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

Answer

A

Explanation

Given
` eta _1 =1/6`
According to the question ,
` eta _1 =(T_1-T_2)/(T_1)`
` (T_1-T_2)/( T_1)=(1)/(6)` …(I )
` eta _2 =(T_1 -(T _2 - 62 ) )/(T_1) implies 2 xx eta _1 = (T_1 +T_2 + 62))/(T_1)`
` 2 xx 1/(6) = ( T _1 - T_2 + 62 )/(T_1) implies 1/3 = (T _1 - T_2 + 62) /( T_1) `.. . . .(ii)
from Eq ,(i) we get
` T_1 - T_2 = (T_1)/(6)`
Substituting this value in Eq .(ii ) we get
` 1/3 = ((T_1)/(6)+62)/(T_1)implies 1/3 = (T _1 + 372 ) /( 6T_1)`
` 6T _1 = 3T _1 + 1116 implies 3 T_1 = 1116 implies T_1 = 372 K`
from Eq . (i ) we get
` ( 372 - T_2)/(372 )=1/6`
` 6 ( 372 - T_2)=372`
`2232 - 6T_2 = 372`
` 6T_2= 2232 - 372`
` 6T_2 = 1860 `
` T_2 = ( 1860 ) /( 6) `
` T_2 = 310 K `