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Physics · Thermodynamics

JEE Main 2024 — 29 January, Shift 1 — Question 44

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

figure

  1. Option A:

    33800 J

  2. Option B:

    2200 J

  3. Option C:

    800 J

    Correct
  4. Option D:

    1200 J

Answer: C

Step-by-step solution

figure

Work done AB=12(8000+6000)\mathrm{AB}=\frac{1}{2}(8000+6000) Dyne /cm2×/ \mathrm{cm}^{2} \times

4 m3=(60004 \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}

=2×103×1105 N cm2×4 m3=2×10−2×N10−4 m2×4 m3=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}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy