A metallic bob weights `50 g` in air. If it is immersed in a liquid at a temperature of `25^@ C`, it weights `45 g` . When the temperature of the liquid is raised to `100^@ C`, it weights `45.1 g`. Calculate the coefficient of cubical expansion of the liquid. Given that coefficient of cubical expansion of the metal is `12 xx 10^(-6) .^@ C^-1`.

Explanation

To solve the problem, we need to find the coefficient of cubical expansion of the liquid using the information provided about the weights of the metallic bob in different conditions.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Weight of the metallic bob in air (W_air) = 50 g
- Weight of the bob in the liquid at 25°C (W_liquid_25) = 45 g
- Weight of the bob in the liquid at 100°C (W_liquid_100) = 45.1 g
- Coefficient of cubical expansion of the metal (γ_metal) = 12 × 10^(-6) °C^(-1)

2. **Calculate the Buoyant Force at 25°C:**
- The buoyant force (F_b) at 25°C can be calculated using the difference in weight:
\[ F_b = W_{air} - W_{liquid_{25}} = 50 \, \text{g} - 45 \, \text{g} = 5 \, \text{g} \]

3. **Calculate the Buoyant Force at 100°C:**
- The buoyant force (F_b') at 100°C:
\[ F_b' = W_{air} - W_{liquid_{100}} = 50 \, \text{g} - 45.1 \, \text{g} = 4.9 \, \text{g} \]

4. **Using Archimedes' Principle:**
- The buoyant force is equal to the weight of the liquid displaced. The weight of the liquid displaced is related to the density of the liquid (ρ_liquid) and the volume of the bob (V_bob):
\[ F_b = \rho_{liquid} \cdot V_{bob} \cdot g \]
- At 25°C:
\[ 5 \, \text{g} = \rho_{liquid} \cdot V_{bob} \cdot g \]
- At 100°C:
\[ 4.9 \, \text{g} = \rho_{liquid}' \cdot V_{bob}' \cdot g \]

5. **Relating the Volumes:**
- The volume of the bob changes with temperature due to thermal expansion:
\[ V_{bob}' = V_{bob} \cdot (1 + \gamma_{metal} \cdot \Delta T) \]
- Where ΔT = (100 - 25) = 75°C.

6. **Setting Up the Equations:**
- For the liquid at 25°C:
\[ 5 = \rho_{liquid} \cdot V_{bob} \cdot g \]
- For the liquid at 100°C:
\[ 4.9 = \rho_{liquid} \cdot V_{bob} \cdot (1 + \gamma_{metal} \cdot 75) \cdot g \]

7. **Dividing the Two Equations:**
- Dividing the buoyant forces gives:
\[ \frac{5}{4.9} = \frac{1}{1 + \gamma_{metal} \cdot 75} \]

8. **Solving for the Coefficient of Cubical Expansion of the Liquid (γ_liquid):**
- Rearranging gives:
\[ 1 + \gamma_{liquid} \cdot 75 = \frac{4.9}{5} \]
- Solving for γ_liquid:
\[ \gamma_{liquid} = \frac{(5 - 4.9)}{4.9 \cdot 75} = \frac{0.1}{4.9 \cdot 75} \]

9. **Calculating the Value:**
- Calculate γ_liquid:
\[ \gamma_{liquid} = \frac{0.1}{367.5} \approx 2.72 \times 10^{-4} \, \text{°C}^{-1} \]

### Final Answer:
The coefficient of cubical expansion of the liquid is approximately \( \gamma_{liquid} \approx 2.72 \times 10^{-4} \, \text{°C}^{-1} \).