A U-tube open to the atmosphere at both ends contains water of density 1000\text{ kg m}^{-… — JAMB UTME

A U-tube open to the atmosphere at both ends contains water of density $1000\text{ kg m}^{-3}$ . An immiscible liquid is poured into one arm of the tube until it forms a column of height $15.0\text{ cm}$ . If the interface between the two liquids lies $12.0\text{ cm}$ below the surface of the water in the opposite arm, what is the density of the liquid in $\text{kg m}^{-3}$ ?

Answer: 800 kg m^-3

Answer

The density of the immiscible liquid is

800\text{ kg m}^{-3}

At the level of the liquid interface, the hydrostatic pressure exerted by the

15.0\text{ cm}

liquid column must equal the hydrostatic pressure exerted by the

12.0\text{ cm}

water column above it. Equating

\rho_1 h_1 = \rho_2 h_2

yields

\rho_{\text{liquid}} = 1000 \times (12.0 / 15.0) = 800\text{ kg m}^{-3}

Step-by-Step Solution

Apply the principal of equal pressure at the same horizontal level in a continuous fluid at rest.

Pressure due to liquid column equals pressure due to water column above the interface line.

Hydrostatic pressure at depth

h

is given by

P = \rho g h

, and points at equal depth in a connected liquid body share identical pressure.

Set up the density-height ratio relationship:

\rho_{\text{liquid}} \cdot h_{\text{liquid}} = \rho_{\text{water}} \cdot h_{\text{water}}

\rho_{\text{liquid}} = \rho_{\text{water}} \times \frac{h_{\text{water}}}{h_{\text{liquid}}}

Acceleration due to gravity

g

cancels out from both sides of the pressure balance equation.

Substitute

h_{\text{liquid}} = 15.0\text{ cm}

h_{\text{water}} = 12.0\text{ cm}

, and

\rho_{\text{water}} = 1000\text{ kg m}^{-3}

\rho_{\text{liquid}} = 1000 \times \frac{12.0}{15.0} = 800\text{ kg m}^{-3}

Heights can remain in centimeters since their unit ratio is dimensionless.

Key Concept

Hydrostatic Pressure Balance in U-Tube Manometers

Estimated Time:1m 30s

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