 \]

### Step 4: Set the volume of the cone equal to the total volume of the spheres
Since the cone is melted to form the spheres, we have:
\[ 1440 \pi = 12 \pi r^3 \]
Cancelling \( \pi \) from both sides:
\[ 1440 = 12 r^3 \]
Dividing both sides by 12:
\[ r^3 = 120 \]

### Step 5: Calculate the radius of the smaller sphere
Taking the cube root:
\[ r = \sqrt[3]{120} \]
We can express \( 120 \) as \( 4 \times 30 \):
\[ r = \sqrt[3]{4 \times 30} = \sqrt[3]{4} \times \sqrt[3]{30} \]
Since \( \sqrt[3]{4} = 2^{2/3} \) and \( \sqrt[3]{30} \) is approximately \( 3.107 \), we can find \( r \) numerically or leave it in this form.

### Final Answer
The radius of the smaller sphere is:
\[ r \approx 4.93 \text{ cm} \quad (\text{exactly } \sqrt[3]{120} \text{ cm}) \]