A solid metallic right circular cone of height 6.75 cm and radius of the base 12 cm is melted and two solid sphees formed form it. If the volume of one of the sphere is 8 times that of the other find the radius of the smaller sphere.

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and break it down into clear steps:

### Step 1: Identify the given data
We have a right circular cone with:
- Height (h) = 6.75 cm
- Radius of the base (r) = 12 cm

We need to find the radius of the smaller sphere formed when the cone is melted.

### Step 2: Calculate the volume of the cone
The formula for the volume of a cone is given by:
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]
Substituting the values:
\[ V_{\text{cone}} = \frac{1}{3} \pi (12)^2 (6.75) \]
Calculating \(12^2\):
\[ 12^2 = 144 \]
Now substituting this back into the volume formula:
\[ V_{\text{cone}} = \frac{1}{3} \pi (144) (6.75) \]
Calculating \(144 \times 6.75\):
\[ 144 \times 6.75 = 972 \]
Thus, the volume of the cone is:
\[ V_{\text{cone}} = \frac{1}{3} \pi (972) = 324 \pi \text{ cm}^3 \]

### Step 3: Set up the equation for the volumes of the spheres
Let the volume of the smaller sphere be \(V_2\) and the volume of the larger sphere be \(V_1\). According to the problem, the volume of one sphere is 8 times that of the other:
\[ V_1 = 8 V_2 \]
The total volume of both spheres is equal to the volume of the cone:
\[ V_1 + V_2 = 324 \pi \]
Substituting \(V_1\) in terms of \(V_2\):
\[ 8 V_2 + V_2 = 324 \pi \]
This simplifies to:
\[ 9 V_2 = 324 \pi \]
Thus, we can find \(V_2\):
\[ V_2 = \frac{324 \pi}{9} = 36 \pi \text{ cm}^3 \]

### Step 4: Use the volume of the smaller sphere to find its radius
The formula for the volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For the smaller sphere:
\[ 36 \pi = \frac{4}{3} \pi r_2^3 \]
Dividing both sides by \(\pi\):
\[ 36 = \frac{4}{3} r_2^3 \]
Multiplying both sides by \(\frac{3}{4}\):
\[ r_2^3 = 36 \times \frac{3}{4} = 27 \]
Taking the cube root of both sides:
\[ r_2 = \sqrt[3]{27} = 3 \text{ cm} \]

### Final Answer
The radius of the smaller sphere is:
\[ \boxed{3 \text{ cm}} \]