Correct Answer - B
In a cyclic process heat absorbed by the system is equal to the work done by the system
`therefore` Hence absorbed =work done =area of the given circle
`=pir^(2)=pixx((Delta,P)/(2))((DeltaV)/(2))`
=`3.14xx((200)/(2))xx10^(3)xx((200)/(2))xx10^(-6)`
`=3.14xx100xx10^(3)xx10xx10^(-6)=31.4 J`