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.

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

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

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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

if hole 1 is h distance below water surface then

and

Also

and

thus

pe = weightxheight = 2000x1000=2,000,000joules