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Heat and Thermodynamics question

2024 · 5 Apr · Shift 1 · Q68
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  5. /2024 · 5 Apr · Shift 1 · Q68

Heat and Thermodynamics question

2024 · 5 Apr · Shift 1 · Q68

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The heat absorbed by a system in going through the given cyclic process is : JEE Main 2024 (Online) 5th April Morning Shift Physics - Heat and Thermodynamics Question 58 English
  1. A
    61.6 J
  2. B
    431.2 J
  3. C
    19.6 J
  4. D
    616 J
View written solutionFree

Correct answer: A

To find the heat absorbed in a cyclic process, we use the first law of thermodynamics.

1. Key idea for a cyclic process

For any cyclic process,

ΔU=0\Delta U = 0ΔU=0

because the system returns to its initial state.

Hence,

Q=WQ = WQ=W

So, the heat absorbed by the system is equal to the net work done in the cycle.


2. Work done in a cyclic process on a PPP-VVV diagram

The net work done in one complete cycle is equal to the area enclosed by the cycle on the PPP-VVV graph.

From the given figure, the enclosed area is triangular.

So,

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

Using the values from the graph:

  • Base =0.4 m3= 0.4\,\text{m}^3=0.4m3
  • Height =308 N/m2= 308\,\text{N/m}^2=308N/m2

Therefore,

W=12×0.4×308W = \frac{1}{2} \times 0.4 \times 308W=21​×0.4×308 W=0.2×308=61.6 JW = 0.2 \times 308 = 61.6\,\text{J}W=0.2×308=61.6J

3. Heat absorbed

Since for a cyclic process,

Q=WQ = WQ=W

we get

Q=61.6 JQ = 61.6\,\text{J}Q=61.6J

4. Checking options

  • A: 61.6 J61.6\,\text{J}61.6J ✅
  • B: 431.2 J431.2\,\text{J}431.2J ❌
  • C: 19.6 J19.6\,\text{J}19.6J ❌
  • D: 616 J616\,\text{J}616J ❌

So, the correct answer is:

61.6 J\boxed{61.6\,\text{J}}61.6J​
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