Question

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

• A. \( 2.4 \times 10^6 \, \text{cm}^3 \)
• B. \( 1.2 \times 10^5 \, \text{cm}^3 \)
• C. \( 6.0 \times 10^4 \, \text{cm}^3 \)
• D. \( 4.8 \times 10^5 \, \text{cm}^3 \)

Hint:

For solids, use \( \gamma = 3\alpha \) to find volumetric expansion. Ensure units for \( V_0 \) are consistent and surface area formulas are accurately applied.

Answer 1

The Correct Option is B

The volumetric expansion is calculated as follows: \[ \Delta V = V_0 \gamma \Delta T, \] where: \( V_0 = a^3 \) (initial volume of the cube), \( \gamma = 3\alpha \) (coefficient of volumetric expansion), \( \Delta T = 10^\circ \, \text{C} \) (temperature change). The total surface area of a cube is given by: \[ \text{Total Surface Area} = 6a^2. \] Given that the total surface area is \( 6a^2 = 24 \, \text{m}^2 \), we solve for the side length \( a \): \[ a^2 = \frac{24}{6} = 4 \, \text{m}^2 \quad \implies \quad a = 2 \, \text{m}. \] The initial volume \( V_0 \) is then: \[ V_0 = a^3 = 2^3 = 8 \, \text{m}^3 = 8 \times 10^6 \, \text{cm}^3. \] Substituting these values into the formula for \( \Delta V \): \[ \Delta V = 8 \times 10^6 \cdot (3 \cdot 5 \times 10^{-4}) \cdot 10. \] The simplified result for the change in volume is: \[ \Delta V = 8 \times 10^6 \cdot 15 \times 10^{-3} = 1.2 \times 10^5 \, \text{cm}^3. \]