A) 6 cal
B) 16 cal
C) 66 cal
D) 14 cal
Correct Answer: A
Solution :
From first law of thermodynamics, \[\sqrt{2}{{I}_{AC}}={{I}_{EF}}\] For path iaf, \[{{I}_{AD}}=4{{I}_{EF}}\] \[{{I}_{AC}}={{I}_{EF}}\] \[{{I}_{AC}}=\sqrt{2}{{I}_{EF}}\] For path ibf, \[x={{x}_{0}}\cos (\omega t-\pi /4)\] or \[a=A\text{ }cos(\omega t+\delta ),\] (\[A={{x}_{0}},\text{ }\delta =-\pi /4\]is same being path independent function)You need to login to perform this action.
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