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(B) The increase in volume is given by $\Delta V = \gamma V_{0} \Delta T$.Since $\gamma = 3\alpha$,we have $\Delta V = (3\alpha) V_{0} \Delta T$.The t

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$1.2 	imes 10^{5} \; cm^{3}$

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(B) The increase in volume is given by $\Delta V = \gamma V_{0} \Delta T$.Since $\gamma = 3\alpha$,we have $\Delta V = (3\alpha) V_{0} \Delta T$.The total surface area of a cube is $6a^{2}$,where $a$ is the side length.Given $6a^{2} = 24 \; m^{2}$,so $a^{2} = 4 \; m^{2}$,which means $a = 2 \; m$.The initial volume $V_{0} = a^{3} = (2)^{3} = 8 \; m^{3}$.Substituting the values: $\Delta V = (3 	imes 5.0 	imes 10^{-4} \; ^{\circ}C^{-1}) 	imes (8 \; m^{3}) 	imes (10 \; ^{\circ}C)$.$\Delta V = 15 	imes 10^{-4} 	imes 80 = 1200 	imes 10^{-4} = 0.12 \; m^{3}$.Since $1 \; m^{3} = 10^{6} \; cm^{3}$,then $\Delta V = 0.12 	imes 10^{6} \; cm^{3} = 1.2 	imes 10^{5} \; cm^{3}$.

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

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