$Given,$

$Bed$ $width$ $(b) = 40m$

$Afflux$ $(y_2 - y_1) = 2m$

$Depth$ $of$ $flow$ $(y_1) = 2.5m$

$Depth$ $of$ $flow$ $of$ $downstream$ $(y_2) = 2+2.5m = 4.5$

$Bed$ $slope$ $S_0$ = ${1 \over 11000}$

$Manning's$ $constant$ $(N)$= $0.03$

$Area$ $of$ $flow$ $at$ $Section$ $ ; $ $A_1 = by_1 = 40 * 2.5 = 100m^2$

$Wetted$ $perimeter$ $P_1 = b+2y_1 = 40+2(2.5) = 45m$

$R_1 = {A_1 \over P_1} = {100 \over 4.5} = 2.22m$

$Using $ $Manning's$ $formula:$ $V_1 = {1 \over N}R_1^{2 \over 3}S_0^{1 \over 2} = {1 \over 0.03} [2.22]^{2 \over 3} [{1 \over 11000}]^{1 \over 2}$

$Therefore, $ $ $ $ $ $V_1 = 0.54$ $m/sec$

$Specific$ $energy$ $at$ $Section;$ $E_1 = y_1 + {V_1^2 \over 2g}$ $ = 2.5 + {0.54^2 \over 2*9.81}$ $ = 2.515 $ $m$ $Area$ $of$ $flow$ $at$ $Section;$ $A_2 = by_2 = 40*4.5 = 180m^2$

$Wetted$ $Perimeter$ $(P_2) = b+2y_2 = 40 + 2(4.5) = 49m$

$R_2 = {A_2 \over P_2} = {180\over 49} = 3.673$ $m$ \ltbr\gt $Using$ $continuity$ $equation:$ $A_1V_1 = A_2V_2$

$i.e$ $ $ $V_2 = {A_1V_1 \over A_2}$

$Therefore,$ $V_2 = {1000*0.54 \over 180} = 0.3$ $m/sec$

$Specific$ $energy$ $at$ $Section;$ $E_2 = y_2 + {V_2^2 \over 2g} = 4.5 + {(0.3)^2 \over 2*9.81} $ $ = 4.504 $ $m$ \ltbr\gt $Average$ $depth$ $of$ $flow$ $(y') = {2.5 + 4.5 \over 2} = 3.5$ $m$ \ltbr\gt $Average$ $area$ $of$ $flow$ $(A') = 40 * 3.5 = 140$ $m^2$

$Average$ $wetted$ $perimeter$ $(P') = 40 + 2(3.5) = 47$ $m$ \ltbr\gt $R' = {A' \over P'} = {140 \over 47} = 2.979$ $m$

$Average$ $velocity$ $of$ $flow$ $(V') = {1 \over N}$ $R^{-2/3}$ $p_e^{1/2}$

$Using$ $continuity$ $equation:$ $A_1V_1 = A'V'$

$i.e.$ $V' = {A_1 V_1 \over A'} $

$Therefore,$ $V' = {1000 * 0.54 \over 140} = 0.386$ $m/sec$

$Using $ $Manning's$ $formula:$ $0.386 = {1 \over 0.03}$ $[2.978]^{2/3}$ $[S_e]^{1/2}$

$Therefore,$ $S_e = 0.00003128 = {1 \over 31965.2}$

$So$ $therefore,$

$Length$ $of$ $back$ $water$ $curve$ $(Δx) = {E_2 - E_1 \over S_0 - S_e} = {4.504 - 2.515 \over {1 \over 11000} - {1 \over 31965.2}}$

$Therefore,$ $Δx = 33358.5$ $m$ $ $ $Ans.$