Respuesta :
Answer:[tex]\frac{r_1}{r_2}=1.565[/tex]
Explanation:
Given
two holes are made with different sizes
Hole 1 is large in size and hole 2 is small
If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.
Now for radius apply Bernoulli's theorem at hole 1 and 2
[tex]P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2[/tex]
[tex]P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2[/tex]
if hole 1 is h distance below water surface then [tex]h_2=6h[/tex]
and [tex]P_1=P_2=P_{atm}[/tex]
Also [tex]v_1=\sqrt{2gh}[/tex]
[tex]v_2=\sqrt{2g(6h)}[/tex]
and [tex]Q=A_1v_1=A_2v_2[/tex]
[tex]A=\pi r^2[/tex]
thus [tex]\dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}[/tex]
[tex]\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}[/tex]
[tex]\frac{r_1}{r_2}=1.565[/tex]
A) The hole located nearest the surface is; hole 2
B) rA/rB = 1.565
We are told that the center of one of the holes is located 6 times as far beneath the surface of the water as the other. Thus;
h2 = 6h1
- A) Since the volume flow rate of water coming out of both holes is the same, then it means that the small hole must be located below the large hole.
- B) Using Bernoulli's theorem at hole 1 and 2, we have;
P1 + ρgh1 = P_atm + ½ρ(v1)² - - - (eq 1)
P2 + ρg6h1 = P_atm + ½ρ(v2)² - - - (eq 2)
We are told that the top is open to the atmosphere and as such;
P1 = P2 = P_atm
Thus,equations 1 and 2 give us;
ρgh1 = ½ρ(v1)²
gh1 = (v1)²
v1 = √(2gh1)
Similarly;
v2 = √(12gh1)
Now, we know from continuity equation and conservation of energy that;
Q = A1v1 = A2v2
Now, A1 = π(r1)²
And A2 = π(r2)²
Thus;
π(r1)²v1 = π(r2)²v2
This reduces to;
(r1)²v1 = (r2)²v2
Plugging in the relevant values;
(r1/r2)² = v2/v1
(r1/r2)² = [√(12gh1)]/[√(2gh1)]
(r1/r2)² = √6
(r1/r2)² = 2.44948974
r1/r2 = √2.44948974
r1/r2 = 1.565
Read more at; https://brainly.com/question/14082066
