A) \[7.5\times {{10}^{5}}\ joule\]
B) \[7.5\times {{10}^{5}}\ erg\]
C) \[12\times {{10}^{5}}\ joule\]
D) \[6\times {{10}^{5}}\ joule\]
Correct Answer: C
Solution :
Work done = Area of PV graph (here trapezium) \[=\frac{1}{2}(1\times {{10}^{5}}+5\times {{10}^{5}})\times (5-1)=12\times {{10}^{5}}J\]
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