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A carnot engine has efficiency `1//5` . Efficiency becomes `1//3` when temperature of sink is decreased by 50 K What is the temperature of sink ?A. 325 KB. 375 KC. 300 KD. 350 K

Answer» Correct Answer - C
The efficiency of Carnot engine , `eta=1-(T_(L))/(T_(H))`
Where , `T_(L)` is temperature of sink and `T_(H)` is temperature of According to question ,
`(1)/(5)=1-(T_(L))/(T_(H))` …(i)
`and " " (1)/(3)=1-(T_(L)-50)/(T_(H))` ....(ii)
Form Eq . (i) , `T_(L)/T_(H)=(4)/(5)rArrT_(H)=(5)/(4)T_(L)`
Substituting value of `T_(H)` in Eq .(ii) , we get
`(1)/(3)=1-(T_(L)-50)/((5)/(4)T_(L))rArr(4(T_(L)-50))/(5T_(L))=(2)/(3)`
`"or" " " T_(L)-50=(2)/(3)xx(5)/(4)T_(L)`
`"or" " " T_(L)-(5)/(6)T_(L)=50`
`therefore" " T_(L)=50xx6=300K `


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