Correct option is (2) Arithmetic mean of temperature of B & D
We know that, the efficiency of engine
\(\eta=1-\frac{T_2}{T_1}\)
where T2 = sink temperature
T1 = source temperature
\(\eta_{AB} = 1-\frac{T_B}{T_A}\)
\(\eta_{AD} = 1-\frac{T_D}{T_A}\)
\(\eta_{AC} = 1-\frac{T_C}{T_A}\)
Given
\(\eta_{AC}=\frac{\eta_{AB}+\eta_{AD}}{2}\)
1 - \(\frac{T_C}{T_A}\) = \(\cfrac{1-\frac{T_B}{T_A}+1-\frac{T_D}{T_A}}2\)
1 - \(\frac{T_C}{T_A}\) = \(\frac12-\frac{T_B}{2A}+\frac12-\frac{T_D}{2T_A}\)
1 - \(\frac{T_C}{T_A}\) = 1 - \(\frac{T_B}{2T_A}-\frac{T_D}{2T_A}\)
\(-\frac{T_C}{T_A}=-\frac1{T_A}(\frac{T_B+T_D}2)\)
TC = \(\frac{T_B+T_D}2\)
Arithmetic mean of temperature of B and D.