The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

Options

A. `1//2`
B. `1//4`
C. `1//3`
D. `2//3`

Answer

C

Explanation

Heat supplied `Q` = Area `FBCFF` = `T(S_(2) - S_(1)) + (1)/(2)(T_(2)- T_(1))(S_(2)-S_(1))`
Work done `W` = Area `ABC` = `(1)/(2)(T_(2)- T_(1))(S_(2) -S_(1))`
`eta = (W)/(Q) = ((1)/(2)(T_(2)- T_(1))(S_(2)-S_(1)))/(T_(1)(S_(2)-S_(1)) + (1)/(2)(T_(2) -T_(1))(S_(2)-S_(1)))`
=`(T_(2)-T_(1))/(T_(2)+T_(1)) = (2T_(0) - T_(0))/(2T_(0) + T_(0)) = (1)/(3)`