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Heat and Thermodynamics question

2021 · 22 Jul · Shift 2 · Q72
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Heat and Thermodynamics question

2021 · 22 Jul · Shift 2 · Q72

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10 ∘^\circ∘ C. Coefficient of liner expansion of material of track is 10 −-− 5/∘^\circ∘ C. The energy stored per meter in the track is ‾\underline{\hspace{2cm}}​ J/m. (Young's modulus of material of track is 1011 Nm −-− 2)
Numerical answer
View written solutionFree

Correct answer: 05

  1. Given data
  • Cross-sectional area: A=0.01 m2A = 0.01\,\text{m}^2A=0.01m2
  • Temperature change: ΔT=10∘C\Delta T = 10^\circ \text{C}ΔT=10∘C
  • Coefficient of linear expansion: α=10−5/∘C\alpha = 10^{-5}/^\circ\text{C}α=10−5/∘C
  • Young's modulus: Y=1011 N/m2Y = 10^{11}\,\text{N/m}^2Y=1011N/m2

We need the strain energy stored per meter length of the track when expansion is prevented.

  1. Thermal strain produced

If free expansion is prevented, the thermal strain developed is

ϵ=αΔT\epsilon = \alpha \Delta Tϵ=αΔT

So,

ϵ=10−5×10=10−4\epsilon = 10^{-5} \times 10 = 10^{-4}ϵ=10−5×10=10−4
  1. Stress developed

The thermal stress is

σ=Yϵ=YαΔT\sigma = Y\epsilon = Y\alpha\Delta Tσ=Yϵ=YαΔT

Thus,

σ=1011×10−4=107 N/m2\sigma = 10^{11} \times 10^{-4} = 10^7\,\text{N/m}^2σ=1011×10−4=107N/m2
  1. Energy density (strain energy per unit volume)

Strain energy per unit volume is

u=12σϵ=12Yϵ2u = \frac{1}{2}\sigma \epsilon = \frac{1}{2}Y\epsilon^2u=21​σϵ=21​Yϵ2

Substituting values:

u=12×1011×(10−4)2 u = \frac{1}{2} \times 10^{11} \times (10^{-4})^2u=21​×1011×(10−4)2 u=12×1011×10−8 u = \frac{1}{2} \times 10^{11} \times 10^{-8}u=21​×1011×10−8 u=12×103=500 J/m3 u = \frac{1}{2} \times 10^3 = 500\,\text{J/m}^3u=21​×103=500J/m3
  1. Energy stored per meter length

Volume of 1 meter length of track:

V=A×1=0.01×1=0.01 m3V = A \times 1 = 0.01 \times 1 = 0.01\,\text{m}^3V=A×1=0.01×1=0.01m3

Hence energy stored per meter:

U=νV=500×0.01=5 JU = \nu V = 500 \times 0.01 = 5\,\text{J}U=νV=500×0.01=5J

So, the energy stored per meter is

5 J/m\boxed{5\,\text{J/m}}5J/m​
  1. Comparison with stored correct answer

Stored correct answer: 050505

Derived answer: 555

These are the same integer value.

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