A carnot engine operating between temperatures `T_(1)` and `T_(2)` has efficiency `1/6`. When `T_(2)` is lowered by 62K, its efficience increases to `(1)/(3)`. Then `T_(1)` and `T_(2)` are respectively:

Options

A. 372 K and 330 K
B. 330 K and 268 K
C. 310 K and 248 K
D. 372 K and 310 K

Answer

D

Explanation

`eta_(1) =1-T_(2)/T_(1) rArr 1/6 =1-T_(2)/T_(1) = 5/6`……… (i)
`eta_(2) =1-(T_(2)-62)/T_(1) rArr 1/3 =1-(T_(2)-62)/T_(1)`………..(ii)
On solving Eqs. (i) and (ii)
`T_(1) = 372` K and `T_(2) = 310 K`