To solve the problem of finding the increase in the volume of a metal box when it is heated, we need to understand the concept of thermal expansion. The box is initially a cube with side length a and the coefficient of linear expansion for the material is given as \alpha.
When the temperature of the box increases by \Delta T, each dimension of the box expands. For a linear expansion, the change in length \Delta L for each side is given by:
\Delta L = a \alpha \Delta T
Therefore, the new length of each side of the cube becomes:
a_{\text{new}} = a + \Delta L = a(1 + \alpha \Delta T)
To find the change in volume, we calculate the new volume of the cube:
V_{\text{new}} = (a_{\text{new}})^3 = \left[a(1 + \alpha \Delta T)\right]^3
Expanding this expression using the binomial theorem (for small \alpha \Delta T), we get:
V_{\text{new}} = a^3 (1 + 3\alpha \Delta T + 3(\alpha \Delta T)^2 + (\alpha \Delta T)^3)
Since \alpha \Delta T is small, the higher-order terms (\alpha \Delta T)^2 and (\alpha \Delta T)^3 can be neglected. Thus, the new volume becomes approximately:
V_{\text{new}} \approx a^3 (1 + 3\alpha \Delta T)
Therefore, the increase in volume \Delta V is:
\Delta V = V_{\text{new}} - V_{\text{original}}
\Delta V = a^3(1 + 3\alpha \Delta T) - a^3
\Delta V = a^3 + 3a^3 \alpha \Delta T - a^3
\Delta V = 3a^3 \alpha \Delta T
Therefore, the correct answer is 3a^3 \alpha \Delta T.
This formula shows that the increase in volume is directly proportional to the original volume, the coefficient of linear expansion, and the temperature change.