Respuesta :
Answer:
Therefore the diameter of the hole is [tex]1.94 \times 10^{-3}[/tex] m.
Step-by-step explanation:
Bernoulli's equation,
[tex]P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2[/tex]
P₁ = P₂= atmospheric presser
[tex]\rho[/tex]= density
[tex]\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2[/tex] [since P₁ = P₂]
[tex]\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)[/tex]
[tex]\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2[/tex]
[tex]\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2[/tex]
[tex]\Rightarrow v^2_2- v^2_1=2g h[/tex] [[tex]h_1-h_2=h[/tex]]
Here [tex]v_1\approx 0[/tex]
[tex]\Rightarrow v^2_2=2g h[/tex]
[tex]\therefore v_2=\sqrt {2gh[/tex]
Here g= 9.8 m/s² , h = 10.4 m
The velocity of water that leaves from the hole [tex]v_2[/tex] = [tex]\sqrt {2\times 9.8\times 10.4}[/tex] m/s
=14.28 m/s.
Given, the rate of flow from the leak is [tex]2.53\times 10^{-3}[/tex] [tex]m^3/min[/tex]
[tex]=\frac{2.53\times 10^{-3}}{60}[/tex] [tex]m^3/s[/tex]
Let the diameter of the hole be d.
Then the cross section area of the hole is [tex]=\pi (\frac d2)^2[/tex]
We know that,
The rate of flow = Cross section area × speed
[tex]\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28[/tex]
[tex]\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}[/tex]
[tex]\Rightarrow d= 1.94 \times 10^{-3}[/tex]
Therefore the diameter of the hole is [tex]1.94 \times 10^{-3}[/tex] m.
