Question Number 52324 by prakash jain last updated on 06/Jan/19
$$\mathrm{Two}\:\mathrm{carnot}\:\mathrm{emgines}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{operated}\:\mathrm{in} \\ $$$$\mathrm{series}.\:\mathrm{engine}\:\mathrm{A}\:\mathrm{receiepved}\:\mathrm{heat}?\mathrm{from} \\ $$$$\mathrm{a}\:\mathrm{reservoir}\:\mathrm{at}\:\mathrm{600}\:\mathrm{K}\:\mathrm{and}\:\mathrm{rejects}\:\mathrm{it}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{reservoir}\:\mathrm{at}\:\mathrm{temp}\:\mathrm{T}.\:\mathrm{B}\:\mathrm{receives} \\ $$$$\mathrm{heat}\:\mathrm{rejected}\:\mathrm{by}\:\mathrm{A}\:\mathrm{and}\:\mathrm{in}\:\mathrm{turn}\:\mathrm{rejects} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{reservoir}\:\mathrm{at}\:\mathrm{100K}. \\ $$$$\mathrm{Find}\:\frac{\eta_{{B}} }{\eta_{{A}} } \\ $$$$\eta\:\mathrm{efficiency} \\ $$
Commented by Tinkutara last updated on 14/Jan/19
Commented by Tinkutara last updated on 14/Jan/19
In my JEE main it appeared with correction!
Answered by ajfour last updated on 06/Jan/19
$$\eta_{{A}} =\mathrm{1}−\frac{{T}}{\mathrm{600}}\:\: \\ $$$$\eta_{{B}} \:=\:\mathrm{1}−\frac{\mathrm{100}}{{T}}\: \\ $$$$\:\:\eta_{{eq}} =\:\mathrm{1}−\frac{\mathrm{100}}{\mathrm{600}}\:=\frac{\mathrm{5}}{\mathrm{6}}\:=\:\left(\eta_{{A}} +\eta_{{B}} \right)−\eta_{{A}} \eta_{{B}} \\ $$$${let}\:\:\frac{\eta_{{B}} }{\eta_{{A}} }\:=\:{r} \\ $$$$…. \\ $$
Commented by prakash jain last updated on 06/Jan/19
Thanks. I thought there is some info missing. But the question came in JEE main 2018 exam
Commented by ajfour last updated on 06/Jan/19
$${yes}\:{Sir},\:{i}\:{inquired}\:{and}\:{got}\:{to}\:{know} \\ $$$${the}\:{same},\:{but}\:{neither}\:{could}\:{solve} \\ $$$${nor}\:{found}\:{solution}\:{given},\:{please} \\ $$$${share}\:{the}\:{solution}\:{if}\:{you}'{ve} \\ $$$${understood}. \\ $$
Commented by ajfour last updated on 06/Jan/19
$$\eta_{{A}} =\:\mathrm{1}−\frac{{Q}_{\mathrm{1}} }{{Q}_{\mathrm{0}} }\:\:,\:\:\eta_{{B}} =\mathrm{1}−\frac{{Q}_{\mathrm{2}} }{{Q}_{\mathrm{1}} } \\ $$$$\eta_{{eq}} \:=\:\mathrm{1}−\frac{{Q}_{\mathrm{2}} }{{Q}_{\mathrm{0}} } \\ $$$$\left(\mathrm{1}−\eta_{{A}} \right)\left(\mathrm{1}−\eta_{{B}} \right)=\frac{{Q}_{\mathrm{2}} }{{Q}_{\mathrm{0}} }\:=\:\left(\mathrm{1}−\eta_{{eq}} \right) \\ $$$$\Rightarrow\:\:\eta_{{eq}} =\left(\eta_{{A}} +\eta_{{B}} \right)−\eta_{{A}} \eta_{{B}} \\ $$$$ \\ $$
Commented by prakash jain last updated on 06/Jan/19
I don't know the answer I just wanted to confirm the same. If you cannot solve a physics question it then no-one else can. My son is appearing for JEE Main next week and asked me this question.