/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Two vertical parallel glass plat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two vertical parallel glass plates are partially submerged in water. The distance between the plates is \(d\) and the length is \(l\). Assume that the water between the plates does not reach the upper edges of the plates and that the wetting is complete. The water will rise to height \((\rho=\) density of water and \(\sigma=\) surface tension of water \()\) (1) \(\frac{2 \sigma}{\rho g d}\) (2) \(\frac{\sigma}{2 \rho g d}\) (3) \(\frac{4 \sigma}{\rho g d}\) (4) \(\frac{5 \sigma}{\rho g d}\)

Short Answer

Expert verified
The correct height is given by option (3): \( \frac{4 \sigma}{\rho g d} \).

Step by step solution

01

Understanding Three Key Effects

When water rises between two close, vertical, parallel glass plates, it's mainly due to capillary action, surface tension, and gravitational pull. The water surface forms a concave meniscus due to alignment with the glass surfaces, and due to complete wetting conditions, water attempts to rise against gravity.
02

Applying Capillary Rise Formula

The height to which liquid rises, denoted as \( h \), can be determined using the formula: \[ h = \frac{2\sigma \cos\theta}{\rho g r} \] where \( \sigma \) is the surface tension, \( \theta \) is the contact angle (which is 0 here for complete wetting), \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( r \) is the radius of the capillary formed by the distance \( d \) between the plates, which is \( \frac{d}{2} \).
03

Simplifying the Derivation for Parallel Plates

Substituting \( \theta = 0 \) (since \( \cos 0 = 1 \)) and \( r = \frac{d}{2} \), the formula for height simplifies to: \[ h = \frac{2 \sigma}{\rho g \cdot \frac{d}{2}} = \frac{4 \sigma}{\rho g d} \].
04

Identifying the Correct Answer

Comparing the derived formula with the given options, the correct expression for the rise of the water between the plates is \( \frac{4 \sigma}{\rho g d} \), which corresponds to the option (3).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Tension
Surface tension is a fascinating property of liquids that allows them to resist an external force. It occurs because of the cohesive forces between liquid molecules. This tension acts like a stretched elastic sheet over the liquid's surface. In the context of water between two glass plates, surface tension causes the water to form a meniscus, where the water level is higher at the edges, owing to the attraction between the water molecules and the glass.

It's important to note that surface tension is responsible for minimizing the surface area of a liquid. For water, this is described by the surface tension energy per unit area, denoted as \( \sigma \). This property enables water to climb up the glass plates, battling gravitational force, and causing it to rise to a certain height. The strong adhesive force between the water molecules and glass compared to the cohesive forces within the water itself is another reason the water rises, as seen in the capillary action.
Capillary Rise
Capillary rise is a phenomenon where a liquid in a thin tube or between narrow spaces climbs higher than its surrounding level. This occurs even without external aid, defying gravity, and mainly results from the liquid's surface tension and the adhesive forces between the liquid and the solid surface.

In the example of water between two glass plates, the height \( h \) to which the water rises can be calculated using the formula:
  • \[ h = \frac{2\sigma \cos\theta}{\rho g r} \]
Where \( \sigma \) is the surface tension, \( \theta \) is the contact angle between the liquid and the plate, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( r \) is the effective radius of the capillary space, measured as \( \frac{d}{2} \) for plates separated by distance \( d \).

With complete wetting, the contact angle \( \theta \) becomes 0, simplifying the equation, as \( \cos(0) = 1 \), further helping us understand how capillary rise challenges gravity to allow the liquid to ascend.
Density of Water
The density of water, represented by \( \rho \), plays a critical role in the phenomenon of capillary rise. Density is a measure of how much mass is in a given volume of a substance. For water, this value is usually around \( 1000 \, \text{kg/m}^3 \) at standard temperature and pressure.

In the capillary rise formula, density appears in the denominator, \( \frac{2\sigma}{\rho g r} \). This indicates that the higher the density of the liquid, the more it resists being pulled up through capillary action, as a greater gravitational force acts upon it.

Therefore, understanding the density of water helps us appreciate why it has a particular height in capillary action scenarios. Lowering the density would make it easier for water or any liquid to climb higher between surfaces, while increasing it would have the opposite effect.
Gravitational Force
Gravitational force is the attraction that pulls objects towards one another, most notably dragging them towards the Earth's center. It acts as a counterforce to surface tension in capillary action. In the context of the water between two plates, it opposes the liquid ascending due to capillary action.

The part that gravity plays in capillary rise is represented by \( g \), the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \). In the capillary rise equation, gravity affects how high the water can rise:
  • \[ h = \frac{2\sigma \cos\theta}{\rho g r} \]
Since gravity appears in the denominator, the stronger the gravitational force, the more it hinders the height that the liquid can achieve.

Understanding the balance between gravitational force and other factors like surface tension and density helps illustrate why, under specific conditions, water can climb a certain distance between plates, displaying the captivating action of capillarity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two glass plates are separated by water. If surface tension of water is 75 dyn/cm and the area of each plate wetted by water is \(8 \mathrm{~cm}^{2}\) and the distance between the plates is \(0.12 \mathrm{~mm}\), then the force applied to separate the two plates is (1) \(10^{2}\) dyn (2) \(10^{4} \mathrm{dyn}\) (3) \(10^{5} \mathrm{dyn}\) (4) \(10^{6} \mathrm{dyn}\)

Water rises to a height of \(2 \mathrm{~cm}\) in a capillary tube, If the tube is tilted \(60^{\circ}\) from the vertical, water will rise in the tube to a length of (1) \(4.0 \mathrm{~cm}\) (2) \(2.0 \mathrm{~cm}\) (3) \(1.0 \mathrm{~cm}\) (4) water will not rise at all

A rain drop starts falling from a height of \(2 \mathrm{~km}\). It falls with a continuously decreasing acceleration and attains its terminal velocity at a height of \(1 \mathrm{~km}\). The ratio of the work done by the gravitational force in the first half and the second half of the drops journey is (1) \(1: 1\) and the times of fall of the drop in the two halves is \(a: 1\) (where \(a>1\) ) (2) \(1: 1\) and the times of fall of the drop in the two halves is \(a: 1\) (where \(a<1\) ) (3) \(a: 1\) (where \(a>1\) ) and the times of fall of the drop in the two halves is \(1: 1\) (4) \(a: 1\) (where \(a<1\) ). and the times of fall of the drop in the two halves is \(1: 1\)

If \(n\) drops of a liquid, each with surface energy \(E\), join to form a single drop, then (1) some energy will be released in the process (2) some energy will be absorbed in the process (3) the energy released or absorbed will be \(E\left(n-n^{2 / 3}\right)\) (4) the energy released or absorbed will be \(n E\left(2^{23}-1\right)\)

A cubical block (of side \(2 \mathrm{~m}\) ) of mass \(20 \mathrm{~kg}\) slides on inclined plane lubricated with the oil of viscosity \(\eta=10^{-1}\) poise with constant velocity of \(10 \mathrm{~m} / \mathrm{s}\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). \(\mathrm{Th}_{\mathrm{h}}\) thickness of layer of liquid is \(\left(10^{-1}\right.\) poise \(\left.=10^{-2} \mathrm{~N} \mathrm{~s} \mathrm{~m}^{2}\right)\) (1) \(2.5 \mathrm{~mm}\) (2) \(4 \mathrm{~mm}\) (3) \(6 \mathrm{~mm}\) (4) \(5.5 \mathrm{~mm}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.