How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm?

Answer

To solve the problem of how many spherical lead shots can be obtained from a solid rectangular lead piece, we will follow these steps:

### Step 1: Calculate the Volume of the Rectangular Lead Piece
The volume \( V \) of a rectangular cuboid can be calculated using the formula:
\[ V = \text{length} \times \text{width} \times \text{height} \]
Given the dimensions of the cuboid are 66 cm, 42 cm, and 21 cm, we can substitute these values into the formula:
\[ V = 66 \, \text{cm} \times 42 \, \text{cm} \times 21 \, \text{cm} \]

### Step 2: Perform the Multiplication
Now, we will perform the multiplication:
\[ V = 66 \times 42 = 2772 \, \text{cm}^2 \]
Next, multiply this result by the height:
\[ V = 2772 \times 21 = 58212 \, \text{cm}^3 \]
So, the volume of the rectangular lead piece is \( 58212 \, \text{cm}^3 \).

### Step 3: Calculate the Volume of One Spherical Lead Shot
The volume \( V_s \) of a sphere can be calculated using the formula:
\[ V_s = \frac{4}{3} \pi r^3 \]
First, we need to find the radius \( r \) of the spherical lead shot. The diameter is given as 4.2 cm, so the radius is:
\[ r = \frac{\text{diameter}}{2} = \frac{4.2 \, \text{cm}}{2} = 2.1 \, \text{cm} \]
Now, substituting the radius into the volume formula:
\[ V_s = \frac{4}{3} \pi (2.1)^3 \]

### Step 4: Calculate \( (2.1)^3 \)
Calculating \( (2.1)^3 \):
\[ (2.1)^3 = 2.1 \times 2.1 \times 2.1 = 9.261 \, \text{cm}^3 \]
Now substituting this back into the volume formula:
\[ V_s = \frac{4}{3} \pi \times 9.261 \]
Using \( \pi \approx 3.14 \):
\[ V_s \approx \frac{4}{3} \times 3.14 \times 9.261 \approx \frac{4 \times 3.14 \times 9.261}{3} \approx 38.79 \, \text{cm}^3 \]

### Step 5: Calculate the Number of Spherical Lead Shots
To find the number of spherical lead shots \( n \), we use the relationship:
\[ n = \frac{\text{Volume of the cuboid}}{\text{Volume of one spherical lead shot}} \]
Substituting the values we calculated:
\[ n = \frac{58212 \, \text{cm}^3}{38.79 \, \text{cm}^3} \approx 1505.3 \]
Since the number of lead shots must be a whole number, we take the integer part:
\[ n = 1505 \]

### Final Answer
The number of spherical lead shots that can be obtained is **1505**.

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