Question

A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary with the wall.

Answer 1

Answer:If the tube is immersed in a liquid which does not wet the glass (mercury), then the liquid level inside the tube decreases. This phenomenon of the rise or fall of liquid in a capillary tube is called capillary action or capillarity. ... The liquid will rise in the capillary tube. The surface of the liquid will be concave.

Answer 2

(a). The height of water raised in the capillary is 1.55 cm.

(b). The angle made by the water surface in the capillary with the wall is 60°.

Explanation:

Given that,

Radius = 1 mm

Let T be the surface tension of the liquid.

Then, for cos θ = 1, height of liquid level

(a). We need to calculate the height of water raised in the capillary

Using formula of height

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

$\cos\theta=1$

Where, T = surface tension

r = radius

[tezx]\rho[/tex] = density of water

g = acceleration due to gravity

Put the value into the formula

$h=\dfrac{2\times0.076}{10^{-3}\times10^{3}\times9.8}$

$h=1.55\ cm$

(b). If the length of the capillary tube is half the answer of part (a),

We need to calculate the angle made by the water surface in the capillary with the wall

Using formula of height

$h'=\dfrac{2T\cos}{r\rho g}$

$\cos\theta=\dfrac{h'r\rho g}{2T}$

Put the value into the formula

$\cos\theta=\dfrac{\dfrac{1.55\times10^{-2}}{2}\times10^{-3}\times1000\times9.8}{2\times0.076}$

$\cos\theta=0.4996$

$\theta=\cos^{-1}(0.4996)$

$\theta=60^{\circ}$

Hence, (a). The height of water raised in the capillary is 1.55 cm.

(b). The angle made by the water surface in the capillary with the wall is 60°.

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Topic : Height of capillary tube

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