/*! 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 23 Two equal mases \(m\) and \(m\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 23

Two equal mases \(m\) and \(m\) are hung from balance whose scale pans differ in vertical height by \(h\). Calculate the error in weighing, if any, in terms of density of earth \(\rho\). (a) \(\frac{2}{3} \pi \rho R^{3} \mathrm{Gm}\) (b) \(\frac{8}{3} \pi \rho G m h\) (c) \(\frac{8}{3} \pi \rho R^{3} G m\) (d) \(\frac{4}{3} \pi \rho G m^{2} h\)

Short Answer

Expert verified
The error is given by option (b): \(\frac{8}{3} \pi \rho Gmh\).

Step by step solution

01

Understand the problem

We need to determine the error in weighing when two equal masses \(m\) and \(m\) are hung from balance scales differing in height by \(h\). The problem involves calculating this error in terms of the density of Earth \(\rho\).
02

Analyze the weight difference

The gravitational force acting on each mass is \(F = mg\), where \(g\) is the acceleration due to gravity. However, at different heights, \(g\) may have a negligible change due to altitude based on Earth's density.
03

Use the concept of gravitational field variation

The gravitational field at a height \(h\) above the Earth's surface can be approximated using: \( g' = g(1 - \frac{2h}{R}) \), where \(R\) is the Earth's radius. This gives rise to a slight difference in gravitational force experienced by the masses.
04

Determine the gravitational field contribution

The change in gravitational field due to this height difference is a factor of \(\Delta g = g - g' = g\frac{2h}{R}\). Therefore, the force difference \(\Delta F\) between two masses at different heights becomes: \( \Delta F = m\Delta g = mg\frac{2h}{R}\).
05

Consider the effect of Earth's density

The change in weight due to Earth's density \(\rho\) can be connected through the volume of Earth considered. Since the shape is spherical, use the formula \(V = \frac{4}{3} \pi R^3\). Substitute \(R^3\) in terms of mass \(\rho = \frac{M}{V}\) to find the relationship.
06

Establish the error expression

Inserting Earth's density \(\rho\) and simplifying leads to: \( \Delta W = mg\frac{2h}{R} \) translating to an error term \( \frac{8}{3} \pi \rho Gmh \), since the gravitational constant \(G\) links mass, distance, and resulting gravitational force.
07

Select the correct option

This expression for error \( \frac{8}{3} \pi \rho Gmh \) corresponds to option (b). Thus, the error due to height difference considering Earth's density is given by this term.

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!

Key Concepts

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

Error in Weighing
When weighing two equal masses at different heights, a difference in gravitational force can occur. This is due to variations in the gravitational field strength caused by the height difference. As a result, the measurement can show a slight error.

The error in weighing arises because the acceleration due to gravity is slightly less at a higher altitude. This means the mass at the higher scale pan experiences less gravitational force than the mass at the lower pan. This discrepancy is generally very small but becomes significant for precise measurements. In the context of our problem, we calculate this as \( \Delta F = mg\frac{2h}{R} \).

This outcome underscores the importance of considering altitude when calibrating scales for highly precise applications.
Density of Earth
The density of Earth, \( \rho \), plays an essential role in understanding gravitational variations. Density refers to the mass per unit volume of Earth, fundamentally affecting how gravitational force diminishes with altitude.

To calculate the Earth's density, we use its mass \(M\) and volume \(V\). The formula is \( \rho = \frac{M}{V} \). For a spherical Earth, the volume is represented by \( V = \frac{4}{3} \pi R^3 \), where \( R \) is Earth's radius. This density helps form a basis for calculating changes in gravity with height.

These changes, though minimal near Earth's surface, influence measurements like weighing, which can be crucial for scientific and engineering applications.
Acceleration Due to Gravity
Acceleration due to gravity, symbolized as \( g \), is the force that attracts objects towards the Earth. It's approximately \( 9.81 \, m/s^2 \) at sea level. However, it's not a fixed value everywhere.

The formula \( g' = g(1 - \frac{2h}{R}) \) helps predict how \( g \) changes slightly with elevation \( h \). This diminutive variation is because gravity lessens as you move further from the Earth's center. When scales at differing heights are used, this change affects the measured weight.

Understanding \( g \) is pivotal in physics as it affects how objects behave under Earth's pull, including falling objects and the operation of pendulums and springs.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a fundamental component of gravitational equations. Its value is approximately \( 6.674 \times 10^{-11} \, N(m/kg)^2 \).

It's crucial when relating mass, force, and distance, particularly in Newton's law of universal gravitation: \( F = G\frac{m_1m_2}{r^2} \). It allows us to calculate the force between two masses \( m_1 \) and \( m_2 \) placed at a distance \( r \).

In our context, \( G \) helps to connect the error in weighing with Earth's density and the height difference. It acts as a bridge between these physical properties, leading to insights into the effect of altitude on weight.

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

A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is (Radius of earth \(=6400 \mathrm{~km}\) and \(g=9.8 \mathrm{~ms}^{-2}\) ) (a) \(11.2 \mathrm{kms}^{-1}\) (b) \(8 \mathrm{kms}^{-1}\) (c) \(3.2 \mathrm{kms}^{-1}\) (d) \(1.5 \mathrm{kms}^{-1}\)

Two identical satellites are at \(R\) and \(7 R\) away from earth surface, the wrong statement is ( \(R=\) Radius of earth) (a) ratio of total energy will be \(y\) (b) ratio of kinetic energies will be \(y\) (c) ratio of potential energies will be \(y\) (d) ratio of total energy will be 4 but ratio of potential and kinetic energy will be 1

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is (a) \(32000 \mathrm{~km}\) (b) \(36000 \mathrm{~km}\) (c) \(6400 \mathrm{~km}\) (d) \(4800 \mathrm{~km}\)

A satellite orbits the earth at a height of \(400 \mathrm{~km}\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 \mathrm{~kg}\), mass of the earth \(=6.0 \times 10^{24} \mathrm{~kg}\), radius of the earth \(=6.4 \times 10^{6} \mathrm{~m}, G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\). (a) \(5.2 \times 10^{10} \mathrm{~J}\) (b) \(3 \times 10^{6} \mathrm{~J}\) (c) \(4 \times 10^{6} \mathrm{~J}\) (d) \(6 \times 10^{9} \mathrm{~J}\)

Two bodies of masses \(100 \mathrm{~kg}\) and \(1000 \mathrm{~kg}\) are separated by a distance of \(1 \mathrm{~m}\). What is the intensity of gravitational field at the mid point of the line joining them? (a) \(6.6 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\) (b) \(2.4 \times 10^{-8} \mathrm{Nkg}^{-1}\) (c) \(2.4 \times 10^{-7} \mathrm{Nkg}^{-1}\) (d) \(2.4 \times 10^{-6} \mathrm{Nkg}^{-1}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Electromagnetism

Read Explanation

Fluids

Read Explanation

Nuclear Physics

Read Explanation

Measurements

Read Explanation

Famous Physicists

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.