A wooden cylindrical block floats vertically with 70% of its volume immersed in a liquid at `0^@C` . When temperature increases to `70^@C` , the block sinks in the liquid. The coefficient of cubical expansion of liquid in `K^(-1)` is

Options

A. `12.6 xx 10^(-3) K^(-1)`
B. `61.2 xx 10^(-3) K^(-1)`
C. `6.12 xx 10^(-3) K^(-1)`
D. `6.00 xx 10^(-3) K^(-1)`

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
At 0°C, a wooden cylindrical block floats with 70% of its volume immersed in a liquid. Let:
- \( V \) = total volume of the block
- \( \rho_W \) = density of the wood
- \( \rho_L \) = density of the liquid at 0°C

2. **Setting Up the Equilibrium Condition**:
The weight of the block is equal to the buoyant force acting on it. The weight of the block is given by:
\[ \text{Weight} = V \cdot \rho_W \cdot g \]
The buoyant force is equal to the weight of the liquid displaced, which is 70% of the volume of the block:
\[ \text{Buoyant Force} = 0.7V \cdot \rho_L \cdot g \]
Setting these equal gives:
\[ V \cdot \rho_W \cdot g = 0.7V \cdot \rho_L \cdot g \]
Cancelling \( V \) and \( g \) from both sides, we get:
\[ \rho_W = 0.7 \rho_L \]

3. **Finding the Density of the Liquid**:
Rearranging the equation gives:
\[ \rho_L = \frac{10}{7} \rho_W \]

4. **Considering the Effect of Temperature Increase**:
When the temperature increases to 70°C, the density of the liquid changes due to thermal expansion. The new density \( \rho_L' \) can be expressed as:
\[ \rho_L' = \frac{\rho_L}{1 + \gamma \Delta T} \]
where \( \Delta T = 70°C - 0°C = 70°C \) and \( \gamma \) is the coefficient of cubical expansion of the liquid.

5. **Setting Up the New Equilibrium Condition**:
At 70°C, the block sinks, meaning the weight of the block equals the buoyant force again:
\[ V \cdot \rho_W = V \cdot \rho_L' \]
Cancelling \( V \) gives:
\[ \rho_W = \rho_L' \]

6. **Substituting for \( \rho_L' \)**:
Substituting \( \rho_L' \) into the equation gives:
\[ \rho_W = \frac{\rho_L}{1 + \gamma \Delta T} \]
Substituting \( \rho_L = \frac{10}{7} \rho_W \):
\[ \rho_W = \frac{\frac{10}{7} \rho_W}{1 + \gamma \cdot 70} \]

7. **Solving for \( \gamma \)**:
Rearranging gives:
\[ 1 + \gamma \cdot 70 = \frac{10}{7} \]
Thus:
\[ \gamma \cdot 70 = \frac{10}{7} - 1 = \frac{3}{7} \]
Therefore:
\[ \gamma = \frac{3}{7 \cdot 70} = \frac{3}{490} \approx 0.00612 \, \text{K}^{-1} \]

### Final Answer:
The coefficient of cubical expansion of the liquid is:
\[ \gamma \approx 6.12 \times 10^{-3} \, \text{K}^{-1} \]