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 \[\Delta BPC\] \[=(2-1)(2\times {{10}^{2}}-1\times {{10}^{2}})+\frac{1}{2}\] \[\times \,(3-2)\times (2\times {{10}^{2}}-1\times {{10}^{2}})\] \[=100+\frac{100}{2}=150\,joule\]