1. Initial dimensions of the cube

   The box is a cube of side length $a$ at temperature $T$.

   So, its initial volume is $V = a^3.$

2. Expansion of each side

   If the coefficient of linear expansion is $\alpha$, then for a small temperature rise $\Delta T$, each side becomes $a' = a(1+\alpha \Delta T).$

3. New volume of the cube

   The new volume is $V' = (a')^3 = \left[a(1+\alpha \Delta T)\right]^3 = a^3(1+\alpha \Delta T)^3.$

4. Use approximation for small expansion

   Since $\Delta T$ is small, $\alpha \Delta T \ll 1$. Hence, $(1+\alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T.$

   Therefore, $V' \approx a^3(1+3\alpha \Delta T).$

5. Increase in volume

   $\Delta V = V' - V$ $\Delta V = a^3(1+3\alpha \Delta T) - a^3$ $\Delta V = 3a^3\alpha \Delta T.$

6. Check options

   • A: $3a^3\alpha \Delta T$ ✅
   • B: $4\pi a^3\alpha \Delta T$ ❌
   • C: $\frac{4}{3}\pi a^3\alpha \Delta T$ ❌
   • D: $4a^3\alpha \Delta T$ ❌

Hence, the correct answer is A.