The capillaries displayed in the figure possess inner radii of $0.5 \text{ mm}$ , $1.0 \text{ mm}$ and $1.5 \text{ mm}$ respectively. The liquid contained within the beaker is water. Determine the heights of the water level in the capillaries. The surface tension of water is $7.5 \times 10^{-2} \text{ N/m}$ .

Question

The capillaries displayed in the figure possess inner radii of

0.5 \text{ mm}

,

1.0 \text{ mm}

and

1.5 \text{ mm}

respectively. The liquid contained within the beaker is water. Determine the heights of the water level in the capillaries. The surface tension of water is

7.5 \times 10^{-2} \text{ N/m}

.

The capillaries displayed in the figure possess inner radii of 0.5 mm , 1.0 mm and 1.5 mm respectively. The liquid

Quantrex Academy · Accepted Answer

Correct answer: A. 3 cm in A , 1.5 cm in B , 1 cm in C

The height of liquid in a capillary tube is given by Jurin's law:

h = \dfrac{2T \cos \theta}{\rho g r}

For water and glass, the angle of contact

\theta \approx 0^{\circ}

, so

\cos \theta = 1

.
Density of water,

\rho = 1000 \text{ kg/m}^3

.
Acceleration due to gravity,

g = 10 \text{ m/s}^2

.
Surface tension,

T = 7.5 \times 10^{-2} \text{ N/m}

.

For capillary A,

r_A = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}

:

h_A = \dfrac{2 \times 7.5 \times 10^{-2} \times 1}{1000 \times 10 \times 0.5 \times 10^{-3}} = \dfrac{15 \times 10^{-2}}{5} = 3 \times 10^{-2} \text{ m} = 3 \text{ cm}

For capillary B,

r_B = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}

:

h_B = \dfrac{2 \times 7.5 \times 10^{-2} \times 1}{1000 \times 10 \times 1.0 \times 10^{-3}} = \dfrac{15 \times 10^{-2}}{10} = 1.5 \times 10^{-2} \text{ m} = 1.5 \text{ cm}

For capillary C,

r_C = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}

:

h_C = \dfrac{2 \times 7.5 \times 10^{-2} \times 1}{1000 \times 10 \times 1.5 \times 10^{-3}} = \dfrac{15 \times 10^{-2}}{15} = 1 \times 10^{-2} \text{ m} = 1 \text{ cm}

The capillaries displayed in the figure possess inner radii of 0.5...

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