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Pls answer this Carnot engine mains question

Q 5. Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B are represent ed by $\eta $ _{A }and $\eta $ _{B} , respectively, then what is the value of $\frac{{\eta}_{B}}{{\eta}_{A}}?$
Option :

$1.\frac{7}{12}\phantom{\rule{0ex}{0ex}}2.\frac{12}{5}\phantom{\rule{0ex}{0ex}}3.\frac{12}{7}\phantom{\rule{0ex}{0ex}}4.\frac{5}{12}$

$1.\frac{7}{12}\phantom{\rule{0ex}{0ex}}2.\frac{12}{5}\phantom{\rule{0ex}{0ex}}3.\frac{12}{7}\phantom{\rule{0ex}{0ex}}4.\frac{5}{12}$

Let the output of both engines be W

Let the 1st engine take Q1 heat as input at temperature T1 = 600K and gives out heat Q2 at temperature T the second engine receive Q2 as input and give out Q3 at temperature T3 = 100k to the sink.

Work done by first engine W = Q1 – Q2

Work done by 2nd engine W = Q2 – Q3

Thus,

Q1 – Q2 = Q2 – Q3

Dividing both sides by Q1

1 – Q2/Q1 = Q2/Q1 – Q3/Q1

=> 1-T/T1 = Q2/ Q1(1- Q3/ Q2)

=> 1-T/T1 = Q2/ Q1(1- T3/ T)

=> 1-T/T1 = T/T1 (1- T3/ T)

=> T1/T – 1 = 1 - T3/ T

=> T1/T + T3/ T = 2

=> 1/T (T1 + T3) = 2

=>T = (T1 + T3)/2

=> T = (600+100)/2 =350 K

${\eta}_{A}=\frac{600-T}{600}u\mathrm{sin}gcarnotengineefficienyformula\phantom{\rule{0ex}{0ex}}{\eta}_{B}=\frac{T-100}{T}\phantom{\rule{0ex}{0ex}}\frac{{\eta}_{B}}{{\eta}_{A}}=\frac{\frac{T-100}{T}}{\frac{600-T}{600}}=\frac{\frac{350-100}{350}}{\frac{600-350}{600}}=\frac{{\displaystyle \frac{250}{350}}}{{\displaystyle \frac{250}{600}}}=\frac{600}{350}=\frac{12}{7}\phantom{\rule{0ex}{0ex}}Regards$

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