Q137 / 160 · AP EAMCET 2023 (16 May Shift 2) · Physics

In the cyclic process given in the $\mathrm{P}-\mathrm{V}$ diagram, the work done is Figure

  • A
    $\pi\left(\frac{P_2-P_1}{2}\right)^2$
  • B
    $\pi\left(\frac{V_2-V_1}{2}\right)^2$
  • C
    $\pi\left(\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right)$
  • Correct answer
    D
    $\frac{\pi}{4}\left(P_2-P_1\right)\left(V_2-V_1\right)$

Solution

From diagram, $\begin{aligned} & \text { major axis, } 2 \mathrm{~b}=\mathrm{V}_2-\mathrm{V}_1 \Rightarrow \mathrm{b}=\frac{\mathrm{V}_2-\mathrm{V}_1}{2} \\ & \text { minor axis, } 2 \mathrm{a}=\mathrm{P}_2-\mathrm{P}_1 \Rightarrow \mathrm{a}=\frac{\mathrm{P}_2-\mathrm{P}_1}{2} \\ & \text { Work done }=\text { Area of cyclic work }=\pi \mathrm{ab} \\ & =\pi \times\left(\frac{\mathrm{P}_2-\mathrm{P}_1}{2}\right) \times\left(\frac{\mathrm{V}_2-\mathrm{V}_1}{2}\right) \\ & =\frac{\pi}{4}\left(\mathrm{P}_2-\mathrm{P}_1\right)\left(\mathrm{V}_2-\mathrm{V}_1\right) \end{aligned}$

Hence, the correct answer is option: 'D'.

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