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Heat and Thermodynamics question

2025 · 7 Apr · Shift 1 · Q75
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  5. /2025 · 7 Apr · Shift 1 · Q75

Heat and Thermodynamics question

2025 · 7 Apr · Shift 1 · Q75

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is ‾×10−1 J\underline{\hspace{2cm}}\times 10^{-1} \mathrm{~J}​×10−1 J. (Take π=3.14\pi=3.14π=3.14 ) JEE Main 2025 (Online) 7th April Morning Shift Physics - Heat and Thermodynamics Question 13 English
Numerical answer
View written solutionFree

Correct answer: 3140

  1. Key idea: In a cyclic process on a PPP–VVV diagram, the net work done by the gas over one complete cycle equals the area enclosed by the cycle.

  2. From the figure, the cycle is a circle/ellipse-like closed curve on the PPP–VVV plane with semi-axes:

    • along volume axis: 2×10−3 m32 \times 10^{-3}\,\text{m}^32×10−3m3
    • along pressure axis: 5×104 Pa5 \times 10^{4}\,\text{Pa}5×104Pa

    Hence enclosed area is W=πabW = \pi abW=πab where a=2×10−3 m3,b=5×104 Paa = 2 \times 10^{-3}\,\text{m}^3, \qquad b = 5 \times 10^4\,\text{Pa}a=2×10−3m3,b=5×104Pa

  3. Substitute values: W=3.14×(2×10−3)×(5×104)W = 3.14 \times (2 \times 10^{-3}) \times (5 \times 10^4)W=3.14×(2×10−3)×(5×104)

  4. Calculate: (2×10−3)(5×104)=10×101=100(2 \times 10^{-3})(5 \times 10^4) = 10 \times 10^1 = 100(2×10−3)(5×104)=10×101=100

    So, W=3.14×100=314 JW = 3.14 \times 100 = 314\,\text{J}W=3.14×100=314J

  5. The question asks in the form: ‾×10−1 J\underline{\hspace{1cm}} \times 10^{-1}\,\text{J}​×10−1J

    Write 314 J314\,\text{J}314J as: 314 J=3140×10−1 J314\,\text{J} = 3140 \times 10^{-1}\,\text{J}314J=3140×10−1J

  6. Therefore, the required integer to fill in the blank is 3140\boxed{3140}3140​

  7. Comparison with stored answer:

    • Stored answer = 314314314
    • Derived answer = 314031403140

    So the stored answer appears to have missed the conversion into the requested "×10−1 J\times 10^{-1}\,\text{J}×10−1J" format.

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