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A Carnot engine operating between temperature T1 and T2 has efficiency 0.4, when T2 lowered by 50K, its efficiency increases to 0.5. Then T1 and T2 are respectively. 

(A) 300 K and 100 K 

(B) 400 K and 200 K 

(C) 600 K and 400 K 

(D) 500 K and 300 K

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Correct option: (D) 500 K and 300 K

Explanation:

Initial efficiency = η = 0.4

When T2 is lowered by 50k, η' = 0.5

η = 1 – (T2 / T1)……. efficiency of cannot engine

Hence   0.4 = 1 – (T2/T1)         (1)

0.5 = 1 – [(T2 – 50) / T1]       (2)

From (2)

0.5 = 1 – (T2 / T1) + (50 / T1)

From (1),

0.5 = 0.4 + (50 / T1)

        0.1 = (50 / T1)

        T1 = 500K

From (1),

1 – 0.4 = [T2 / (500)] hence T2 = 500 × 0.6 = 300K 

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