Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 84

A stone of weight \(W\) has specific gravity \(2.50 .\) (a) When the stone is suspended from a scale and submerged in water, what is the scale reading in terms of its weight in air? (b) What is the scale reading for the stone when it is submerged in oil (specific gravity \(=0.90\) )?

Short Answer

Expert verified
(a) The scale reading for the stone submerged in water is \(W_\text{apparent, water} = \frac{1.5W}{2.5}\). (b) The scale reading for the stone submerged in oil is \(W_\text{apparent, oil} = \frac{11W}{25}\).

Step by step solution

01

Find the stone's volume

Using the given specific gravity of the stone (2.50), we can find its density: \(d_\text{stone} = 2.50 \times d_\text{water} = 2.50 \times 1000 \ \text{kg}/\text{m}^3 = 2500 \ \text{kg}/\text{m}^3\). Now, we can find the stone's volume using its weight and density: \(V = \frac{W}{d_\text{stone} \times g} = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2}\).
02

Calculate the buoyant force in water

In water, the buoyant force can be found using the stone's volume and the density of water: \(F_\text{buoyant, water} = V \times d_\text{water} \times g = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2} \times 1000 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2 = \frac{W}{2.5}\).
03

Calculate the apparent weight in water

To find the apparent weight of the stone in water, we can subtract the buoyant force from its actual weight: \(W_\text{apparent, water} = W - F_\text{buoyant, water} = W - \frac{W}{2.5} = \frac{2.5W - W}{2.5} = \frac{1.5W}{2.5}\).
04

Calculate the buoyant force in oil

For the stone submerged in oil, we need to use the given specific gravity of the oil (0.9) to find its density: \(d_\text{oil} = 0.9 \times d_\text{water} = 0.9 \times 1000 \ \text{kg}/\text{m}^3 = 900 \ \text{kg}/\text{m}^3\). Now, calculate the buoyant force in oil: \(F_\text{buoyant, oil} = V \times d_\text{oil} \times g = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2} \times 900 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2 = \frac{W}{2.5} \times \frac{9}{10}\).
05

Calculate the apparent weight in oil

To find the apparent weight of the stone in oil, subtract the buoyant force from its actual weight: \(W_\text{apparent, oil} = W - F_\text{buoyant, oil} = W - \frac{W}{2.5} \times \frac{9}{10} = \frac{W}{2.5} + \frac{W}{25} = \frac{W(10 + 1)}{25}\). (a) The scale reading for the stone submerged in water is \(W_\text{apparent, water} = \frac{1.5W}{2.5}\). (b) The scale reading for the stone submerged in oil is \(W_\text{apparent, oil} = \frac{11W}{25}\).

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

A block of wood, with density \(780 \mathrm{kg} / \mathrm{m}^{3},\) has a cubic shape with sides \(0.330 \mathrm{m}\) long. A rope of negligible mass is used to tie a piece of lead to the bottom of the wood. The lead pulls the wood into the water until it is just completely covered with water. What is the mass of the lead? [Hint: Don't forget to consider the buoyant force on both the wood and the lead.]
A lid is put on a box that is \(15 \mathrm{cm}\) long, \(13 \mathrm{cm}\) wide, and \(8.0 \mathrm{cm}\) tall and the box is then evacuated until its inner pressure is \(0.80 \times 10^{5} \mathrm{Pa} .\) How much force is required to lift the lid (a) at sea level; (b) in Denver, on a day when the atmospheric pressure is \(67.5 \mathrm{kPa}\) ( \(\frac{2}{3}\) the value at sea level)?
An aluminum cylinder weighs \(1.03 \mathrm{N}\). When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is $3.90 \times 10^{-5} \mathrm{m}^{3} .$ If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is \(0.730 \mathrm{N}\). What is the specific gravity of the alcohol? (tutorial: ball in beaker).
An air bubble of 1.0 -mm radius is rising in a container with vegetable oil of specific gravity 0.85 and viscosity \(0.12 \mathrm{Pa} \cdot \mathrm{s} .\) The container of oil and the air bubble are at \(20^{\circ} \mathrm{C} .\) What is its terminal velocity?
In the Netherlands, a dike holds back the sea from a town below sea level. The dike springs a leak \(3.0 \mathrm{m}\) below the water surface. If the area of the hole in the dike is \(1.0 \mathrm{cm}^{2},\) what force must the Dutch boy exert to save the town?
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Classical Mechanics

Read Explanation

Energy Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Turning Points in 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.

Sign-up for free