IIT JEE , AIEEE Forum - Physics , Thermal Physics

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Q3)

This is nice question in my opinion.This is how I did it.

For process ACB,we can say the net work done is work done in going from A to C and then C to B. Now work in going A to C is -ve as V decreases and in going from C to B it increases as V increases.And area under CB is greater than the area under AC, so net work done is +ve.

For process ADB, no doubt no work is done as V is constant, it is hence 0.

For process AEB, following the same lines as we did for process ACB, we can say in going from A to E,work is +ve and in going from E to B ,work is-ve. And area under EB is greater than area under AE => net work done is -ve.

Now Heat = deltaU + Work

deltaU is same for all and work is least for process AEB followed by process ADB and lastly process ACB.

SO option D is correct.

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