Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 143

A not-too-honest citizen is thinking of making bogus gold bars by first making a hollow iridium \((S=22.5)\) ingot and plating it with a thin layer of gold \((S=19.3)\) of negligible weight and volume. The bogus bar is to have a mass of 100 lbm. What must be the volumes of the bogus bar and of the air space inside the iridium so that an inspector would conclude it was real gold after weighing it in air and water to determine its density? Could lead \((S=11.35)\) or platinum \((S=21.45)\) be used instead of iridium? Would either be a good idea?

Short Answer

Expert verified
The volume of the fake gold bar hence is approximately \( \frac{100}{19.3} \) ft³ and the volume of the airspace inside the iridium is approximately \( \frac{100}{19.3} - \frac{100}{22.5} \) ft³. Depending on whether the calculated apparent volume for lead/platinum surpasses their real volume, it could be concluded if they can be used instead of iridium.

Step by step solution

01

Identify Knowns and Unknowns

We know that the specific gravity (S) of iridium is 22.5, of gold is 19.3, of lead is 11.35, and of platinum is 21.45. The mass of the bar to be made is 100 lbm. We need to find the volume of both the fake iridium bar and the hollow space inside it.
02

Calculate the Volume of the Fake Gold Bar

The apparent volume of the bar (V1) in air can be calculated by the formula V1 = mass/specific gravity. Where mass is 100 lbm and specific gravity of gold is 19.3. So, \( V1 = \frac{100}{19.3} \) ft³.
03

Calculate the Volume of Iridium Used

The volume of iridium or real volume (V2) needed is determined by the formula V2 = mass/specific gravity, where specific gravity is that of iridium (22.5). So, \( V2 = \frac{100}{22.5} \) ft³.
04

Determine the Volume of the Hollow Space

The volume of the hollow space inside the iridium can thus be calculated by subtracting the volume of iridium used from the apparent volume of the bar. i.e. \(V = V1 - V2 = \frac{100}{19.3} - \frac{100}{22.5} \) ft³.
05

Decide if Lead or Platinum Can be Used

To decide whether lead or platinum could be used instead, we calculate the volume V1 for these two metals using their respective specific gravities. If V1 obtained is greater than their real volume, they could be used instead. But the volume of the hollow space should remain the same, which is \( \frac{100}{19.3} - \frac{100}{22.5} \) ft³.

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

Often young children drink milk \(\left(\rho=1030 \mathrm{kg} / \mathrm{m}^{3}\right)\) through a straw. Determine the maximun length of a vertical straw that a child can use to empty a milk container, assuming that the child can develop \(75 \mathrm{mm}\) Hg of suction, and use this answer to determine if you think this is a reasonable estimate of the suction that a child can develop.

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

A spherical balloon filled with helium at \(40^{\circ} \mathrm{F}\) and 20 psia has a 25 -ft diameter. What load can it support in atmospheric air at \(40^{\circ} \mathrm{F}\) and 14.696 psia? Neglect the balloon's weight.

A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the balloon tilts to the right. On the other hand, the child tends to be forced forward as the car decelerates and to the left as the car makes a right-hand turn. Explain these observed effects on the balloon and child.

Because of elevation differences, the water pressure in the second floor of your house is lower than it is in the first floor. For tall buildings this pressure difference can become unacceptable. Discuss possible ways to design the water distribution system in very tall buildings so that the hydrostatic pressure difference is within acceptable limits.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Physics of Motion

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Quantum Physics

Read Explanation

Kinematics Physics

Read Explanation

Magnetism

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