Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 996

The relation between surface tension T. Surface area \(\mathrm{A}\) and surface energy \(\mathrm{E}\) is given by. (A) \(\mathrm{T}=(\mathrm{E} / \mathrm{A})\) (B) \(\mathrm{T}=\mathrm{EA}\) (C) \(\mathrm{E}=(\mathrm{T} / \mathrm{A})\) (D) \(\mathrm{T}=(\mathrm{A} / \mathrm{E})\)

Short Answer

Expert verified
The correct relation between surface tension (T), surface area (A), and surface energy (E) is given by option (B): \(T=EA\).

Step by step solution

01

Recall the definition of surface tension

Surface tension is a property of liquid surfaces which makes them behave as if they are covered with a stretched elastic membrane. This tension arises due to an imbalance in the molecular cohesive forces at the surface. Surface energy is the extra energy that molecules at the surface possess due to the imbalance of force.
02

Relate surface tension to surface energy and surface area

The relation between surface tension, surface energy, and surface area can be established by considering a change in surface area. The work done to increase the surface area is equal to the increase in surface energy. Mathematically, this can be represented as: \[ dE = T dA \] This equation states that the differential work (\(dE\)) required to change the surface area by a small amount (\(dA\)) is equal to the product of the surface tension (T) and change in surface area (\(dA\)). Integrating both sides of the equation will give us the relation between surface energy, surface tension, and surface area: \[ E = T A \]
03

Identify the correct option

Now that we have derived the relationship between surface tension (T), surface area (A), and surface energy (E): \[ E = TA \] We can see that the correct relation corresponds to option (B): \(T=EA\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The upper end of a wire of radius \(4 \mathrm{~mm}\) and length $100 \mathrm{~cm}$ is clamped and its other end is twisted through an angle of \(30^{\circ}\). Then what is the angle of shear? (A) \(12^{\circ}\) (B) \(0.12^{\circ}\) (C) \(1.2^{\circ}\) (D) \(0.012^{\circ}\)

The force required to separate two glass plates of area $10^{-2} \mathrm{~m}^{2}\( with a film of water \)0.05 \mathrm{~mm}$ thick between them is (surface tension of water is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) ) (A) \(28 \mathrm{~N}\) (B) \(14 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(38 \mathrm{~N}\)

The work done in blowing a soap bubble of \(10 \mathrm{~cm}\) radius is [surface tension of soap solution is \(\\{(3 / 100) \mathrm{N} / \mathrm{m}\\}\) ] (A) \(75.36 \times 10^{-4}\) Joule (B) \(37.68 \times 10^{-4}\) Joule (C) \(150.72 \times 10^{-4}\) Joule (D) \(75.36\) Joule

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}$ the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

If the length of a cylinder on heating increases by \(2 \%\) the area of its base will increase by. (A) \(0.5 \%\) (B) \(2 \%\) (C) \(1 \%\) (D) \(4 \%\)
See all solutions

Recommended explanations on English Textbooks

Semiotics

Read Explanation

Rhetoric

Read Explanation

Multiple Choice Questions

Read Explanation

Single Paragraph Essay

Read Explanation

Argumentative Essay

Read Explanation

5 Paragraph Essay

Read Explanation
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.

Sign-up for free