Question

By how much depth will the surface of a liquid be depressed in a glass tube of radius \(0.2 \mathrm{~mm}\) if the angle of contact of the liquid is \(135^{\circ}\) and the surface tension is \(0.547 \mathrm{Nm}^{-1}\) ? (Density of the liquid is \(1.28 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}\) ) (A) \(3 \mathrm{~cm}\) (B) \(4 \mathrm{~cm}\) (C) \(5 \mathrm{~cm}\) (D) \(2 \mathrm{~cm}\)

Short answer

The depth of the liquid surface depression in the glass tube is approximately 2 cm. The correct answer is (D).

Step-by-step solution

Convert given values into SI units

First, we need to convert the given radius into meters and the surface tension into newtons per meter. Radius: \(0.2 \mathrm{~mm} = 0.2 \times 10^{-3} \mathrm{~m}\). The surface tension is already in the right unit.

Plug in the given values into Jurin's Law equation

Using - s = 0.547 Nm⁻¹ - θ = 135° - ρ = 1.28 x 10⁴ kg/m³ - g = 9.81 m/s² - r = 0.2 x 10⁻³ m We write the Jurin's Law equation with these values: \(h = \frac{2(0.547)\cos135^{\circ}}{(1.28 \times 10^{4})(9.81)(0.2 \times 10^{-3})}\)

Evaluate the depression depth (h)

To evaluate the depression depth (h), we first find the value of cos(135°): \(\cos135^{\circ} = -\frac{\sqrt{2}}{2}\) Now, substitute this value back into the expression for h: \(h = \frac{2(0.547)(-\frac{\sqrt{2}}{2})}{(1.28 \times 10^{4})(9.81)(0.2 \times 10^{-3})}\) Solve for h: \(h \approx -0.02 \mathrm{~m}\)

Convert h to a positive value and into centimeters

Since we are looking for the depression depth, we should take the absolute value of h: |h| = 0.02 m Now, convert h into centimeters: |h| = 0.02 m * (100 cm/m) = 2 cm. So, the depth of the liquid surface depression in the glass tube is approximately 2 cm. The correct answer is (D).

Key concepts

Surface Tension

Surface tension is a fascinating phenomenon often seen as the "skin" on the surface of liquids. It is caused by the cohesive forces between liquid molecules, which pull each other together. Imagine tiny invisible springs at the surface, trying to minimize the surface area.

• This force is what allows small objects, like a paperclip, to float on water if placed carefully.
• Measured in Newtons per meter (N/m), it indicates the energy required to increase the surface area of the liquid by one square meter.

In the given problem, surface tension is what causes the liquid in the tube to either rise or fall, depending on other factors like the angle of contact and the liquid's cohesive properties.

Jurin's Law

Jurin's Law helps us understand how liquids move in capillary tubes based on various factors. It describes the height to which a liquid will climb or fall within a tube due to surface tension. According to Jurin's Law:
\[ h = \frac{2s \cos \theta}{\rho g r}\]

• \(h\) is the height the liquid rises or falls.
• \(s\) represents surface tension.
• \(\theta\) is the angle of contact.
• \(\rho\) is the density of the liquid.
• \(g\) stands for gravitational acceleration, approximately \(9.81 \text{ m/s}^2\).
• \(r\) is the radius of the tube.

In the exercise, Jurin's Law is applied to calculate how much the liquid's surface descends due to an angle of contact greater than 90 degrees, indicating depression rather than ascent.

Angle of Contact

The angle of contact is the angle formed between the tangent to the liquid surface and the solid surface at the point of contact. This angle can affect whether a liquid wets a surface or forms droplets.

• An angle less than 90 degrees means the liquid wets the surface, leading to ascent in capillarity experiments.
• If the angle is greater than 90 degrees, as in the exercise (135°), it suggests that the liquid does not wet the surface, leading to depression in a capillary tube.
• This angle is crucial for understanding the behavior of liquids on different surfaces.

For this problem, knowing the angle of contact as 135° helps us predict that the liquid will be depressed in the tube, not elevated.

Density of Liquids

Density influences how a liquid interacts with gravity and affects capillary action. Defined as mass per unit volume, it is measured in kilograms per cubic meter (kg/m³). Higher density means a heavier liquid per unit of volume.

• The liquid's density influences how much force is needed to counteract gravity when subjected to surface tension.
• In the problem, the liquid has a density of \(1.28 \times 10^4\) kg/m³, a crucial factor in determining how much the surface tension can depress the liquid in the tube.
• Coupled with surface tension and gravitational force, it helps predict the movement of the liquid in a confined space.

The interplay between density and these other forces determines if a liquid will rise or fall, playing a significant role in applications where fluid movement is critical.