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If the temperature of a source increases...

If the temperature of a source increases, then the efficiency of a heat engine

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The correct Answer is:
60
`eta=1 - (T_L)/(T_H) " "...(i)`
and `3 eta = 1 - (T_(L))/(2T_H) " "...(ii)`
From (i) and (ii)
`3((1-T_L)/T_H)=(1-T_L/(2T_H))`
`implies 3 -(3T_L)/(T_H) =1 - T_L/(2T_H) implies 2 = (3T_L)/(T_H) - (T_L)/(2T_H)`
`implies 2 = (6T_L-T_L)/(2T_H)=(5T_L)/(2T_H)`
`:. T_L/T_H=(4/5):. eta =1 -(4/5) =(1/5)`
`:. 3 eta =(3/5)xx100 =60%`
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