Correct Answer - Option 1 : 260
Concept:
Discharge over a Triangular notch or V-Notch is given by:
\(Q = \frac{8}{{15}}{C_d}\sqrt {2g} \tan \frac{θ }{2}{H^{\frac{5}{2}}}\)
Where,
Q = Discharge, H = Height of water surface
Cd = Coefficient of discharge
Calculation:
Given,
θ = 90°, Cd = 0.58, H = 0.10 m
∵ \(Q = \frac{8}{{15}}{C_d}\sqrt {2g} \tan \frac{θ }{2}{H^{\frac{5}{2}}}\)
⇒ \(Q = \frac{8}{{15}} × 0.58 × \sqrt {2 × 9.81} × \tan \left( {\frac{{{{90}^o}}}{2}} \right) × {0.10^{5/2}}\)
Q = 4.332 × 10-3 m3/sec
Q = 4.332 × 10-3 × 103 × 60 = 259.92 liter/min
Q ≈ 260 liter/min
