A cannot engine has efficiency `(1)/(6)`. If temperature of sink is decreased by `62^(@)C` then its efficiency becomes `(1)/(3)` then the temperature of source and sink:

Options

A. `956^(@)C, 37^(@)C`
B. `80^(@)C, 37^(@)C`
C. `99^(@)C, 37^(@)C`
D. `90^(@)C, 37^(@)C`

Answer

C

Explanation

`eta= 1/6= (T_(1)-T_(2))/(T_(1))`….(i)
`eta'= 1/3= (T_(1)-(T_(2)-62))/(T_(1))` ….(2)
From (i), `6T_(1)-6T_(2)= T_(1)`
`5T_(1)=6T_(2), T_(1)=6/5T_(2)`….(3)
Put in (2), `1/3=(6/5T_(2)-T_(2)+62)/(6/5T_(2))`
`3/5T_(2)+186= 6/5T_(2)`
`186=3/5T_(2),T_(2)=(186xx5)/3=310K`
From (3), `T_(1)=6/5xx310=372K`
`:. T_(1)= 372-273= 99^(@)C`
`T_(2)=310-273= 37^(@)C`