Question

A container contains two immiscible liquids of density 1 and 2 ( 2 > 1 ) . A capillary of radius r is inserted in the liquid so that its bottom reaches up to denser liquid. Denser liquid rises in capillary and attains height equal to h which is also equal to column length of lighter liquid. Assuming zero contact angle find the surface tension of the heavier liquid.

Answer 1

Answer:

Surface tension of the liquid is given as

$S = \frac{rgh(\rho_2 - \rho_1)}{2}$

Explanation:

Pressure inside the liquid at open end is given as

$P_o - P = \frac{2S}{r}$

so we have

$P = P_o - \frac{2S}{r}$

now we have

$P + \rho_2 g h = P_o + \rho_1 g h$

so we have

$P_o - \frac{2S}{r} + \rho_2 gh = P_o + \rho_1 gh$

now we have

$(\rho_2 - \rho_1)gh= \frac{2S}{r}$

now the surface tension of liquid is given as

$S = \frac{rgh(\rho_2 - \rho_1)}{2}$

#Learn

Topic : Surface Tension

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