A heat engine has an efficiency n.Temperature of source and sink are each decreased by 100K. The efficiency of the engine

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Decreases

Remains constant

Increases

Becomes one

answer is C.

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Detailed Solution

$\begin{array}{l} e f f i c i e n c y o f h e a t e n g i n e = η = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η = \frac{T_{1} - T_{2}}{T_{1}} \\ \Rightarrow η = \frac{T_{1} - T_{2}}{T_{1}} \\ t e m p e r a t u r e o f s o u r c e r e d u c e d b y 100 = T_{1} - 100 \\ t e m p e r a t u r e o f \sin k r e d u c e d b y 100 = T_{2} - 100 \\ n e w e f f i c i e n c y i s η^{1} = \frac{T_{1} - 100 - (T_{2} - 100)}{T_{1} - 100} = \frac{T_{1} - T_{2}}{T_{1} - 100} > η \\ η^{1} = \frac{T_{1} - T_{2} - 200}{T_{1} - 100} \\ W h e n T_{1} a n d T_{2} b o t h a r e d e c r e a s e d b y 100 K e a c h \\ T_{1} - T_{2} r e m a i n s c o n s \tan t \\ T_{1} d e c r e a s e s η^{1} i n c r e a s e s \\ e f f i c i e n c y i n c r e a s e s \end{array}$