Consider a reversible engine of efficiency `1/6` when the temperature of the sink is reduced by ` 62 ^@C ` , its efficiency gets doubled . The temperature of the source and sink respectively are

Options

A. 372K and 310K
B. 273K and 300K
C. `99^@C and 10^@C`
D. `200^@C and 37^@C`

Answer

A

Explanation

Answer(1) Efficiency of engine, `eta = 1 - T_1/T_2`
`T_1` = Sink temperature .
`T_2` = source temperature .
`eta = (T_2 - T_1)/T_2 = 1/6` .
`Rightarrow 6T_2 - 6T_1 = T_2` `therefore T_2 = 1.2 T_1` .
If temperature of sink is reduced by `62^@C,` efficiency gets doubled. .
`therefore eta = (T_2 - (T_1 - 62))/T_2 = 2(1/6)` ...........(1).
`Rightarrow 6T_2 - 6T_1 + 372 = 2T_2` .
`therefore 4T_2 - 6T_1 + 372 =0` ....(2)
By solving (1) and (2) equations,
`T_1 = 310 K` .
`therefore T_2 = 1.2(310) = 372 K`