Correct Answer - C
Given: `P_(1)=P_(2)=2P_(3)=P`
Since `1rarr2 and 3rarr4` are isobaric porcesses
`therefore` Heat received = `nC_(P)(T_(F)-T_(i))`
Accrording to question,
`C_(P)(T_(2)-T_(0))=C_(P)(T_(0)-T_(3))`
For process `2rarr3` i.e isochoric process
`(P_(2))/(t_(2))=(P_(5))/(t_(5))=(P_(3))/(T_(3))`
`therefore (P)/(T_(2))=(P_(5))/(T_(0))=(P//2)/(T_(3))`(From eqn(i))
or `T_(2)=2T_(3)`
and `P_(5)=(P)/(2)(T_(0))/(t_(3))`
soliving eqn (ii) using eqn (iii) we get
`P_(5)=(3)/(4)P`
`therefore` pressure at point 5 is `(3P)/(4)`