The temperature of the source of Carnot's heat engine is `T_(1)`, and that of sink is `T_(2)` having efficienty 0.5. The efficiency of another Carnot's heat engine is also 0.5. The temperature of source and sink of the second engine are respectively.

Options

A. `2T_(1),2T_(2)`
B. `2T_(1),(T_(2))/2`
C. `T_(1)+5,T_(2)-5`
D. `T_(1)+10,T_(2)-10`

Answer

A

Explanation

We know that `eta=1-(T_(2))/(T_(1))` It is given that `eta_(1)=eta_(2)`
`:.` Ratio of `(T_(2))/(T_(1))` is same.