Radius of each circular opening,
$$
\begin{aligned}
r_1 & =1 \mathrm{~mm} \\
& =10^{-3} \mathrm{~m}
\end{aligned}
$$
$\therefore$ Cross-sectional area of each opening,
$$
A_1=\pi r_1^2=\pi \times 10^{-6} \mathrm{~m}^2
$$
Radius of the pipe,
$$
r_2=2 \mathrm{~cm}=0.02 \mathrm{~m}
$$
$\therefore$ Cross-sectional area of pipe,
$$
\begin{aligned}
A_2 & =\pi r_2^2 \\
& =\pi \times(0.02)^2 \mathrm{~m}^2
\end{aligned}
$$
The speed of water in the pipe,
$$
\begin{aligned}
v_2 & =25 \mathrm{~cm} / \mathrm{s} \\
& =0.25 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Let the speed of water when it exits from the circular openings is $v_1 \mathrm{~m} / \mathrm{s}$.
Number of circular openings, $n=25$
As, the flow rate is always constant.
$$
\begin{aligned}
\therefore \quad 25 A_1 v_1 & =A_2 v_2 \\
\Rightarrow \quad v_1 & =\frac{A_2}{25 A_1} v_2 \\
& =\frac{\pi \times(0.02)^2}{\pi \times 10^{-6}} \times \frac{0.25}{25}=4 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$