/*! 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 2 A wooden stick \(2 m\) long is f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 2

A wooden stick \(2 m\) long is floating on the surface of water. The surface tension of water \(0.07 \mathrm{~N} / \mathrm{m}\). By putting soap solution on one side of the sticks the surface tension is reduced to \(0.06 \mathrm{~N} / \mathrm{m}\). The net force on the stick will be (a) \(0.07 N\) (b) \(0.06 \mathrm{~N}\) (c) \(0.01 N\) (d) \(0.02 N\)

Short Answer

Expert verified
The net force on the stick is \(0.02 N\), so the answer is (d) \(0.02 N\).

Step by step solution

01

Identify and Understand the Variables

In this problem, we need to identify the given variables and the formula we need. The length of the stick is \(2 m\), surface tension of water is \(0.07 \mathrm{~N} / \mathrm{m}\) and that of soap solution is \(0.06 \mathrm{~N} / \mathrm{m}\). To find the net force, we apply the concept of difference in surface tensions.
02

Using the concept of Surface Tension

Surface tension is defined as the force acting at right angles to any line of unit length drawn on the surface. Thus, the force due to surface tension on either side of the stick equals the surface tension times the length of the stick.
03

Calculation of the Force due to Water

The force on one side of the stick due to water is given by multiplying the surface tension of water by the length of the stick: \(F1 = \text{Surface tension on water side} \times \text{length of stick} = 0.07 \mathrm{~N/m} \times 2 \mathrm{m} = 0.14 \mathrm{N}\).
04

Calculation of the Force due to Soap Solution

Similarly, the force on the other side of the stick due to soap solution is: \(F2 = \text{Surface tension on soap side} \times \text{length of stick} = 0.06 \mathrm{~N/m} \times 2 \mathrm{m} = 0.12 \mathrm{N}\).
05

Net Force Calculation

The net force on the stick will be the difference between the force on the water side and the force on the soap side that is \(F = F1 - F2 = 0.14 \mathrm{N} - 0.12 \mathrm{N} = 0.02 \mathrm{N}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 work done in blowing a soap bubble of \(10 \mathrm{~cm}\) radius is (surface tension of the soap solution is \(\left.\frac{3}{100} N / m\right)\) [MP PMT 1995; MH CET 2002] (a) \(75.36 \times 10^{-4} J\) (b) \(37.68 \times 10^{-4} J\) (c) \(150.72 \times 10^{-4} J\) (d) \(75.36 J\)

The radii of two soap bubbles are \(r_{1}\) and \(r_{2}\). In isothermal conditions, two meet together in vacuum. Then the radius of the resultant bubble is given by \(\quad\) [RPET 1999; MP PMT 2001; EAMCET 2003] (a) \(R=\left(r_{1}+r_{2}\right) / 2\) (b) \(R=r_{1}\left(r_{1} r_{2}+r_{2}\right)\) (c) \(R^{2}=r_{1}^{2}+r_{2}^{2}\) (d) \(R=r_{1}+r_{2}\)

An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true [Roorkee 2000] (a) Bubble rises upwards because pressure at the bottom is less than that at the top (b) Bubble rises upwards because pressure at the bottom is greater than that at the top (c) As the bubble rises, its size increases (d) As the bubble rises, its size decreases

A drop of mercury of radius \(2 \mathrm{~mm}\) is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is \(0.465 \mathrm{~J} / \mathrm{m}^{2}\) ) (a) \(23.4 \mu J\) (b) \(18.5 \mu \mathrm{J}\) (c) \(26.8 \mu \mathrm{J}\) (d) \(16.8 \mu J\)

The pressure inside a small air bubble of radius \(0.1 \mathrm{~mm}\) situated just below the surface of water will be equal to (Take surface tension of water \(70 \times 10^{-3} \mathrm{Nm}^{-1}\) and atmospheric pressure \(=1.013 \times 10^{5} \mathrm{Nm}^{-2}\) ) [AMU (Med.) 2002] (a) \(2.054 \times 10^{3} \mathrm{~Pa}\) (b) \(1.027 \times 10^{3} \mathrm{~Pa}\) (c) \(1.027 \times 10^{5} \mathrm{~Pa}\) (d) \(2.054 \times 10^{5} \mathrm{~Pa}\)
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Electrostatics

Read Explanation

Classical Mechanics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Quantum Physics

Read Explanation

Engineering Physics

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.