3 A block of wood floats in fresh water with $$ 67 \% $$ of its volume inside, water. In some oil, the block floats with $$ 90 \% $$ of its volume inside the oil. What is the density of the wood and the oil? Density of water is $$ 1000 \mathrm{~kg} / \mathrm{m}^{3} $$.

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UpStudy AI · Accepted Answer

**Step 1.** For an object floating, the buoyant force equals the weight of the object, so
$$
\text{Buoyant Force} = \rho_{\text{fluid}}\, V_{\text{submerged}}\, g = \rho_{\text{wood}}\, V\, g \,.
$$

**Step 2.** In water, the block has $$67\%$$ of its volume submerged. Thus:
$$
\rho_{\text{water}} \times 0.67\, V \times g = \rho_{\text{wood}} \, V \times g \,.
$$
Cancelling $$V$$ and $$g$$ (which are non-zero) gives:
$$
\rho_{\text{wood}} = 0.67 \, \rho_{\text{water}}\,.
$$
Since $$\rho_{\text{water}} = 1000\,\mathrm{kg/m^3}$$, we have:
$$
\rho_{\text{wood}} = 0.67 \times 1000 = 670\,\mathrm{kg/m^3}\,.
$$

**Step 3.** In oil, the block has $$90\%$$ of its volume submerged. Therefore:
$$
\rho_{\text{oil}} \times 0.9 \, V \times g = \rho_{\text{wood}} \, V \times g \,.
$$
Cancelling $$V$$ and $$g$$ gives:
$$
0.9\, \rho_{\text{oil}} = \rho_{\text{wood}} \,.
$$
Substitute $$\rho_{\text{wood}} = 670\,\mathrm{kg/m^3}$$:
$$
\rho_{\text{oil}} = \frac{670}{0.9} \approx 744.44\,\mathrm{kg/m^3}\,.
$$

**Final Answer:**
- The density of the wood is $$\rho_{\text{wood}} = 670\,\mathrm{kg/m^3}$$.
- The density of the oil is $$\rho_{\text{oil}} \approx 744.44\,\mathrm{kg/m^3}$$.
The wood has a density of 670 kg/m³ and the oil has a density of approximately 744.44 kg/m³.

3 A block of wood floats in fresh water with \( 67 \% \) of its volume inside, water. In some oil, the block floats with \( 90 \% \) of its volume inside the oil. What is the density of the wood and the oil? Density of water is \( 1000 \mathrm{~kg} / \mathrm{m}^{3} \).

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