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Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q49
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  5. /2022 · 25 Jun · Shift 2 · Q49

Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q49

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
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 α\alphaα= 5.0 ×\times× 10 −-− 4 ∘^\circ∘ C −-− 1).
  1. A
    2.4 ×\times× 106 cm3
  2. B
    1.2 ×\times× 105 cm3
  3. C
    6.0 ×\times× 104 cm3
  4. D
    4.8 ×\times× 105 cm3
View written solutionFree

Correct answer: B

  1. Find the side of the cube

For a cube of side aaa, total surface area is 6a2=246a^2 = 246a2=24 So, a2=4⇒a=2 ma^2 = 4 \Rightarrow a = 2\,\text{m}a2=4⇒a=2m

  1. Find the initial volume

V=a3=23=8 m3V = a^3 = 2^3 = 8\,\text{m}^3V=a3=23=8m3

  1. Use volume expansion relation

For an isotropic solid, β=3α\beta = 3\alphaβ=3α Given, α=5.0×10−4 ∘C−1\alpha = 5.0 \times 10^{-4}\, ^\circ \text{C}^{-1}α=5.0×10−4∘C−1 So, β=3×5.0×10−4=1.5×10−3 ∘C−1\beta = 3 \times 5.0 \times 10^{-4} = 1.5 \times 10^{-3}\, ^\circ \text{C}^{-1}β=3×5.0×10−4=1.5×10−3∘C−1

Now, ΔV=βVΔT\Delta V = \beta V \Delta TΔV=βVΔT Substitute values: ΔV=(1.5×10−3)(8)(10)\Delta V = (1.5 \times 10^{-3})(8)(10)ΔV=(1.5×10−3)(8)(10) ΔV=0.12 m3\Delta V = 0.12\,\text{m}^3ΔV=0.12m3

  1. Convert to cm3^33

Since 1 m3=106 cm31\,\text{m}^3 = 10^6\,\text{cm}^31m3=106cm3 therefore, 0.12 m3=0.12×106=1.2×105 cm30.12\,\text{m}^3 = 0.12 \times 10^6 = 1.2 \times 10^5\,\text{cm}^30.12m3=0.12×106=1.2×105cm3

  1. Match with the options

1.2×105 cm3\boxed{1.2 \times 10^5\,\text{cm}^3}1.2×105cm3​

So the correct option is B.

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