The efficiency of carnot engine working between temperatures `200^@ C` and `0^@ C` is `eta_1`. The efficiency of this engine is `eta_2` when it work between temperatures `0^@ C `and `-200^@ C`. Then `eta_1/eta_2`=_____

Options

A. `1.0`
B. `0.721`
C. `0.577`
D. `0.34`

Answer

C

Explanation

`T_1`=200+273 =473 K
`T_2`=0+273=273 K
`T._1`=0+273 =273 K
`T._2`=-200+273 =73 K
`eta_1=200/473`
`eta_2=200/273`
`therefore eta_1/eta_2=273/473`=0.577