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.
Step-by-step explanation
Let the original area be A
Let the speed at the narrowed portion be u
By conservation lawwe have,
![\begin{gathered} v* A=u*(1)/(4)A \\ \Rightarrow2*4=u \\ \Rightarrow u=8\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/2ykwedogde1odcacqjrbocvijf6e2tciiu.png)
Conclusion
The speed at the narrowed portion is 8 m/s