Thermodynamics

JEE Main level, numerical answer, medium

71 of 190

Question 71 of 190

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\times 10^{-1} \mathrm{~J}.
(Take
π=3.14\pi=3.14)

Show answer and solution

Answer: 314


Area of circle,
W=π4 d1 d2\mathrm{W}=\frac{\pi}{4} \mathrm{~d}_1 \mathrm{~d}_2
W=π4(500−300)×103(350−150)×10−6\mathrm{W}=\frac{\pi}{4}(500-300) \times 10^3(350-150) \times 10^{-6}
W = 31.4 Joule

W=314×10−1Joule\mathrm{W}=314 \times 10^{-1} \mathrm{Joule}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.