Given,
The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m
The initial speed of the water, v₁=12.5 m/s
The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m
From the continuity equation,
[tex]\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}[/tex]Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.
On substituting the known values,
[tex]\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}[/tex]Thus, the flow rate of the water at the later position is 5.99 m/s