Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Hole 1 is the larger hole, and hole 2 is the smaller hole. The center of one of these holes is located 6 times as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole (1 or 2) is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole, rA/rB.
To determine which hole (1 or 2) is located nearest the surface of the water, we can consider the pressure at each hole.
The pressure at a particular depth in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Since the volume flow rate of the water coming out of both holes is the same, the velocities of the water at both holes will be equal. Therefore, we can equate the pressure at both holes:
P1 = P2
ρgh1 = ρgh2
Since both holes are in the same water tank, the density and gravity will be the same. So we can cancel out ρ and g from both sides of the equation:
h1 = h2
Given that the center of one hole is located 6 times as far beneath the surface of the water as the other, we can express this relationship as:
h1 = 6h2
Since h1 = h2, we can substitute this into the equation:
6h2 = h2
6 = 1
This is not possible, so it means that our assumption that both holes are at the same depth was incorrect. Therefore, the hole located nearest the surface of the water is hole 2.
For part (b), we need to calculate the ratio of the radius of the larger hole (hole 1) to the radius of the smaller hole (hole 2), denoted as rA/rB.
We know that the volume flow rate of the water coming out of both holes is the same. The volume flow rate through a hole is given by Q = Av, where Q is the volume flow rate, A is the cross-sectional area of the hole, and v is the velocity of the water.
Since the volume flow rate is the same for both holes, we can equate their expressions:
A1v1 = A2v2
The cross-sectional area of a hole is given by A = πr^2, where A is the cross-sectional area and r is the radius.
Applying this to both holes, we have:
πrA^2v1 = πrB^2v2
Since the volume flow rate velocities are equal, we can cancel them out:
rA^2 = rB^2
Taking the square root of both sides, we have:
rA = rB
Therefore, the ratio of the radius of the larger hole to the radius of the smaller hole is 1:1, or rA/rB = 1.