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Question

In the PV diagram shown in figure ABC is a semicircle. The work done in the process ABC 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 π2atmL
Given,ABC is a semicircle of radius 1

We know that, work done is area under the PV graph

For AB,ΔVAB=veΔWAB=ve
For BC,ΔVBC=+veΔWBC=+ve

Also,
|WAB|<|WBC|

Hence, total work done is positive.

ΔWABC=Area of semicircle
=12π(1)2

=π2 atmL
Hence, option (B) is correct.


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