An engine has an efficiency of `1/6`. When the temperature of sink is reduced by `62^(@)C`, its efficiency is doubled. Temperature of the source is

Options

A. `124^(@)C`
B. `37^(@)C`
C. `62^(@)C`
D. `99^(@)C`

Answer

D

Explanation

`eta=1-T_(2)/T_(1)" "therefore1/6=1-T_(2)/T_(1)`
`thereforeT_(2)/T_(1)=1-1/6=5/6" "thereforeT_(1)=6/5T_(2)`
In the second case,
`eta=2xx1/6=1/3andT'_(2)=T_(2)-62`
`therefore1/3=1-(T_(2)-62)/T_(1)`
`therefore(T_(2)-62)/T_(1)=1-1/3=2/3`
`thereforeT_(2)-62=(2T_(1))/3=2/3xx6/5T_(2)=4/5T_(2)..."from"(1)`
`T_(2)-4/5T_(2)=62" "therefore1/5T_(2)=62`
`thereforeT_(2)=310K`
`thereforeT_(1)=6/5T_(2)=6/5xx310=372K`
`thereforeT_(1)=(372-273)=99^(@)C`