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A U-tube contains two liquids, X and Y, with densities hoX = 1000 ext kg/m3 and hoY = 800 ext kg/m3. Liquid Y has a height of 20 cm. If the pressures at the same horizontal level in the U-tube are equ

Physics

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Answer

b

16 cm\text{16 cm}

Step-by-step solution

3 steps

Part a

Explain why height h is less than 20 cm.

Liquid X is denser than liquid Y (ρX=1000 kg/m3\rho_X = 1000 \text{ kg/m}^3 and ρY=800 kg/m3\rho_Y = 800 \text{ kg/m}^3). For the pressures to be equal at the same horizontal level in the U-tube, the column of the denser liquid (X) must be shorter than the column of the less dense liquid (Y). Therefore, height hh must be less than 20 cm.

Part b

16 cm\text{16 cm}

Determine the height h.

  1. Identify the given values.

    Density of liquid X, ρX=1000 kg/m3\rho_X = 1000 \text{ kg/m}^3
    Density of liquid Y, ρY=800 kg/m3\rho_Y = 800 \text{ kg/m}^3
    Height of liquid Y, HY=20 cm=0.20 mH_Y = 20 \text{ cm} = 0.20 \text{ m}
    Height of liquid X, HX=hH_X = h

  2. Apply the principle of equal pressure at the same horizontal level.

    The pressure exerted by a liquid column is given by P=ρgHP = \rho g H. At the interface level, the pressure from both columns must be equal.

    PX=PYP_X = P_Y ρXgHX=ρYgHY\rho_X g H_X = \rho_Y g H_Y
  3. Solve for hh.

    Cancel gg from both sides:

    ρXHX=ρYHY\rho_X H_X = \rho_Y H_Y

    Substitute the known values:

    1000 kg/m3×h=800 kg/m3×0.20 m1000 \text{ kg/m}^3 \times h = 800 \text{ kg/m}^3 \times 0.20 \text{ m} 1000h=1601000 h = 160 h=1601000 mh = \frac{160}{1000} \text{ m} h=0.16 mh = 0.16 \text{ m}

    Convert the height to centimeters:

    h=0.16×100 cmh = 0.16 \times 100 \text{ cm} h=16 cmh = 16 \text{ cm}

    The height hh is 16 cm\boxed{\text{16 cm}}.

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