Sol.
1. Given Data:
- Area of hose, $A_1=30 \mathrm{~cm}^2=30 \times 10^{-4} \mathrm{~m}^2$
- Speed of water in hose, $v_1=50 \mathrm{~cm} / \mathrm{s}=0.5 \mathrm{~m} / \mathrm{s}$
- Number of perforations, $n=10$
- Area of each perforation, $a=15 \mathrm{~mm}^2=15 \times 10^{-6} \mathrm{~m}^2$
- Total area of perforations, $A_2=n \times a=10 \times 15 \times 10^{-6} \mathrm{~m}^2=1.5 \times 10^{-4} \mathrm{~m}^2$
2. Calculation: Using the Equation of Continuity:

$$
\begin{aligned}
& A_1 v_1=A_2 v_2 \\
& \left(30 \times 10^{-4}\right) \times(0.5)=\left(1.5 \times 10^{-4}\right) \times v_2 \\
& 15 \times 10^{-4}=1.5 \times 10^{-4} \times v_2 \\
& v_2=\frac{15 \times 10^{-4}}{1.5 \times 10^{-4}}
\end{aligned}
$$
$v_2=10 \mathrm{~m} / \mathrm{s}$