$I.$ Area $ABCD =$ Work done on the gas
$II.$ Area $ABCD =$ Net heat absorbed
$III.$ Change in the internal energy in cycle $= 0$
Which of these are correct
since initial and final conditions are the same, hence internal energy has not changed which is
a function of $T$ that is temperature. $\Delta U=0$
$Q=W+\Delta U=W$
Hence, the area also shows heat absorbed as it is equal to work done by hte gas.
Answer- $(I I, I I I)$
