I feel that the answer should be C as kinetic energy term (for piston) is missing. A & B are correct.
How does Net force =0(equilibrium) imply velocity = 0? Am i missing something?
It's asking for the Incorrect options
C is an incorrect option as per me because the kinetic energy term is missing. Thus answer should be C.
It would be nice to know your thought process on this question @Aashish_Patel @Chirag_Hegde.
They have said equilibrium so ∆KE=0
Equilibrium means net force = 0. It does not mean that the velocity is zero.
What you mean is equilibrium position, what the question means is final rest state imo
It would be a very special case in which the final velocity of pistion=0 and the accelaration of the piston =0 and would depend on the initial set of conditions and the nature of the gas taken. You can not have this for all cases.
A,B,C are correct. So, it'd be (D) None of these.
But, None Of These is the "correct" option and not the 'incorrect' as they asked. So, there's a bit of language problem in question.
Give detailed solution
There is an ambiguity in the C option. The kinetic energy term is missing. Hence c must be the answer.
Alright, just a few min.
@Akshat_Joshi Which K.E. bro?
Wtot = \Delta KE
As the piston moves up to aquire equilibrium(net force=0), it will indeed posses some velocity. This kinetic energy will come from the loss in the internal energy of the gas.
It is assumed that the process will be slow (in my solution atleast).
And yeah, I agree that piston being at equilibrium and \triangle K.E. being zero couldn't be easily assumed.
Actually, in a quasi-static process each step is an equilibrium which is not the case here.
It's not mentioned if the process is quasi-static or not, so better assume it, right?