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 \[\Delta BPC\] \[=(2-1)(2\times {{10}^{2}}-1\times {{10}^{2}})+\frac{1}{2}\] \[\times \,(3-2)\times (2\times {{10}^{2}}-1\times {{10}^{2}})\] \[=100+\frac{100}{2}=150\,joule\]You need to login to perform this action.
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