Answer:
[tex]v_2=3.2\ m/s[/tex]
Explanation:
given,
diameter of inlet = 11.2 cm
r₁ = 5.6 cm
r₁ = 0.056 m ∵ 1 cm = 0.01 m
speed of inlet = 5 m/s
diameter of outlet = 17.5 cm
r₂ = 8.75 cm
r₂ = 0.0875 m ∵ 1 cm = 0.01 m
speed of outlet = ?
using continuity equation
A₁ v₁ = A₂ v₂
π r₁² v₁ = π r₂² v₂
[tex]v_2= \dfrac{r_1^2}{r_2^2} v_1[/tex]
[tex]v_2= \dfrac{0.056^2}{0.0875^2} v_1[/tex]
[tex]v_2= 0.64 v_1[/tex]
[tex]v_2= 0.64\times 5[/tex]
[tex]v_2=3.2\ m/s[/tex]
Velocity of liquid at the outlet of the tube is equal to 3.2 m/s
