Thermal Properties of Matter

JEE Main level, single correct, medium

24 of 126

Question 24 of 126

A solid metallic cube having total surface area 24m224{m}^{2} is uniformly heated. If its temperature is increased by 10∘C10^{\circ}C, calculate the increase in volume of the cube. (Given α=5.0×10−4C∘−1\alpha =5.0\times {10}^{-4}{C^{\circ}}^{-1}).

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Answer: (B) 1.2×105cm31.2\times {10}^{5}{cm}^{3}

Total surface area of a cube is 6a2=24m26{a}^{2}=24{m}^{2}⇒a2=4\Rightarrow {a}^{2}=4 ⇒a=2m\Rightarrow a=2m
Volume of a cube V=a3=8m3V={a}^{3}=8{m}^{3}
Change in volume is given by,
ΔV=V0γΔT\Delta V={V}_{0}\gamma \Delta T, where γ=3α\gamma =3\alpha
ΔV=[8]×[1.5×10−4]×[10]m3\Delta V=\left[8\right]\times \left[1.5\times {10}^{-4}\right]\times \left[10\right]{m}^{3}⇒ΔV=1.2×105cm3\Rightarrow \Delta V=1.2\times {10}^{5}{cm}^{3}

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