A U-shaped glass tube is open to the atmosphere at both ends. The left arm of the tube has… — Armed Services Vocational Aptitude Battery (ASVAB)

A U-shaped glass tube is open to the atmosphere at both ends. The left arm of the tube has a cross-sectional area of $10\text{ cm}^2$ , and the right arm has a cross-sectional area of $5\text{ cm}^2$ . The tube initially contains water with a density of $1.0\text{ g/cm}^3$ . A student carefully pours a column of oil measuring $20\text{ cm}$ in length, with a density of $0.8\text{ g/cm}^3$ , into the narrower right arm. Assuming the oil-water interface remains in the vertical section of the right arm, what is the vertical height difference between the top surface of the oil and the top surface of the water once the fluids come to rest?

A
$16\text{ cm}$
B
$0\text{ cm}$
$4\text{ cm}$ Answer
D
$8\text{ cm}$

Answer

4\text{ cm}

Hydrostatic pressure is determined solely by fluid density and depth (

P = \rho g h

), entirely independent of the container's cross-sectional area. Equating the pressures at the fluid interface gives a water column height of

16\text{ cm}

. Subtracting this from the

20\text{ cm}

oil column reveals that the oil surface rests

4\text{ cm}

higher than the water surface.

Step-by-Step Solution

Identify the reference point of equal pressure.

The horizontal plane at the oil-water interface in the right arm represents the point where pressure from both sides must be equal.

In a static fluid system, pressure at any given horizontal depth within the same continuous fluid (water) is uniform.

Set up the hydrostatic pressure equation.

\rho_{\text{water}} \times g \times h_{\text{water}} = \rho_{\text{oil}} \times g \times h_{\text{oil}}

. The cross-sectional areas (

10\text{ cm}^2

and

5\text{ cm}^2

) are irrelevant and ignored.

Hydrostatic pressure (

P = \rho g h

) depends only on fluid density and depth, not on the container's shape or the total volume of fluid.

Calculate the required height of the water column.

1.0\text{ g/cm}^3 \times h_{\text{water}} = 0.8\text{ g/cm}^3 \times 20\text{ cm}

, which simplifies to

h_{\text{water}} = 16\text{ cm}

This establishes how high the water surface in the left arm must rise above the interface to balance the downward pressure of the oil.

Calculate the difference in surface heights.

20\text{ cm} - 16\text{ cm} = 4\text{ cm}

The question asks for the vertical difference between the top of the oil and the top of the water, not just the absolute height of the water column.

Key Concept

Hydrostatic pressure is independent of container shape or cross-sectional area, depending solely on fluid density and vertical column height.

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