question_answer

A cyclic process is shown in the figure. Work done during the cyclic process ABCDA is:

A) 160J    

B)               150J 

C) 600J                

D) 900J

Correct Answer: B

Solution :

[b]   Work done during cyclic process ABCDA = Area enclosed by ABCDA = Area DCPA + Area of \[\Delta BPC\] \[=(2-1)(2\times {{10}^{2}}-1\times {{10}^{2}})+\frac{1}{2}\] \[\times \,(3-2)\times (2\times {{10}^{2}}-1\times {{10}^{2}})\] \[=100+\frac{100}{2}=150\,joule\]