A solid metallic cube having total surface area $$24$$ m$$^2$$ is uniformly heated. If its temperature is increased by $$10°$$C, calculate the increase in volume of the cube. (Given $$\alpha = 5.0 \times 10^{-4}$$ °C$$^{-1}$$).

The total surface area of the cube is $$24$$ m$$^2$$.

A cube has 6 faces, so the area of each face is:

$$\frac{24}{6} = 4 \text{ m}^2$$

The side length of the cube is:

$$a = \sqrt{4} = 2 \text{ m}$$

The original volume of the cube is:

$$V = a^3 = 2^3 = 8 \text{ m}^3$$

The coefficient of volume expansion is:

$$\gamma = 3\alpha = 3 \times 5.0 \times 10^{-4} = 1.5 \times 10^{-3} \text{ °C}^{-1}$$

The increase in volume is:

$$\Delta V = V \gamma \Delta T = 8 \times 1.5 \times 10^{-3} \times 10$$

$$\Delta V = 0.12 \text{ m}^3$$

Converting to cm$$^3$$ ($$1 \text{ m}^3 = 10^6 \text{ cm}^3$$):

$$\Delta V = 0.12 \times 10^6 = 1.2 \times 10^5 \text{ cm}^3$$

Hence, the correct answer is Option B.