A small circular hole 6.00 mm in diameter is cut in the side of a large water tank. The top of the tank is open to the air. The water is escaping from the hole at a speed of 10 m/s. How far below the water surface is the hole

The water surface is 5.10 m below the hole.

Given data:

The diameter of circular hole is, d = 6.00 mm = 0.006 m.

The speed of water escaping through hole is, v = 10 m/s.

Here, we will apply the concept of kinematic equation of motion. As per the third kinematic equation of motion,

here,

u is the initial speed of water.

g is the gravitational acceleration.

h is the vertical distance of hole from the water surface..

Solving as,

Thus, we can conclude that the water surface is 5.10 m below the hole.

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5.097 m

Explanation:

This involves Bernoulli's equation, so we will use;

P + ρgh + ½ρv² = constant

Applying this equation to both points, we have;

P_i + ρ_water*g*h_i + ½*ρ_water*v_i² = P_f + ρ_water*g*h_f + ½*ρ_water*v_f²

Since open to air, P_i and P_f will cancel out.

Also, v_i = 0 and h_f = 0

ρ_water = 1 g/ml

v_f is given as 10 m/s

h_i is depth of the hole below the water surface.

Thus, plugging in values, we have;

1*9.81*h_i + 0 = 0 + ½*1*10²

9.81h_i = 50

h_i = 50/9.81

h_i = 5.097 m

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