When a cube of wood floats in water,`60%` of its volume is submerged. When the same cube floats in an unknown fluid `85%` of its volume is submerged. Find the densities of wood and the unknown fluid.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the concept of buoyancy
When an object floats in a fluid, the buoyant force acting on it is equal to the weight of the object. This principle is described by Archimedes' principle.

### Step 2: Set up the equations for the wooden cube in water
Let:
- \( \rho_{wood} \) = density of wood
- \( \rho_{water} = 1000 \, \text{kg/m}^3 \) (density of water)
- \( V \) = total volume of the cube
- \( V_d \) = volume of the cube submerged in water

Since 60% of the cube is submerged in water:
\[ V_d = 0.6V \]

Using the buoyancy force, we have:
\[ \text{Buoyant Force} = \rho_{water} \cdot g \cdot V_d \]
\[ \text{Weight of the cube} = \rho_{wood} \cdot g \cdot V \]

Setting these two forces equal gives us:
\[ \rho_{water} \cdot g \cdot (0.6V) = \rho_{wood} \cdot g \cdot V \]

### Step 3: Simplify the equation
Cancelling \( g \) and \( V \) from both sides:
\[ \rho_{water} \cdot 0.6 = \rho_{wood} \]

### Step 4: Substitute the known value of water density
Substituting \( \rho_{water} = 1000 \, \text{kg/m}^3 \):
\[ 1000 \cdot 0.6 = \rho_{wood} \]
\[ \rho_{wood} = 600 \, \text{kg/m}^3 \]

### Step 5: Set up the equations for the wooden cube in the unknown fluid
Let:
- \( \rho_{liquid} \) = density of the unknown fluid
- \( V_d \) = volume of the cube submerged in the unknown fluid

Since 85% of the cube is submerged in the unknown fluid:
\[ V_d = 0.85V \]

Using the buoyancy force for the unknown fluid, we have:
\[ \text{Buoyant Force} = \rho_{liquid} \cdot g \cdot V_d \]
\[ \text{Weight of the cube} = \rho_{wood} \cdot g \cdot V \]

Setting these two forces equal gives us:
\[ \rho_{liquid} \cdot g \cdot (0.85V) = \rho_{wood} \cdot g \cdot V \]

### Step 6: Simplify the equation
Cancelling \( g \) and \( V \) from both sides:
\[ \rho_{liquid} \cdot 0.85 = \rho_{wood} \]

### Step 7: Substitute the known value of wood density
Substituting \( \rho_{wood} = 600 \, \text{kg/m}^3 \):
\[ \rho_{liquid} \cdot 0.85 = 600 \]
\[ \rho_{liquid} = \frac{600}{0.85} \]

### Step 8: Calculate the density of the unknown fluid
Calculating \( \rho_{liquid} \):
\[ \rho_{liquid} = 705.88 \, \text{kg/m}^3 \]
Rounding off gives:
\[ \rho_{liquid} \approx 706 \, \text{kg/m}^3 \]

### Final Answers
- Density of wood, \( \rho_{wood} = 600 \, \text{kg/m}^3 \)
- Density of unknown fluid, \( \rho_{liquid} \approx 706 \, \text{kg/m}^3 \)