Solution:
Let b'>b$b$ and y'>y$y$ be the width and depth of the section.
Now,
flow area, A=b⋅y=Qv=8.400.7=12m2'>A=bâ??y=Qv=8.400.7=12m2$A=bâ??y=\frac{Q}{v}=\frac{8.40}{0.7}=12m^2$
or, b=12y'>b=12y$b=\frac{12}{y}$
We have, from chezy's equation,
v=cR⋅So=1nR23S012'>v=cRâ??Soâ??â??â??â??â??â??=1nR23S120$v=c\sqrt{Râ??S_o}=\frac{1}{n}{R}^{\frac{2}{3}}{S}_0^{\frac{1}{2}}$
or, R23=v⋅nS012=1.0844353'>R23=vâ??nS120=1.0844353$R^{\frac{2}{3}}=\frac{vâ??n}{S_0^{\frac{1}{2}}}=1.0844353$
or, R=1.12929'>R=1.12929$R=1.12929$
Again, we have,
R=AP=b⋅yb+2y=1212y+2y=12y12+2y2=6y6+y2'>R=AP=bâ??yb+2y=1212y+2y=12y12+2y2=6y6+y2$R=\frac{A}{P}=bâ??\frac{y}{b+2y}=\frac{12}{\frac{12}{y}+2y}=\frac{12y}{12}+2y^2=\frac{6y}{6}+y^2$
or,y2-5.3131y+6=0'>y2â??5.3131y+6=0$y^2-5.3131y+6=0$
solving this, we get,
y=3.685mand1.628m'>y=3.685mand1.628m$y=3.685m\text{and}1.628m$
Now,
When, y=3.685m,b=3.256m'>y=3.685m,b=3.256m$y=3.685m,b=3.256m$
When y=1.628m,b=7.371m'>y=1.628m,b=7.371m$y=1.628m,b=7.371m$
Both the solutions are valid and satisfy the requirement.

Solution:
Let b'>b$b$ and y'>y$y$ be the width and depth of the section.
Now,
flow area, A=b⋅y=Qv=8.400.7=12m2'>A=bâ??y=Qv=8.400.7=12m2$A=bâ??y=\frac{Q}{v}=\frac{8.40}{0.7}=12m^2$
or, b=12y'>b=12y$b=\frac{12}{y}$
We have, from chezy's equation,
v=cR⋅So=1nR23S012'>v=cRâ??Soâ??â??â??â??â??â??=1nR23S120$v=c\sqrt{Râ??S_o}=\frac{1}{n}{R}^{\frac{2}{3}}{S}_0^{\frac{1}{2}}$
or, R23=v⋅nS012=1.0844353'>R23=vâ??nS120=1.0844353$R^{\frac{2}{3}}=\frac{vâ??n}{S_0^{\frac{1}{2}}}=1.0844353$
or, R=1.12929'>R=1.12929$R=1.12929$
Again, we have,
R=AP=b⋅yb+2y=1212y+2y=12y12+2y2=6y6+y2'>R=AP=bâ??yb+2y=1212y+2y=12y12+2y2=6y6+y2$R=\frac{A}{P}=bâ??\frac{y}{b+2y}=\frac{12}{\frac{12}{y}+2y}=\frac{12y}{12}+2y^2=\frac{6y}{6}+y^2$
or,y2-5.3131y+6=0'>y2â??5.3131y+6=0$y^2-5.3131y+6=0$
solving this, we get,
y=3.685mand1.628m'>y=3.685mand1.628m$y=3.685m\text{and}1.628m$
Now,
When, y=3.685m,b=3.256m'>y=3.685m,b=3.256m$y=3.685m,b=3.256m$
When y=1.628m,b=7.371m'>y=1.628m,b=7.371m$y=1.628m,b=7.371m$
Both the solutions are valid and satisfy the requirement.

Solution:
 Width of the rectangular channel, b=7.5m'>b=7.5m$b=7.5m$
 Depth of flow, y=2.25m'>y=2.25my=2.25my=2.25m
Bed slope, S0'>S0S0S0 =11000'>110001100011000
Chezy's constant, C=55'>C=55C=55C=55
Now, we have,
Rate of flow, Q=Aâ??V=Aâ??CRâ??S0.......(I)'>Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)
Where, Flow area, A=bâ??y=7.5â??2.25=16.875m2'>A=bâ??y=7.5â??2.25=16.875m2A=bâ??y=7.5â??2.25=16.875m2A=bâ??y=7.5â??2.25=16.875m2
Wetted perimeter, P=b+2y=7.5+2â??2.25=12.0'>P=b+2y=7.5+2â??2.25=12.0P=b+2y=7.5+2â??2.25=12.0P=b+2y=7.5+2â??2.25=12.0
Also,
hydraulic radius, R=AP=16.87512.0=1.406'>R=AP=16.87512.0=1.406R=AP=16.87512.0=1.406R=AP=16.87512.0=1.406
Thus from equation (I)'>(I)(I)(I),
Q=Aâ??V=Aâ??CRâ??S0=16.875'>Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875â??'>â??â??â??551.406â??11000'>551.406â??11000â??â??â??â??â??â??â??â??â??â??â??551.406â??11000â??â??â??â??â??â??â??â??â??â??â??551.406â??11000â??â??â??â??â??â??â??â??â??â??â??
or, Q=34.76m3sec'>Q=34.76m3secQ=34.76m3secQ=34.76m3sec
Also, Conveyance,K=Aâ??cR=16.875'>K=Aâ??cRâ??â??â??=16.875K=Aâ??cRâ??â??â??=16.875K=Aâ??cRâ??â??â??=16.875â??'>â??â??â??55'>555555â??'>â??â??â??1.406'>1.406â??â??â??â??â??1.406â??â??â??â??â??1.406â??â??â??â??â??=1100.5'>=1100.5=1100.5=1100.5
Again, Froude's number, Fr=Vgâ??y=Câ??Râ??S09.81â??2.25=0.438'>Fr=Vgâ??yâ??â??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438Fr=Vgâ??yâ??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438Fr=Vgâ??yâ??â??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438
Since froudes number,Fr'>FrFrFr is less than 1'>111, the flow in the channel is of tranquil in nature.

Solution:
 Width of the rectangular channel, b=7.5m'>b=7.5m$b=7.5m$
 Depth of flow, y=2.25m'>y=2.25my=2.25my=2.25m
Bed slope, S0'>S0S0S0 =11000'>110001100011000
Chezy's constant, C=55'>C=55C=55C=55
Now, we have,
Rate of flow, Q=Aâ??V=Aâ??CRâ??S0.......(I)'>Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??.......(I)
Where, Flow area, A=bâ??y=7.5â??2.25=16.875m2'>A=bâ??y=7.5â??2.25=16.875m2A=bâ??y=7.5â??2.25=16.875m2A=bâ??y=7.5â??2.25=16.875m2
Wetted perimeter, P=b+2y=7.5+2â??2.25=12.0'>P=b+2y=7.5+2â??2.25=12.0P=b+2y=7.5+2â??2.25=12.0P=b+2y=7.5+2â??2.25=12.0
Also,
hydraulic radius, R=AP=16.87512.0=1.406'>R=AP=16.87512.0=1.406R=AP=16.87512.0=1.406R=AP=16.87512.0=1.406
Thus from equation (I)'>(I)(I)(I),
Q=Aâ??V=Aâ??CRâ??S0=16.875'>Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875Q=Aâ??V=Aâ??CRâ??S0â??â??â??â??â??â??=16.875â??'>â??â??â??551.406â??11000'>551.406â??11000â??â??â??â??â??â??â??â??â??â??â??551.406â??11000â??â??â??â??â??â??â??â??â??â??â??551.406â??11000â??â??â??â??â??â??â??â??â??â??â??
or, Q=34.76m3sec'>Q=34.76m3secQ=34.76m3secQ=34.76m3sec
Also, Conveyance,K=Aâ??cR=16.875'>K=Aâ??cRâ??â??â??=16.875K=Aâ??cRâ??â??â??=16.875K=Aâ??cRâ??â??â??=16.875â??'>â??â??â??55'>555555â??'>â??â??â??1.406'>1.406â??â??â??â??â??1.406â??â??â??â??â??1.406â??â??â??â??â??=1100.5'>=1100.5=1100.5=1100.5
Again, Froude's number, Fr=Vgâ??y=Câ??Râ??S09.81â??2.25=0.438'>Fr=Vgâ??yâ??â??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438Fr=Vgâ??yâ??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438Fr=Vgâ??yâ??â??â??â??â??=Câ??Râ??S0â??â??â??â??â??â??9.81â??2.25â??â??â??â??â??â??â??â??â??=0.438
Since froudes number,Fr'>FrFrFr is less than 1'>111, the flow in the channel is of tranquil in nature.