Given,

$Q = 50 cumecs$

$S = {1 \over 4000}$

$m = 1.1$

$n = 0.023$

$As$ $we$ $know$ $the$ $ equation$ $ for$ $ critical$ $ velocity$ $ is, $

$V_0 = 0.55 $ $my^{0.64}$

$Assume$ $a$ $depth$ $equal$ $ to$ $2m,$

$V = 0.55 * 1.1 * 2^{0.64} = 0.605 * 1.558 = 0.942 m/sec$

$A = {Q \over V_0} = {50 \over 0.942} = 53.1 m^2$

$Assume$ $side $ $slopes$ $as $ ${1 \over 2} : 1\left[\begin{array}{lll}{1 \over 2}H = 1V\end{array}\right]$

$Now, $

$A = y\left[\begin{array}{lll}b + y*{1 \over 2}\end{array}\right]$

$Therefore, 53.1 = 2(b+1)$

$or, $ $b = 25.55m$

$And,$

$P = b + 2 \sqrt {1 + {1 \over 4}}*y$

$P = b + 2{\sqrt {5} \over 3}y = 25.55 + \sqrt 5*2 = 30.03$

$R = {A \over P} = {53.1 \over 30.03} = 1.77m$

$But\ from\ Kutter's\ formula, $

$V = y\left[\begin{array}{lll}{{1 \over n} + (23 + {0.00155 \over S}) \over 1 + (23 + {0.00155\over S} {n \over {\sqrt R}})} \sqrt {RS} \end{array}\right]$

$V = y\left[\begin{array}{lll}{{1 \over 0.023} + (23 + {0.00155 \over 1/4000}) \over 1 + (23 + {0.00155\over 1/4000} {0.023 \over {\sqrt 1.77}})} \sqrt {1.77 * {1 \over 4000}} \end{array}\right]$

$V = y\left[\begin{array}{lll}{43.5 + (23 + 6.2) \over 1 + ({29.2 * 0.023)\over 1.33}} \sqrt {1.33 * {1 \over 63.3}} \end{array}\right]$

${72.7 \over 1+0.505} * 1.33 * {1 \over 63.3} = 1.016\ m/sec\ \gt\ 0.942$

$In\ order\ to\ increase\ the\ critical\ velocity\ (V_0 ),\ we\ have\ to\ increase\ the\ depth$

Use 3m depth:

$V = 0.605 * (3)^{0.64} = 0.605 * 2.02 = 1.22 m/sec$

$A = {50 \over 1.22} = 40.8 m^2$

$40.8 = 3(b + {1 \over 2}*3)$

$or,$ $b = 12.1 m$

$P = 12.1 + 2{\sqrt {5} \over 2}3 = 12.1 + 6.72 = 18.82$

$R = {A \over P} = {40.8 \over 18.82} = 2.17m$

$After$ $calculating$ $the$ $V$ $we get, $ $V = 1.16$ $m/sec$ $\lt$ $1.22$ $or$ $V \lt V_0$

Use 2.5m depth:

$V = 0.605 * 2.05^{0.64} = 1.087 m/sec$

$A = {50 \over 1.087} = 46 m^2$

$46 = 2.5(b + {1 \over 2}*2.5)$

$or,$ $b = 17.15 m$

$P = 17.15 + \sqrt {5} * 2.5 = 17.15 + 5.58 = 22.73$

$R = {4 \over 22.73} = 2.02m$

$After$ $calculating$ $the$ $V$ $we get, $ $V = 1.1$ $m/sec$ $\gt$ $1.087$ $or$ $V \gt V_0$

$Actual$ $velocity$ $V$ $tallies$ $with$ $V_0.$

Use 2.7m depth:

$V = 0.605 * 1.189 = 1.147 m/sec$

$A = {50 \over 1.147} = 43.5 m^2$

$43.5 = 2.8(b + {1 \over 2}*2.8)$

$or,$ $b = 14.14 m$

$P = 14.14 + \sqrt {5} * 2.8 = 14.14 + 6.26 = 20.40$

$R = {43.5 \over 20.4} = 2.13m$

$After$ $calculating$ $the$ $V$ $we get, $ $V = 1.148$ $m/sec$ $=$ $1.147$ $or$ $V = V_0$

$Actual$ $velocity$ $V$ $tallies$ $with$ $V_0.$

$Hence,$ $use$ $the$ $depth$ $equal$ $to$ $2.7m$ $and$ $base$ $width$ $14.14m$ $with$ $slopes$ ${1 \over 2}: 1$ $of$ $trapezoidal$ $section.$