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A solid metallic cube having total surface area $24 m^{2}$ is uniformly heated. If its temperature is increased by $10^{0} C .$ Calculate the increase in volume of the cube (Given $α = 5.0 \times {10^{- 4}}^{_{0}} C^{- 1}$ )

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$4.8 \times 10^{5} c m^{3}$

$1.2 \times 10^{5} c m^{3}$

$2.4 \times 10^{6} c m^{3}$

$6.0 \times 10^{4} c m^{3}$

answer is B.

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Detailed Solution

$γ = 3 α = 3 \times 5 \times 10^{- 4} = 15 \times 10^{- 4} / {}^{0}C$

Given total surface area $A = 24 m^{2}$

$A = 6 a^{2} \Rightarrow 24 = 6 a^{2}$

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

$Δ V = V γ Δ t$

$\Rightarrow 8 \times 15 \times 10^{- 4} \times 10$

$\Rightarrow 120 \times 10^{- 3} m^{3}$

$\Rightarrow 120 \times 10^{- 3} \times 10^{6} {cm}^{3}$

$\Rightarrow 1.2 \times 10^{5} {cm}^{3}$

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