Question

Suppose you have a rectangular wooden block with dimensions 8.0 cm x 8.0 cm x 9.0 cm that has a density of 0.85 x 10^3 kg/m^3. The block has a cylindrical hole inside it so that a lead cylinder 5.0 cm in diameter and 7.5 cm high can be fitted completely inside. What is the volume of the lead cylinder that fits inside (in m^3)? What is the mass of the lead cylinder if the density of lead is 1.13 x 10^4? (Density kg/m^3)? What is the volume of the wood in the wooden block (excluding the volume of the cylindrical hole)?

Answer 1

Question#23227

Suppose you have a rectangular wooden block with dimensions $8.0 \, \text{cm} \times 8.0 \, \text{cm} \times 9.0 \, \text{cm}$ that has a density of $0.85 \times 10^{3} \, \text{kg/m}^3$. The block has a cylindrical hole inside it so that a lead cylinder $5.0 \, \text{cm}$ in diameter and $7.5 \, \text{cm}$ high can be fitted completely inside. What is the volume of the lead cylinder that fits inside (in $\text{m}^3$)? What is the mass of the lead cylinder if the density of lead is $1.13 \times 10^{4}$? (Density $\text{kg/m}^3$)? What is the volume of the wood in the wooden block (excluding the volume of the cylindrical hole)?

Solution:

Let:

$$L = 8 \, \text{cm} = 0.08 \, \text{m}$$ $$B = 8 \, \text{cm} = 0.08 \, \text{m}$$ $$H = 9 \, \text{cm} = 0.09 \, \text{m}$$ $$D = 5 \, \text{cm} = 0.05 \, \text{m}$$ $$h = 7.5 \, \text{cm} = 0.075 \, \text{m}$$ $$\rho_{\text{lead}} = 1.13 \times 10^{4} \, \text{kg/m}^3$$ $$m_{\text{lead}} = ?$$ $$V_{\text{wood}} = ?$$

The mass of lead is:

$m = \rho_{\text{lead}} \times V_{\text{lead}}$, where $V_{\text{lead}}$ is the volume of the lead cylinder

$$V_{\text{lead}} = \frac{1}{4} \pi D^2 h$$ $$V_{\text{lead}} = \frac{1}{4} \cdot 3.142 \cdot 0.05^2 \cdot 0.075 = 0.000147 \, \text{m}^3$$ $$m = 11300 \cdot 0.000147 = 1.664 \, \text{kg}$$

The volume of the wood is:

$V_{\text{wood}} = V - V_{\text{lead}}$, where $V$ is the volume of the rectangular block

$$V = L \times B \times H$$ $$V_{\text{wood}} = L \times B \times H - V_{\text{lead}}$$ $$V_{\text{wood}} = 0.08 \cdot 0.08 \cdot 0.09 - 0.000147 = 0.000429 \, \text{m}^3$$

Answer:

The mass of lead is $1.664 \, \text{kg}$, the volume of wood is $0.000429 \, \text{m}^3 (429 \, \text{cm}^3)$.