A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block of float in water. Specific gravity of wood is 0.8 and that of lead is 11.3.

Explanation

To solve the problem of finding the mass of the lead piece that will allow the wooden block to float in water, we can follow these steps:

### Step 1: Understand the Given Data
- Weight of the wooden block (MW) = 200 g
- Specific gravity of wood (SGW) = 0.8
- Specific gravity of lead (SGL) = 11.3
- Density of water (ρ_water) = 1 g/cm³

### Step 2: Calculate the Densities
Using the specific gravity, we can calculate the densities of wood and lead:
- Density of wood (ρ_wood) = SGW × ρ_water = 0.8 × 1 g/cm³ = 0.8 g/cm³
- Density of lead (ρ_lead) = SGL × ρ_water = 11.3 × 1 g/cm³ = 11.3 g/cm³

### Step 3: Set Up the Floating Condition
For the wooden block to float, the buoyant force must equal the total weight of the wooden block and the lead piece. The buoyant force (B) is given by:
\[ B = \text{Density of fluid} \times g \times V_{displaced} \]
Where \( V_{displaced} \) is the volume of water displaced.

### Step 4: Calculate the Volume of the Wooden Block
The volume of the wooden block can be calculated using its mass and density:
\[ V_{wood} = \frac{M_W}{\rho_{wood}} = \frac{200 \text{ g}}{0.8 \text{ g/cm}^3} = 250 \text{ cm}^3 \]

### Step 5: Write the Equation for Buoyant Force
The buoyant force must equal the weight of the wooden block and the lead piece:
\[ B = M_W \cdot g + M_L \cdot g \]
Where \( M_L \) is the mass of the lead piece.

### Step 6: Substitute Buoyant Force
Since the buoyant force can also be expressed in terms of the volume of the wooden block:
\[ \rho_{water} \cdot g \cdot V_{wood} = M_W \cdot g + M_L \cdot g \]
Cancelling \( g \) from both sides:
\[ \rho_{water} \cdot V_{wood} = M_W + M_L \]

### Step 7: Substitute Values
Substituting the values we have:
\[ 1 \cdot 250 = 200 + M_L \]
\[ 250 = 200 + M_L \]

### Step 8: Solve for Mass of Lead
Rearranging the equation gives:
\[ M_L = 250 - 200 = 50 \text{ g} \]

### Step 9: Adjust for Specific Gravity of Lead
Now, we need to account for the specific gravity of lead. The effective weight of the lead in water is given by:
\[ M_L \cdot \left(1 - \frac{1}{SGL}\right) \]
Substituting the specific gravity of lead:
\[ M_L \cdot \left(1 - \frac{1}{11.3}\right) = 50 \]
Solving this gives:
\[ M_L \cdot \frac{10.3}{11.3} = 50 \]
\[ M_L = 50 \cdot \frac{11.3}{10.3} \approx 54.8 \text{ g} \]

### Final Answer
The mass of the lead piece that will just allow the block to float in water is approximately **54.8 grams**.

---