Correct Answer - Option 4 : H

^{1.5}
__Concept:__

Discharge over an Ogee spillway is given by:

\(Q = \;\frac{2}{3}{C_d}\sqrt {2g} LH_e^{\frac{3}{2}}\)

He = Total head upstream of the crest.

L = crest width

Cd = coefficient of discharge

C = constant, which is equal to \(\frac{2}{3}{C_d}\sqrt {2g}\)<!--[if gte msEquation 12]>^{ }^{H}^{3}^{2}<![endif]--><!--[if !msEquation]-->

**∴ Discharge Q is proportional to H**^{1.5}<!--[endif]-->