Answer:10.4 times of initial velocity
Explanation:
Given
Diameter reduced by 69 %
it approaches with velocity [tex]v_0[/tex]
suppose its velocity is v during blocked passage
suppose d is the initial diameter and [tex]d_2[/tex] diameter is
[tex]d_2=d(1-0.69)[/tex]
[tex]d_2=0.31 d[/tex]
[tex]A_2=\frac{\pi d_2^2}{4}[/tex]
As flow is constant
[tex]Q_1=Q_2[/tex]
[tex]d^2v_0=d_2^2v[/tex]
[tex]v=10.40 v_0[/tex]
