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 find the radius of the smaller sphere formed from melting a solid metallic right circular cone, we can follow these steps:

### Step 1: Calculate the volume of the cone
The formula for the volume \( V \) of a right circular cone is given by:

\[ V = \frac{1}{3} \pi r^2 h \]

Where:
- \( r \) is the radius of the base of the cone (12 cm)
- \( h \) is the height of the cone (6.75 cm)

Substituting the values:

\[ V = \frac{1}{3} \pi (12)^2 (6.75) \]

Calculating \( (12)^2 \):

\[ (12)^2 = 144 \]

Now substituting back:

\[ V = \frac{1}{3} \pi (144) (6.75) \]

Calculating \( 144 \times 6.75 \):

\[ 144 \times 6.75 = 972 \]

So,

\[ V = \frac{1}{3} \pi (972) = 324 \pi \text{ cm}^3 \]

### Step 2: Set up the volume relationship between the spheres
Let the volume of the smaller sphere be \( V_1 \) and the volume of the larger sphere be \( V_2 \). According to the problem, \( V_2 = 8 V_1 \).

The total volume of the two spheres is equal to the volume of the cone:

\[ V_1 + V_2 = V \]

Substituting \( V_2 \):

\[ V_1 + 8V_1 = 324 \pi \]

This simplifies to:

\[ 9V_1 = 324 \pi \]

### Step 3: Solve for \( V_1 \)
Dividing both sides by 9:

\[ V_1 = \frac{324 \pi}{9} = 36 \pi \text{ cm}^3 \]

### Step 4: Calculate the radius of the smaller sphere
The volume of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

Setting \( V_1 = 36 \pi \):

\[ \frac{4}{3} \pi r_1^3 = 36 \pi \]

Dividing both sides by \( \pi \):

\[ \frac{4}{3} r_1^3 = 36 \]

Multiplying both sides by \( \frac{3}{4} \):

\[ r_1^3 = 36 \times \frac{3}{4} = 27 \]

### Step 5: Find \( r_1 \)
Taking the cube root of both sides:

\[ r_1 = \sqrt[3]{27} = 3 \text{ cm} \]

### Final Answer
The radius of the smaller sphere is \( 3 \) cm.

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