Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 980

A beaker of radius \(15 \mathrm{~cm}\) is filled with liquid of surface tension \(0.075 \mathrm{~N} / \mathrm{m}\). Force across an imaginary diameter on the surface of liquid is (A) \(0.075 \mathrm{~N}\) (B) \(1.5 \times 10^{-2} \mathrm{~N}\) (C) \(0.225 \mathrm{~N}\) (D) \(2.25 \times 10^{-2} \mathrm{~N}\)

Short Answer

Expert verified
The force across the imaginary diameter on the surface of the liquid is 0.225 N, so the correct answer is (C) \(0.225 \mathrm{~N}\).

Step by step solution

01

Identify the formula for force due to surface tension

We have surface tension (T) and radius of the beaker (R). We will use the formula for the force (F) exerted by the surface tension of the liquid acting across the imaginary diameter: \(F = 2 \cdot T \cdot L\), where L is the length along which the force is acting. Since we are considering an imaginary diameter, L is equal to the diameter of the beaker.
02

Calculate the diameter of the beaker

We are given the radius of the beaker, R = 15 cm. The diameter (D) is twice the value of R. So, the diameter is calculated as follows: \(D = 2 \cdot R = 2 \cdot 15 \mathrm{~cm} = 30 \mathrm{~cm}\). We'll convert the diameter to meters by dividing by 100: \(D = 0.30 \mathrm{~m}\).
03

Calculate the force due to surface tension

To find the force, we'll plug the values of surface tension and diameter into the formula we identified in step 1: \(F = 2 \cdot T \cdot L = 2 \cdot 0.075 \mathrm{~N/m} \cdot 0.30 \mathrm{~m} = 0.45 \mathrm{~N}\). However, since the force is acting across an imaginary diameter, we need to divide the calculated force by 2: \(F = \frac{0.45 \mathrm{~N}}{2} = 0.225 \mathrm{~N}\). The force across the imaginary diameter on the surface of the liquid is 0.225 N, so the correct answer is (C) \(0.225 \mathrm{~N}\).

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

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) \(0.4 \mathrm{~mm}\)

The density \(\rho\) of coater of bulk modulus \(B\) at a depth \(y\) in the ocean is related to the density at surface \(\rho_{0}\) by the relation. (A) $\rho=\rho_{0}\left[1-\left\\{\left(\rho_{0} \mathrm{gy}\right\\} / \mathrm{B}\right\\}\right]$ (B) $\rho=\rho_{0}\left[1+\left\\{\left(\rho_{0} \mathrm{gy}\right\\} / \mathrm{B}\right\\}\right]$ (C) $\rho=\rho_{0}\left[1+\left\\{\left(\rho_{0} \mathrm{gyh}\right\\} / \mathrm{B}\right\\}\right]$ (D) $\rho=\rho_{0}\left[1-\left\\{\mathrm{B} /\left(\rho_{0} \mathrm{~g} \mathrm{y}\right\\}\right]\right.$

When \(100 \mathrm{~N}\) tensile force is applied to a rod of $10^{-6} \mathrm{~m}^{2}\( cross-sectional area, its length increases by \)1 \%$ so young's modulus of material is \(\ldots \ldots \ldots \ldots\) (A) \(10^{12} \mathrm{~Pa}\) (B) \(10^{11} \mathrm{~Pa}\) (C) \(10^{10} \mathrm{~Pa}\) (D) \(10^{2} \mathrm{~Pa}\)

Assertion and Reason: Read the assertion and reason carefully to mark the correct option out of the option given below (A) If both assertion and reason are true and 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 assertion and reason both are false. Assertion: The stretching of a coil is determined by its shear modulus. Reason: Shear modulus change only shape of a body keeping its dimensions unchanged. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

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
See all solutions

Recommended explanations on English Textbooks

Email

Read Explanation

Syntax

Read Explanation

Discourse

Read Explanation

Essay Plan

Read Explanation

Global English

Read Explanation

Morphology

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