A) ABC
B) AD
C) BAD
D) BC
Correct Answer: D
Solution :
: Work done = (Potential difference)\[Q\] \[{{W}_{ABC}}=[({{V}_{A}}-{{V}_{B}})+({{V}_{B}}-{{V}_{C}})]Q=({{V}_{A}}-{{V}_{C}})Q\] \[=2Q\] \[{{W}_{AD}}=({{V}_{A}}-{{V}_{D}})Q=(-1-2)Q=-3Q\] \[{{W}_{BAD}}=[({{V}_{B}}-{{V}_{A}})+({{V}_{A}}-{{V}_{D}})]Q=({{V}_{B}}-{{V}_{D}})Q\] \[=(1-2)Q=-1Q\] \[{{W}_{BC}}=({{V}_{A}}-{{V}_{C}})Q=(1+3)Q=4Q\] \[\therefore \]Maximum work is done along path BC.You need to login to perform this action.
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