Understanding Metal Weight and Volume Calculation

This problem involves finding the weight of a certain volume of metal when we know the weight of a different volume of the same metal. This is a classic example of a direct proportion problem, meaning that as the volume increases, the weight increases proportionally, assuming the density of the metal remains constant.

Key Information Provided

• Initial Volume ($V_1$): 75 m³
• Initial Weight ($W_1$): 150 gm
• Target Volume ($V_2$): 300 m³
• Target Weight ($W_2$): Unknown

Applying Direct Proportion

In a direct proportion, the ratio between two quantities remains constant. We can set up a relationship between weight and volume:

The ratio of weight to volume is constant for the same metal.

This can be written as:

$$ \frac{\text{Weight}_1}{\text{Volume}_1} = \frac{\text{Weight}_2}{\text{Volume}_2} $$

Substituting the given values:

$$ \frac{150 \text{ gm}}{75 \text{ m}^3} = \frac{W_2}{300 \text{ m}^3} $$

Step-by-Step Calculation for Target Weight

To find the unknown weight ($W_2$), we can rearrange the equation:

$$ W_2 = \frac{150 \text{ gm}}{75 \text{ m}^3} \times 300 \text{ m}^3 $$

First, calculate the weight per cubic meter:

$$ \text{Weight per m}^3 = \frac{150 \text{ gm}}{75 \text{ m}^3} = 2 \text{ gm/m}^3 $$

Now, multiply this rate by the target volume:

$$ W_2 = 2 \text{ gm/m}^3 \times 300 \text{ m}^3 $$

$$ W_2 = 600 \text{ gm} $$

Conclusion

Therefore, the weight of 300 m³ volume of the same metal is 600 gm. This confirms that when the volume quadruples (from 75 m³ to 300 m³), the weight also quadruples (from 150 gm to 600 gm), as expected in a direct proportion.