JEE Advanced 2013 Paper 2

(P) (C$_2$H$_5$)$_3$N + CH$_3$COOH → CH$_3$COO$^{-}$(aq) + (C$_2$H$_5$)$_3$NH$^{+}$(aq)
As CH$_3$COOH is a weak acid,its conductivity is already less. On addition of weak base, acid-base reaction takes place and new ions are created.So conductivity increases.
(Q) KI (0.1M) + AgNO$_3$ (0.01 M) → AgI ↓ (ppt) + KNO$_3$ (aq)
As the only reaction taking place is precipitation of AgI and in place of Ag$^{+}$,K$^{+}$ is coming in the solution, conductivity remain nearly constant and then increases.
(R) CH$_3$COOH + KOH → CH$_3$COOK (aq) + H$_2$O
OH$^{-}$(aq) is getting replaced by CH$_3$COO$^{-}$,which has poorer conductivity.So conductivity decreases and then after the end point, due to common ion effect, no further creation of ions take place. So, conductivity remain nearly same.
(S) NaOH + HI → NaI (aq) + H$_2$O
As H$^{+}$ is getting replaced by Na$^{+}$ conductivity decreases and after end point due to OH$^{-}$, it increases.
So answer is : (P) - (3) ; (Q) - (4) ; (R) - (2) ; (S) - (1). Answer is (A).