A) 107 p J
B) 104 p J
C) 102p J
D) \[{{10}^{-3}}\,\pi J\]
Correct Answer: C
Solution :
In a cyclic, DU = 0 From FLOT, D Q = DU + DW = 0 + DW = Area of closed curve Þ DQ = pr2 \[\pi {{\left( \frac{20}{2} \right)}^{2}}k{{P}_{a}}\times litre\] \[=100\,\pi \times {{10}^{3}}\times {{10}^{-3}}J=100\,\pi J\]
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