Question

Find the height to which water at 4°C will rise in a capillary tube of 10-3 m diameter. Take g = 9.8 ms-1 . Angle of contact, θ = 0° and T = 0.072 Nm-1 .​

Answer 1

Answer

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Answer 2

Step-by-step Explanation

Given: Diameter (D) of the capillary tube = $10^{-3}$ m

Surface Tension (T) = $0.072 \ Nm^{-1}$

The angle of contact ($\theta$) = $0^{o}$

Acceleration due to gravity (g) = $9.8 \ ms^{-2}$

The temperature of water = $4^{o} \ C$

To Find: The height (h) of rising of water in the capillary tube

Solution:

• Formula to find the height of the rise

The following expression is used to find the height (h) of rising of water in the capillary tube;

$h = \frac{2 T cos(\theta)}{r \rho g}$

Where, $\rho$ is the density of water, and $r$ is the radius of the capillary tube.

• Calculating the height of the rise in the capillary tube

Since the diameter of the capillary tube is $D = 10^{-3} m$, the radius of the tube will be;

$r = \frac{D}{2} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} m$

And, at $4^{o}C$, the density of water is $\rho = 10^{3} \ kgm^{-3}$

Substituting all the required values in the above formula, we get;

$\Rightarrow h = \frac{2 \times 0.072 \times cos(0^{o} )}{ 0.5 \times 10^{-3} \times 10^{3} \times 9.8}\\\\\Rightarrow h = \frac{2 \times 0.072}{ 0.5 \times 9.8}\\\\\Rightarrow h = 0.0294 \ m$

Hence, the water will rise in a capillary tube to a height of $h = 0.0294 \ m$