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Q.

A carnot engine operating between temperatures T1  and T2  has efficiency 16 . When T2  is lowered by 62 K, its efficiency increases to 13 . Then T1  and T2  are, respectively:

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a

372 K and 330 K

b

330 K and 268 K

c

310 K and 248 K

d

372 K and 310 K

answer is D.

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Detailed Solution

η1=T1−T2T1=16 η2=T1−(T2−62)T1=13 ⇒T1−T2T1+62T1=13 16+62T1=13 62T1=16 ∴ T1=62×6=372K T1−T2T1=16 1−T2T1=16 T2372=56  ⇒T2=310K
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