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Question

In the $$P-V$$ diagram shown in figure $$ABC$$ is a semicircle. The work done in the process $$A \rightarrow B \rightarrow C$$ is
1132897_e3e3b3c0095745c29c596531c05edf9a.png


A
Zero
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B
π2atmL
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C
π2atmL
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D
4 atm L
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Solution

The correct option is C $$\dfrac {\pi}{2}atm-L$$
$$Given,\\\quad ABC\ is\ a\ semicircle\ of\ radius\ 1$$

$$We\ know\ that,\ work\ done\ is\ area\ under\ the\ P-V\ graph$$

$$For\ A\rightarrow B,\\ \Rightarrow \Delta V_{AB}=-ve\\ \Rightarrow \Delta W_{AB}=-ve\\$$
$$For\ B\rightarrow C,\\ \Rightarrow \Delta V_{BC}=+ve\\ \Rightarrow \Delta W_{BC}=+ve$$

$$Also,$$
$$\qquad \left | W_{AB} \right |<\left | W_{BC} \right |$$

$$Hence,\ total\ work\ done\ is\ positive.$$

$$\Delta W_{ABC}=Area\ of\ semicircle$$
$$\qquad\ \ \ \ \ \  =\dfrac{1}{2}\pi(1)^2$$

$$\qquad\ \ \ \ \ \  =\dfrac{\pi}{2}\ atm-L$$
$$Hence,\ option\ (B)\ is\ correct.$$


Physics

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