A) 20 J
B) -20 J
C) -10 J
D) 10 J
Correct Answer: C
Solution :
[c] : Work done by the gas = Area of \[\Delta ABC\]\[=\frac{1}{2}\times (AB)(BC)\] \[=\frac{1}{2}\times \frac{(450-200)}{{{10}^{6}}}\times (200-120)\times 1000\] \[=\frac{(250)(80)\times 1000}{2\times {{10}^{6}}}=10J\] Work done by the cycle is taken to be negative if the cycle is anticlockwise. \[\therefore \]\[W=-10J\]
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