Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 985

A thin liquid film formed between a u-shaped wire and a light slider supports a weight of \(1.5 \times 10^{-2} \mathrm{~N}\) (see figure). The length of the slider is \(30 \mathrm{~cm}\) and its weight negligible. The surface tension of the liquid film is. (A) \(0.0125 \mathrm{Nm}^{-1}\) (B) \(0.1 \mathrm{Nm}^{-1}\) (C) \(0.05 \mathrm{Nm}^{-1}\) (D) \(0.025 \mathrm{Nm}^{-1}\)

Short Answer

Expert verified
The surface tension of the liquid film is \(0.025\ \mathrm{Nm}^{-1}\) (D).

Step by step solution

01

Convert the length of the slider from centimeters to meters

Before proceeding with the calculation, we need to convert the given slider length from centimeters to meters. Length of slider in meters = Length of the slider in centimeters/100 For a \(30\ \mathrm{cm}\) slider: Length of slider in meters = \(30/100 = 0.3\ \mathrm{m}\)
02

Set up the equation for surface tension

Now that we have the length of the slider in meters, we can set up the equation for surface tension using the force, length, and number of interfaces: \[F = T\cdot L\cdot n\]
03

Calculate the surface tension

Substitute the given force, length in meters, and the number of interfaces into the equation: \(1.5 \times 10^{-2}\ \mathrm{N} = T\cdot (0.3\ \mathrm{m})\cdot2\) Divide both sides of the equation by \(0.6\ \mathrm{m}\): \(T = \frac{1.5 \times 10^{-2}\ \mathrm{N}}{0.6\ \mathrm{m}}\) Calculate the value of T: \(T = 0.025\ \mathrm{Nm}^{-1}\)
04

Choose the correct answer

The surface tension of the liquid film is \(0.025\ \mathrm{Nm^{-1}}\), which corresponds to answer choice (D).

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 temperature at which the vapour pressure of a liquid becomes equals of the external pressure is its. (A) Melting point (B) sublimation point (C) Critical temperature (D) Boiling point

Mercury thermometers can be used to measure temperatures up to (A) \(100^{\circ} \mathrm{C}\) (B) \(212^{\circ} \mathrm{C}\) (C) \(360^{\circ} \mathrm{C}\) (D) \(500^{\circ} \mathrm{C}\)

Read the assertion and reason carefully and mark the correct option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. Assertion: Tiny drops of liquid resist deforming forces better than bigger drops. Reason: Excess pressure inside a drop is directly proportional to surface tension. (A) a (B) \(b\) (C) (D) d

In a capillary tube water rises by \(1.2 \mathrm{~mm}\). The height of water that will rise in another capillary tube having half the radius of the first is (A) \(1.2 \mathrm{~mm}\) (B) \(2.4 \mathrm{~mm}\) (C) \(0.6 \mathrm{~mm}\) (D) \(0.4 \mathrm{~mm}\)

A spherical drop of coater has radius \(1 \mathrm{~mm}\) if surface tension of contex is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) difference of pressures between inside and outside of the spherical drop is (A) \(35 \mathrm{~N} / \mathrm{m}^{2}\) (B) \(70 \mathrm{~N} / \mathrm{m}^{2}\) (C) \(140 \mathrm{~N} / \mathrm{m}^{2}\) (D) zero
See all solutions

Recommended explanations on English Textbooks

Language and Social Groups

Read Explanation

Creative Story

Read Explanation

Synthesis Essay

Read Explanation

Research and Composition

Read Explanation

Rhetoric

Read Explanation

Essay Plan

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