Question

The ordinates of 2-h UH for a basin is as follows

$\begin{array}{ccccccccc}\text{Time (hr)}& 0& 2& 4& 6& 8& 10& 12& 14\\ \text{2-h UH}& 0& 5& 18& 28& 36& 42& 32& 16\\ \left(m^3/sec\right)& & & & & & & & \end{array}$

The ordinate of 4-h UH derived from above UH at 10th hour_____ (in $\frac{m^3}{sec}$) is

• A. 72
• B. 78
• C. 39
• D. 32

Answer 1

The correct option is C 39

$\begin{array}{ccccc}\text{Time}& \text{Ordinate of}& \text{Ordinate of}& \text{Combined}& \text{Ordinate of}\\ & \text{2-h UH}& \text{2-h UH}& \text{Hydrograph due to}& \text{4-h UH}\\ & \text{(col.1)}& \text{lagged by 2h}& \text{2 cm eff. rainfall}& \text{(col.3/2)}\\ & & \text{(col.2)}& \text{(col.3)}& \\ 0& 0& -& 0& 0\\ 2& 5& 0& 5& 2.5\\ 4& 18& 5& 23& 11.5\\ 6& 28& 18& 46& 23\\ 8& 36& 28& 64& 32\\ 10& 42& 36& 78& \text{39}\\ 12& 32& 42& 74& 37\\ 14& 16& 32& 48& 24\\ 16& -& 16& 16& 8\end{array}$

Ordinate of 4-h UH at 10th hr
$=39\frac{m^3}{sec}$