Q64 / 90 · JEE Main 2022 (25 Jun Shift 2) · Physics

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 $α = 5.0 \times 10^{- 4} {{}^{\circ}C}^{- 1}$ ).

Solution

Total surface area of a cube is $6 a^{2} = 24 m^{2}$ $\Rightarrow a^{2} = 4$ $\Rightarrow a = 2 m$

Volume of a cube $V = a^{3} = 8 m^{3}$

Change in volume is given by,

$Δ V = V_{0} γ Δ T$ , where $γ = 3 α$

$Δ V = [8] \times [1.5 \times 10^{- 4}] \times [10] m^{3}$ $\Rightarrow Δ V = 1.2 \times 10^{5} {cm}^{3}$

Hence, the correct answer is option: 'B'.