Correct Answer - B
In a cyclic process,
`Q_(n et)=W_(n et)`
`:.` `Q_(AB)+Q_(BC)+Q_(CA)=` area under the graph
`:.` `600+200+Q_(CA)=1/2(3xx10^-4)(5xx10^5)`
`=75J`
`:.` `Q_(CA)=-725J=W_(CA)+DeltaU_(CA)`
`=-725=` (-Area under the graph)`+DeltaU_(CA)`
`=-(1/2xx11xx10^5)(3xx10^-4)+DeltaU_(CA)`
Solving we get, `DeltaU_(CA)=-560J`