Step 1: Recall formula for discharge.
\[
Q = \frac{\Delta \cdot A}{t}
\]
where, $\Delta =$ depth of water (m), $A =$ area (m$^2$), $t =$ time (s).
Step 2: Convert values.
Area = $2600 \, ha = 2600 \times 10^4 = 2.6 \times 10^7 \, m^2$.
Depth = $17 \, cm = 0.17 \, m$.
Time = $30 \, days = 30 \times 24 \times 3600 = 2.592 \times 10^6 \, s$.
Step 3: Substitute in formula.
\[
Q = \frac{0.17 \times 2.6 \times 10^7}{2.592 \times 10^6}
\]
\[
Q = 1.71 \, m^3/s
\]
Step 4: Conclusion.
Thus, the required discharge capacity is $1.71 \, m^3/s$.
The solution(s) of the ordinary differential equation $y'' + y = 0$, is:
(A) $\cos x$
(B) $\sin x$
(C) $1 + \cos x$
(D) $1 + \sin x$
Choose the most appropriate answer from the options given below:
For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:
The value of $\iint_S \vec{F} \cdot \vec{N} \, ds$ where $\vec{F} = 2x^2y \hat{i} - y^2 \hat{j} + 4xz^2 \hat{k}$ and $S$ is the closed surface of the region in the first octant bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0, x = 2, y = 0, z = 0$, is:
The value of the integral $\displaystyle \oint_C \frac{z^3 - 6}{2z - i} \, dz$, where $C: |z| \leq 1$, is: