How many solid speherical balls of radus 3.5 cm can be recast by melting a metallic cone of height 35 cm and base radius 7 cm?

Answer

To solve the problem of how many solid spherical balls of radius 3.5 cm can be recast from a metallic cone of height 35 cm and base radius 7 cm, we will follow these steps:

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

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

Where:
- \( r \) is the radius of the base of the cone (7 cm),
- \( h \) is the height of the cone (35 cm).

Substituting the values:

\[ V = \frac{1}{3} \pi (7)^2 (35) \]

Calculating \( (7)^2 \):

\[ (7)^2 = 49 \]

Now substituting back into the volume formula:

\[ V = \frac{1}{3} \pi (49)(35) \]

Calculating \( 49 \times 35 \):

\[ 49 \times 35 = 1715 \]

So, the volume of the cone becomes:

\[ V = \frac{1}{3} \pi (1715) = \frac{1715 \pi}{3} \text{ cm}^3 \]

### Step 2: Calculate the volume of one spherical ball.
The formula for the volume \( V \) of a sphere is given by:

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

Where \( r \) is the radius of the sphere (3.5 cm).

Substituting the value:

\[ V = \frac{4}{3} \pi (3.5)^3 \]

Calculating \( (3.5)^3 \):

\[ (3.5)^3 = 42.875 \]

Now substituting back into the volume formula:

\[ V = \frac{4}{3} \pi (42.875) = \frac{171.5 \pi}{3} \text{ cm}^3 \]

### Step 3: Find the number of spherical balls that can be formed.
To find the number of spherical balls that can be formed from the volume of the cone, we divide the volume of the cone by the volume of one spherical ball:

\[ \text{Number of balls} = \frac{\text{Volume of cone}}{\text{Volume of one ball}} = \frac{\frac{1715 \pi}{3}}{\frac{171.5 \pi}{3}} \]

The \( \pi \) and \( \frac{1}{3} \) cancel out:

\[ \text{Number of balls} = \frac{1715}{171.5} \]

Calculating \( \frac{1715}{171.5} \):

\[ \text{Number of balls} = 10 \]

### Final Answer:
Thus, **10 solid spherical balls of radius 3.5 cm can be recast from the metallic cone.**

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