Question

A capillary tube of radius 0.25mm is submerged vertically in water so that 25 mm of its length is outside wayer the radius of curvature of the meniscus will be

Answer 1

Given:

The radius of the capillary tube = 0.25 mm

The capillary tube is submerged vertically so that 25 mm of its length is outside the water

To find:

The radius of curvature of the meniscus

Solution:

We will first find the height to which the water will rise in the capillary tube and for that the following formula is used:

$\boxed{h\:=\:\frac{2\:T\:cos\:\theta}{\rho\:g\:r} }$

where

h = rise in the height of the liquid

T = surface tension of water = 72 dynes/cm = 0.072 N/m

θ = angle of contact of the liquid with the capillary tube = 0°

ρ = density of liquid = 10³ kg/m³

g = acceleration due to gravity = 10 m/s²

r = radius of capillary tube = 0.25 mm = 0.00025 m

∴ $h = \frac{2\:\times\:0.072\:\times\:cos\:0}{10^3\:\times\:10\:\times\:0.00025} = \frac{2\:\times\:0.072\:\times\:1}{10^3\:\times\:10\:\times\:0.00025} = \frac{0.144}{2.5} = 0.0576\: m = 57.6 \:mm$

So, we can see that the water will rise up to a height 57.6 mm in the capillary tube which is greater than the given height of 25 mm of the tube, therefore the meniscus will be formed just at the top of the capillary tube.

Let "R" be the radius of curvature of the meniscus.

To find the value of R, we will use the formula as,

$R\:=\:\frac{2\:T}{\rho\:g\:h}$

substituting the values of ρ = 10³ kg/m³, g = 10 m/s², h = 25 mm = 0.025 m & Τ = 0.072 N/m

⇒ $R\:=\:\frac{2\:\times\:0.072}{10^3\:\times\:10\:\times\:0.025}$

⇒ $R\:=\:\frac{0.144}{250}$

⇒ R = 0.000576 m

∵ 1 m = 1000 mm

⇒ R = 0.576 mm ≈ 0.6 mm ....(rounded to the nearest tenth)

Thus, the radius of curvature of the meniscus will be 0.6 mm.

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