A block of wood floats in water two-thirds of its volume submerged. In oil the block of floats with 0.90 of its volume submerged. Find the density of (a) wood and (b) oil, if density of water I `10^(3) kg//m^(3)`.

Answer

To solve the problem, we will use Archimedes' principle, which states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces.

**Step 1: Understanding the situation in water.**
The block of wood is floating in water with two-thirds of its volume submerged. Let the volume of the block be \( V \).

The volume of water displaced by the block is:
\[ V_{\text{displaced, water}} = \frac{2}{3} V \]

**Step 2: Applying Archimedes' principle for the block in water.**
The weight of the water displaced is equal to the weight of the block:
\[ \text{Weight of water displaced} = \text{Weight of the block} \]

Using the formula for weight:
\[ \text{Weight of water displaced} = V_{\text{displaced, water}} \times \text{density of water} \times g \]
\[ \text{Weight of the block} = V \times \text{density of wood} \times g \]

Setting these equal gives:
\[ \frac{2}{3} V \times \rho_{\text{water}} \times g = V \times \rho_{\text{wood}} \times g \]

**Step 3: Canceling common terms.**
We can cancel \( V \) and \( g \) from both sides:
\[ \frac{2}{3} \rho_{\text{water}} = \rho_{\text{wood}} \]

**Step 4: Substituting the density of water.**
Given that the density of water \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \):
\[ \rho_{\text{wood}} = \frac{2}{3} \times 1000 = 667 \, \text{kg/m}^3 \]

**Step 5: Finding the density of oil.**
Now, the block floats in oil with 0.90 of its volume submerged. The volume of oil displaced is:
\[ V_{\text{displaced, oil}} = 0.90 V \]

Using Archimedes' principle again:
\[ \text{Weight of oil displaced} = \text{Weight of the block} \]
\[ 0.90 V \times \rho_{\text{oil}} \times g = V \times \rho_{\text{wood}} \times g \]

**Step 6: Canceling common terms again.**
Cancel \( V \) and \( g \):
\[ 0.90 \rho_{\text{oil}} = \rho_{\text{wood}} \]

**Step 7: Substituting the density of wood.**
Substituting \( \rho_{\text{wood}} = 667 \, \text{kg/m}^3 \):
\[ 0.90 \rho_{\text{oil}} = 667 \]
\[ \rho_{\text{oil}} = \frac{667}{0.90} \approx 740 \, \text{kg/m}^3 \]

**Final Answers:**
(a) Density of wood \( \rho_{\text{wood}} = 667 \, \text{kg/m}^3 \)
(b) Density of oil \( \rho_{\text{oil}} \approx 740 \, \text{kg/m}^3 \)

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