Question

A capillary tube of radius 5 x 10-4m is immersed in a beaker filled with mercury. The mercury level inside the tube is found to be 8 x 10-3m below the level of reservoir. Determine the angle of contact between mercury & glass. Surface tension of mercury is 0.465 N/m. & its density is 13.6 x 10 kg/m' (g 9.8 m/s2)​

Answer 1

Answer:

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Answer 2

Step-by-step Explanation

Given: Radius of the capillary tube ($r$) = $5 \times 10^{-4}$ m

Height of the mercury level $(h)$ = $8 \times 10^{-3}$ m

Surface Tension of mercury $(S)$ = $0.465 \ Nm^{-1}$

The density of mercury $\rho$ = $13.6 \ kg \ m^{-3}$

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

To Find: The angle of contact between mercury and glass ($\theta$)

Solution:

• Formula to find the angle of contact

The expression to find the angle of contact is the following:

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

Since mercury does not wet the sides of the capillary tube, the meniscus will be convex upwards. Therefore, the angle of contact $(\theta)$ will be obtuse, and the height of the mercury level $(h)$ will be negative.

• Calculating the angle of contact

Substituting the given values in the above expression, we get;

$\Rigtharrow cos(\theta) = \frac{(-8 \times 10^{-3}) \times 5 \times 10^{-4}) \times 13.6 \times 10 \times 9.8}{2 \times 0.465} \\\\\Rigtharrow cos(\theta) = -0.005735\\\\\Rigtharrow \theta = cos^{-1} (-0.005735) = 90.3284^{o}$

Hence, the angle of contact between mercury and glass is $90.3284^{o}$