A piece of wood of uniform cross section and height 15 cm floats vertically with its height 10 cm in water and 12 cm in spirit. Find the density of (i) wood and (ii) spirit

Answer

To solve the problem, we will use Archimedes' principle, which states that the weight of the fluid displaced by a floating object is equal to the weight of the object itself. We will find the densities of both the wood and the spirit step by step.

### Step 1: Understand the problem
We have a piece of wood with a total height of 15 cm. It floats in water with 10 cm submerged and in spirit with 12 cm submerged. We need to find the density of the wood and the density of the spirit.

### Step 2: Calculate the density of wood using water
Using Archimedes' principle for the case of water:
- The volume of the wood (V_wood) = height × cross-sectional area = 15 cm × A
- The volume of the wood submerged in water (V_water) = 10 cm × A
- The density of water (ρ_water) = 1 g/cm³

According to Archimedes' principle:
\[ \text{Weight of water displaced} = \text{Weight of wood} \]

This can be expressed as:
\[ \rho_{water} \cdot V_{water} \cdot g = \rho_{wood} \cdot V_{wood} \cdot g \]

Since g (acceleration due to gravity) cancels out, we have:
\[ \rho_{water} \cdot (10 \cdot A) = \rho_{wood} \cdot (15 \cdot A) \]

Dividing both sides by A:
\[ \rho_{water} \cdot 10 = \rho_{wood} \cdot 15 \]

Substituting the value of ρ_water:
\[ 1 \cdot 10 = \rho_{wood} \cdot 15 \]
\[ \rho_{wood} = \frac{10}{15} = \frac{2}{3} \text{ g/cm}^3 \]
\[ \rho_{wood} = 0.67 \text{ g/cm}^3 \]

### Step 3: Calculate the density of spirit using spirit
Now, we will find the density of the spirit using the same principle:
- The volume of the wood submerged in spirit (V_spirit) = 12 cm × A
- Let the density of spirit be ρ_spirit.

Using Archimedes' principle again:
\[ \rho_{wood} \cdot V_{wood} = \rho_{spirit} \cdot V_{spirit} \]

Substituting the known values:
\[ \rho_{wood} \cdot (15 \cdot A) = \rho_{spirit} \cdot (12 \cdot A) \]

Dividing both sides by A:
\[ \rho_{wood} \cdot 15 = \rho_{spirit} \cdot 12 \]

Substituting the value of ρ_wood:
\[ 0.67 \cdot 15 = \rho_{spirit} \cdot 12 \]
\[ 10.05 = \rho_{spirit} \cdot 12 \]
\[ \rho_{spirit} = \frac{10.05}{12} \]
\[ \rho_{spirit} = 0.8375 \text{ g/cm}^3 \]
\[ \rho_{spirit} \approx 0.83 \text{ g/cm}^3 \]

### Final Answers
(i) Density of wood = 0.67 g/cm³
(ii) Density of spirit = 0.83 g/cm³