Question

Calculate (a) the maximum capillary rise of water between two vertical glassplates spaced 0.15 mm apart and (b) the minimum capillary drop if the sameplates, now spaced 0.20 mm apart, are in mercury. Assume room temperature.

Answer 1

Answer:

The the maximum capillary rise of water between two vertical glass plates is 9.79 cm

The the maximum capillary rise of mercury between two vertical glass plates is -2.77 cm

Explanation:

Given that,

Distance = 0.15 mm

Suppose, The surface tension of water is 72 dynes/cm at 25°C  for pure water and perfectly clean glass, the angle of contact is 0°.

(a). We need to calculate the maximum capillary rise of water between two vertical glass plates

Using formula of rise of liquid in capillary tube

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

Put the value into the formula

$h=\dfrac{2\times72\cos0}{0.015\times1\times980}$

$h=9.79\ cm$

(b). If the same plates, now spaced 0.20 mm apart, are in mercury

The density of mercury is 13.6 g/cm³, the angle of contact of mercury with glass is 139°, and the surface tension of mercury is 490 dyne/cm and g = 980 cm/s².

We need to calculate the minimum capillary drop

$h=\dfrac{2\times490\cos(139)}{0.02\times13.6\times980}$

$h=-2.77\ cm$

Hence, The the maximum capillary rise of water between two vertical glass plates is 9.79 cm

The the maximum capillary rise of mercury between two vertical glass plates is -2.77 cm