A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10 ° C, calculate the increase in volume of the cube. (Given α = 5.0 × 10 - 4 ⁻ 1). (a) 2.4 × 106 cm3 (b) 1.2 × 105 cm3 (c) 6.0 × 104 cm3 (d) 4.8 × 105 cm3

Answer: (b). 6a² = 24 ⇒ a = 2 m, V = 8 m³. Δ V = V(3α)Δ T = 8 × 3 × 5×10⁻⁴ × 10 = 0.12 m³ 0.12 m³ = 1.2×10⁵ cm³, answer (b).

A solid metallic cube having total surface area 24 m2 is uniformly… | JEE Main 2022 (25 Jun, Shift 2)

Q44JEE Main 2022 · 25 JUN · SHIFT 2Medium

Thermometry & Thermal Expansion

A solid metallic cube having total surface area 24 m2 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 $^\circ\text{C}^{-}$ 1).

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Answer: (b)

$6a^2 = 24 \Rightarrow a = 2$ m, $V = 8\text{ m}^3$ .
$\Delta V = V(3\alpha)\Delta T = 8 \times 3 \times 5\times10^{-4} \times 10 = 0.12\text{ m}^3$
$0.12\text{ m}^3 = 1.2\times10^5\text{ cm}^3$ , answer (b).

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Thermometry & Thermal Expansion question from JEE Main 2022 (25 Jun, Shift 2)

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