A

Step 1: Calculate the loss of weight of the metal due to buoyancy in the liquid at 27º C.
The weight of the metal in air, W_air = 46 g.
The weight in the liquid at 27º C, W_liquid_27 = 30 g.
Loss of weight due to buoyancy (Buoyant Force), F_b = W_air - W_liquid_27 = 46 g - 30 g = 16 g.
Step 2: Calculate the volume of the metal piece using the buoyant force equation:
F_b = V_{metal} \cdot \rho_{liquid} \cdot g.
Here, \rho_{liquid} = 1.24 g/cm^3 (Specific gravity at 27º C).
Thus, using g = 9.81 m/s² (cancelling g in units), we have:
V_{metal} = \frac{F_b}{\rho_{liquid}} = \frac{16 g}{1.24 g/cm^3} = 12.903 cm^3.
Step 3: Calculate the linear expansion due to temperature change. The weight in the liquid at 42º C is 30.5 g. Buoyancy in this condition gives:
New buoyant force at 42º C = W_air - W_liquid_42 = 46 g - 30.5 g = 15.5 g.
Using the specific gravity at 42º C (1.20):
New volume for buoyancy under 42º C condition = \frac{15.5 g}{1.20 g/cm^3} = 12.917 cm^3.
Step 4: Change in volume with temperature change from 27º C to 42º C is: \[ \Delta V = V_{final} - V_{initial} = 12.917 cm^3 - 12.903 cm^3 = 0.014 cm^3. \]
Step 5: The temperature change, \Delta T = (42 - 27) = 15º C.
Step 6: Using the volume expansion formula: \[ \Delta V = \beta \cdot V_{initial} \cdot \Delta T \]
Rearranging gives: \[ \beta = \frac{\Delta V}{V_{initial} \cdot \Delta T}. \]
Step 7: Plugging values into the expansion formula:
\[\beta = \frac{0.014 cm^3}{12.903 cm^3 \cdot 15} = \frac{0.014}{193.545} = 0.0000723. \]
Step 8: Since for solids, the linear expansion coefficient \( \alpha \) is related to \( \beta \) by: \( \alpha = \frac{\beta}{3}. \)
Therefore: \[ \alpha = \frac{0.0000723}{3} = 0.0000241. \]
Thus, the coefficient of linear expansion of the metal is approximately \( 2.41 \times 10^{-5} \).
Therefore, A.