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Heat and Thermodynamics question

2019 · 12 Jan · Shift 1 · Q46
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Heat and Thermodynamics question

2019 · 12 Jan · Shift 1 · Q46

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
For the given cyclic process CAB as shown for a gas, the work done is : JEE Main 2019 (Online) 12th January Morning Slot Physics - Heat and Thermodynamics Question 346 English
  1. A
    1 J
  2. B
    10 J
  3. C
    5 J
  4. D
    30 J
View written solutionFree

Correct answer: B

  1. Interpret the cyclic process on the graph

    In a cyclic process on a PPP-VVV diagram, the work done by the gas is equal to the area enclosed by the cycle.

  2. Identify the path C→A→BC \to A \to BC→A→B

    From the figure, the cycle CABCABCAB forms a triangular region on the PPP-VVV diagram.

    The work done in one cycle is: W=Area enclosed by triangleW = \text{Area enclosed by triangle}W=Area enclosed by triangle

  3. Use triangle area formula

    W=12×(base)×(height)W = \frac{1}{2} \times (\text{base}) \times (\text{height})W=21​×(base)×(height)

    From the graph, the horizontal difference in volume is: ΔV=2 m3\Delta V = 2\,\text{m}^3ΔV=2m3

    and the vertical difference in pressure is: ΔP=10 N/m2\Delta P = 10\,\text{N/m}^2ΔP=10N/m2

    Therefore, W=12×2×10=10 JW = \frac{1}{2} \times 2 \times 10 = 10\,\text{J}W=21​×2×10=10J

  4. Final answer

    10 J\boxed{10\,\text{J}}10J​

  5. Compare with stored correct answer

    Stored correct answer is B.

    Our derived answer is also 10 J, i.e. option B.

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