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Heat and Thermodynamics question

2023 · 8 Apr · Shift 2 · Q63
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  5. /2023 · 8 Apr · Shift 2 · Q63

Heat and Thermodynamics question

2023 · 8 Apr · Shift 2 · Q63

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A steel rod of length 1 m1 \mathrm{~m}1 m and cross sectional area 10−4 m210^{-4} \mathrm{~m}^{2}10−4 m2 is heated from 0∘C0^{\circ} \mathrm{C}0∘C to 200∘C200^{\circ} \mathrm{C}200∘C without being allowed to extend or bend. The compressive tension produced in the rod is ‾\underline{\hspace{2cm}}​×104 N\times 10^{4} \mathrm{~N}×104 N. (Given Young's modulus of steel =2×1011Nm−2=2 \times 10^{11} \mathrm{Nm}^{-2}=2×1011Nm−2, coefficient of linear expansion =10−5 K−1=10^{-5} \mathrm{~K}^{-1}=10−5 K−1 )
Numerical answer
View written solutionFree

Correct answer: 4

  1. Given data

    • Length of rod: L=1 mL = 1\,\text{m}L=1m
    • Cross-sectional area: A=10−4 m2A = 10^{-4}\,\text{m}^2A=10−4m2
    • Temperature rise: ΔT=200∘C−0∘C=200 K\Delta T = 200^\circ\text{C} - 0^\circ\text{C} = 200\,\text{K}ΔT=200∘C−0∘C=200K
    • Young's modulus: Y=2×1011 N m−2Y = 2 \times 10^{11}\,\text{N m}^{-2}Y=2×1011N m−2
    • Coefficient of linear expansion: α=10−5 K−1\alpha = 10^{-5}\,\text{K}^{-1}α=10−5K−1

  2. Concept used

    Since the rod is not allowed to expand, thermal expansion is prevented. Hence thermal strain is converted into mechanical strain.

    Thermal strain: ϵ=αΔT\epsilon = \alpha \Delta Tϵ=αΔT

    Stress developed: σ=Yϵ=YαΔT\sigma = Y\epsilon = Y\alpha\Delta Tσ=Yϵ=YαΔT

  3. Calculate the stress

    σ=(2×1011)(10−5)(200)\sigma = (2 \times 10^{11})(10^{-5})(200)σ=(2×1011)(10−5)(200)

    First, 10−5×200=2×10−310^{-5} \times 200 = 2 \times 10^{-3}10−5×200=2×10−3

    So, σ=2×1011×2×10−3=4×108 N m−2\sigma = 2 \times 10^{11} \times 2 \times 10^{-3} = 4 \times 10^8\,\text{N m}^{-2}σ=2×1011×2×10−3=4×108N m−2

  4. Calculate the force produced

    F=σAF = \sigma AF=σA F=(4×108)(10−4)=4×104 NF = (4 \times 10^8)(10^{-4}) = 4 \times 10^4\,\text{N}F=(4×108)(10−4)=4×104N

  5. Final answer

    The compressive force produced is: 4×104 N4 \times 10^4\,\text{N}4×104N

    So the required integer is: 4\boxed{4}4​

  6. Comparison with stored answer

    Stored correct answer = 444

    Our derived answer = 444

    Hence, they agree.

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