AITS & Test SerieshardNUMERICAL

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Question

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Answer: 2.50

💡 Solution & Explanation

total total q w U 0    (cyclic process)   total q 700 0   total q 700 J   Now, 1 3 5 2 4 6 Q Q Q Q Q Q 700       5 500 800 Q 0 0 0 700       5 Q 600 J  Also, 1 3 5 2 4 6 S S S S S S 0        (cyclic process)   5 600 500 800 0 0 0 0 250 200 T        5 T 100K  1T 250 K  (given) So, 1 5 T

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