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Question) An ideal heat engine working between T₁ and T₂ has efficiency $η$ . When T₁ is doubled and T₂ is halved, efficiency becomes / remains $(T_{1} > T_{2})$ .
$(3) \frac{η + 3}{4}$

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Solution

Dear Student,
$A c c o r d i n g t o q u e s t i o n I n i t i a l l y η = 1 - \frac{T_{2}}{T_{1}} - - - 1 a n d a f t e r c h a n g e η' = 1 - \frac{T_{2} / 2}{2 T_{1}} = 1 - \frac{T_{2}}{4 T_{1}} = 1 - \frac{1 - η}{4} η' = \frac{3 + η}{4} (t h i s i s t h e f i n a l e f f i c i e n c y)$
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