A block of wood of density `500 kg//m^(3)` has mass `m` kg in air. A lead block which has apparent weight of `28 kg` in water is attached to the block of wood, and both of them are submerged in water. If their combined apparent weight in water is `20 kg`, find the value of `m`. Take density of water `=1000kg//m^(3)`

Explanation

To solve the problem, we need to analyze the forces acting on the blocks when they are submerged in water.

### Step-by-Step Solution:

1. **Understanding Apparent Weight**:
The apparent weight of an object submerged in a fluid is given by the formula:
\[ \text{Apparent Weight} = \text{Weight in air} - \text{Buoyant Force} \]
The buoyant force can be calculated using Archimedes' principle, which states:
\[ \text{Buoyant Force} = \text{Density of fluid} \times \text{Volume of object} \times g \]
where \( g \) is the acceleration due to gravity.

2. **Mass of the Lead Block**:
Let the mass of the lead block be \( m_L \). The apparent weight of the lead block in water is given as 28 kg. Therefore, we can write:
\[ m_L - \text{Buoyant Force of Lead} = 28 \text{ kg} \]
The buoyant force on the lead block can be expressed as:
\[ \text{Buoyant Force of Lead} = \rho_{water} \times V_L \times g \]
where \( V_L \) is the volume of the lead block and \( \rho_{water} = 1000 \text{ kg/m}^3 \).

3. **Weight of the Lead Block**:
The weight of the lead block in air is:
\[ m_L \times g \]
Thus, we can express the equation for the lead block as:
\[ m_L - 1000 \times V_L \times g = 28 \times g \]
Dividing through by \( g \):
\[ m_L - 1000 \times V_L = 28 \]
Rearranging gives:
\[ m_L = 1000 \times V_L + 28 \]

4. **Combined Apparent Weight**:
The combined apparent weight of the wood and lead blocks when submerged is given as 20 kg:
\[ (m_L + m) - \text{Buoyant Force of Wood} - \text{Buoyant Force of Lead} = 20 \]
The buoyant force of the wood block can be expressed as:
\[ \text{Buoyant Force of Wood} = \rho_{water} \times V_W \times g \]
where \( V_W \) is the volume of the wood block.

5. **Weight Equation**:
Thus, we can write:
\[ (m_L + m) - (1000 \times V_W + 1000 \times V_L) = 20 \]
Rearranging gives:
\[ m_L + m = 20 + 1000 \times (V_W + V_L) \]

6. **Volume of the Blocks**:
The volume of the wood block can be calculated using its density:
\[ V_W = \frac{m}{\rho_{wood}} = \frac{m}{500} \]
Substitute \( V_W \) into the equation:
\[ m_L + m = 20 + 1000 \left( \frac{m}{500} + V_L \right) \]
Simplifying gives:
\[ m_L + m = 20 + 2m + 1000V_L \]
Rearranging gives:
\[ m_L - m = 20 + 1000V_L \]

7. **Substituting for \( m_L \)**:
Substitute \( m_L = 1000V_L + 28 \) into the equation:
\[ (1000V_L + 28) - m = 20 + 1000V_L \]
This simplifies to:
\[ 28 - m = 20 \]
Therefore:
\[ m = 28 - 20 = 8 \text{ kg} \]

### Final Answer:
The mass \( m \) of the block of wood is **8 kg**.