A U-tube open at both ends is partially filled with water (Figure (a)). Oil (𝜌 = 750 kg/m3) is then poured into the right arm and forms a column L = 6.96 cm high (Figure (b)).Three figures portray a U-shaped tube with vertical arms on the left and right of equal height.In figure a, the liquid inside the tube is labeled water, and the columns of liquid in each arm are the same height. The pressure outside the tube is labeled P0.In figure b, an additional liquid labeled oil has been added to the right arm. The column of oil sits atop the column of water in the right arm, and the upper surface of the oil is at a greater height than the surface of water in the left arm. The difference in height is labeled h. The height from the bottom of the column of oil to the top of this column is labeled L.Figure c depicts the same situation as figure b, except that air blowing over the top of the left arm affects the heights of the surfaces of liquids in each arm. The velocity of air over the top of the left arm is labeled vector v. A shield covers the top of the right arm. The height of the top of the column of water in the left arm is equal to the height of the top of the column of oil in the right arm. The height from the bottom of the column of oil to the top of this column is labeled L.(a) Determine the difference h in the heights of the two liquid surfaces. (The density of water is 1.00 ✕ 103 kg/m3.) Your response differs from the correct answer by more than 100%. cm(b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Figure (c)). Determine the speed of the air being blown across the left arm. Assume the density of air is 1.29 kg/m3. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

(a) To find the difference in the heights of the two liquid surfaces, we need to use the principle of hydrostatic pressure. The pressure at the bottom of the oil column is equal to the pressure at the same level in the water column.

The pressure at the bottom of the oil column is the pressure at the top of the oil column (which is atmospheric pressure, P0) plus the pressure due to the oil column, which is the oil density (ρ_oil) times the height of the oil column (L) times gravity (g). So, P_bottom = P0 + ρ_oil * g * L.

The pressure at the same level in the water column is the pressure at the top of the water column (which is also P0) plus the pressure due to the water column, which is the water density (ρ_water) times the height of the water column (h + L) times gravity (g). So, P_bottom = P0 + ρ_water * g * (h + L).

Setting these two expressions for P_bottom equal to each other and solving for h gives:

h = (ρ_oil * L - ρ_oil * h) / ρ_water

Substituting the given values for ρ_oil (750 kg/m^3), L (6.96 cm = 0.0696 m), and ρ_water (1.00 x 10^3 kg/m^3), and using g = 9.81 m/s^2, we find:

h = (750 kg/m^3 * 0.0696 m - 750 kg/m^3 * h) / (1.00 x 10^3 kg/m^3)

Solving this equation for h gives h = 0.0522 m, or 5.22 cm.

(b) To find the speed of the air being blown across the left arm, we need to use Bernoulli's equation, which states that the total pressure (static pressure plus dynamic pressure) is constant along a streamline.

The static pressure at the top of the left arm is the atmospheric pressure minus the pressure due to the water column, which is ρ_water * g * h. The dynamic pressure is 1/2 * ρ_air * v^2, where ρ_air is the density of the air and v is the speed of the air.

Setting the total pressure equal to the atmospheric pressure and solving for v gives:

v = sqrt((2 * (P0 - ρ_water * g * h)) / ρ_air)

Substituting the given values for ρ_water (1.00 x 10^3 kg/m^3), h (0.0522 m), and ρ_air (1.29 kg/m^3), and using g = 9.81 m/s^2, we find:

v = sqrt((2 * (P0 - 1.00 x 10^3 kg/m^3 * 9.81 m/s^2 * 0.0522 m)) / 1.29 kg/m^3)

This equation can be solved for v once the value of P0 (the atmospheric pressure) is known.