A block of wood weighs 12 kg and has a relative density 0.6. It is to be in water with 0.9 of its volume immersed. What weight of a metal is needed if the metal is on the top of wood ? [Relative density of metal = 14]

Options

A. 2 kg
B. 4 kg
C. 6 kg
D. 8 kg

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Given Data
- Weight of the block of wood (W_wood) = 12 kg
- Relative density of wood (RD_wood) = 0.6
- Relative density of water (RD_water) = 1 (since the density of water is taken as 1 g/cm³)
- Volume of wood immersed in water = 0.9 of its total volume (V_wood)
- Relative density of metal (RD_metal) = 14

### Step 2: Calculate the Density of Wood
Relative density is defined as the ratio of the density of a substance to the density of water. Thus, we can find the density of wood (ρ_wood) using the formula:
\[ \text{Relative Density} = \frac{\text{Density of Wood}}{\text{Density of Water}} \]
Substituting the known values:
\[ 0.6 = \frac{\rho_{wood}}{1 \, \text{g/cm}^3} \]
Thus, the density of wood is:
\[ \rho_{wood} = 0.6 \, \text{g/cm}^3 = 600 \, \text{kg/m}^3 \]

### Step 3: Calculate the Volume of Wood
The weight of the wood is given by:
\[ W_{wood} = \rho_{wood} \cdot V_{wood} \cdot g \]
Where g is the acceleration due to gravity (which will cancel out later). Rearranging gives:
\[ V_{wood} = \frac{W_{wood}}{\rho_{wood} \cdot g} \]
Substituting the known values:
\[ V_{wood} = \frac{12 \, \text{kg}}{600 \, \text{kg/m}^3} = 0.02 \, \text{m}^3 \]

### Step 4: Calculate the Volume Immersed
Given that 0.9 of the volume of wood is immersed in water:
\[ V_{immersed} = 0.9 \cdot V_{wood} = 0.9 \cdot 0.02 \, \text{m}^3 = 0.018 \, \text{m}^3 \]

### Step 5: Calculate the Buoyant Force
The buoyant force (F_b) acting on the wood is equal to the weight of the water displaced:
\[ F_b = \rho_{water} \cdot V_{immersed} \cdot g \]
Substituting the values:
\[ F_b = 1000 \, \text{kg/m}^3 \cdot 0.018 \, \text{m}^3 \cdot g = 18 \, \text{kg} \cdot g \]

### Step 6: Set Up the Equilibrium Condition
At equilibrium, the buoyant force equals the total weight of the wood and the metal:
\[ F_b = W_{wood} + W_{metal} \]
Substituting the known values:
\[ 18 \, \text{kg} \cdot g = 12 \, \text{kg} \cdot g + m \cdot g \]
Cancelling g from both sides:
\[ 18 = 12 + m \]
Thus, solving for m gives:
\[ m = 18 - 12 = 6 \, \text{kg} \]

### Final Answer
The weight of the metal needed is **6 kg**.

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