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Mechanical Properties Of Fluids

Question

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Water rises to a height 4.8cm in a certain capillary tube. When the same tube is dipped in some another liquid it falls by 12.5cm. Compare the surface tension of water and liquid. Density of liquid is 9.6 gm/cc and angle of contact of liquid is 120°.

Solution

Given,

Rise in water = 4.8 cm
Fall in water when tube is dipped in some other liquid = 12.5 cm
Density of liquid = 9.6 gm/cc
Angle of contact of liquid = 120^o

The height to which the liquid rises or falls in the capillary tube is given by the relation,

h = \frac{2 T cosθ}{rpg}

The above relation can be simplified to

T = \frac{rgph}{2 \cos θ} . . . . (1)

Let T₁ and T₂ be the surface tension of water and liquid respectively.

By using the equation (1), we have

\frac{T_{1}}{T_{2}} = \frac{ρ_{1} h_{1}}{ρ_{2} h_{2}} \frac{{cosθ}_{2}}{{cosθ}_{1}}

Here, ρ_{1} = 1 gm / cc, h_{1} = 4.8 cm, θ_{1} = 0 ° ρ_{2} = 9.6 gm / cc, h_{2} = - 12.5 cm, θ_{2} = 120 ° ∴ \frac{T_{1}}{T_{2}} = \frac{1 \times 4.8}{9.6 \times (- 1.25)} \frac{\cos 120 °}{\cos 0 °} = \frac{1}{5}

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