Given,
Initial speed is v=2 m/s
The cross sectional area reduces to 1/4 of the original area
To find
The speed of water at the narrowed portion.
Explanation
Let the original area be A
Let the speed at the narrowed portion be u
By conservation lawwe have,
[tex]\begin{gathered} v\times A=u\times\frac{1}{4}A \\ \Rightarrow2\times4=u \\ \Rightarrow u=8\text{ m/s} \end{gathered}[/tex]Conclusion
The speed at the narrowed portion is 8 m/s