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A Carnot engine, whose sink is at 300 K, has an efficiency of 40%. By how much should the temperature of source be increased, so as to increase the efficiency by 50% of original efficiency?

(a) 380 K

(b) 275 K

(c) 325 K

(d) 250 K

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Right answer is (d) 250 K

Explanation: Ƞ=1-T2/T1

0.4=1- 300/T1

T1=300/0.6=500 K

Increase in temperature of the source= (T1)^‘-T1=750-500=250K.

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