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QJEEM24Ph705b: JEE Main 2024 Physics - Class 11 JEEM24Ph705b Solution: A thermodynamic system is taken from an original state to an…

A thermodynamic system is taken from an original state A\mathrm{A} to an intermediate state BB by a linear process as shown in the figure. It's volume is then reduced to the original value from B\mathrm{B} to C\mathrm{C} by an isobaric process. The total work done by the gas from AA to BB and BB to CC would be :

JEE Main 2024 (Online) 29th January Morning Shift Physics - Heat and Thermodynamics Question 56 English

Multiple Choice Options

  • A 800 J
  • B 2200 J
  • C 33800 J
  • D 1200 J

Step-by-Step Verified Solution

JEE Main 2024 (Online) 29th January Morning Shift Physics - Heat and Thermodynamics Question 56 English Explanation

Work done AB=12(8000+4000)\mathrm{AB}=\frac{1}{2}(8000+4000) Dyne /cm2×4 m3=(6000/\mathrm{cm}^2 \times 4 \mathrm{~m}^3=\left(6000\right. Dyne /cm2)×4 m3\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3

Work done BC=−(4000\mathrm{BC}=-\left(4000\right. Dyne /cm2)×4 m3\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3

Total work done =2000=2000 Dyne /cm2×4 m3/\mathrm{cm}^2 \times 4 \mathrm{~m}^3

=2×103amp;×1105Ncm2×4 m3amp;=2×10−2×N10−4 m2×4 m3amp;=2×102×4Nm=800 J\begin{aligned}=2 \times 10^3 & \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 \\& =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 \\& =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J}\end{aligned}

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