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Chapter 9: Problem 35

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).

Short Answer

Expert verified
Answer: The specific gravity of the alcohol is approximately 0.7846.

Step by step solution

01

Understand Archimedes' Principle

Archimedes' Principle states that the buoyant force (upward force) on a submerged object is equal to the weight of the fluid displaced by that object. In this case, the fluid is the alcohol, and the object is the aluminum cylinder.
02

Calculate the buoyant force

The buoyant force can be calculated as the difference between the weight of the cylinder in air and the weight of the cylinder when it is submerged in alcohol. Buoyant force (\(F_B\)) = Weight in air - Submerged weight = \(1.03 \mathrm{N}\) - \(0.730 \mathrm{N}\) = \(0.300 \mathrm{N}\)
03

Find the weight of the displaced alcohol

The buoyant force is equal to the weight of the fluid displaced, so the weight of the displaced alcohol is also \(0.300 \mathrm{N}\).
04

Calculate the mass of the displaced alcohol

To calculate the mass of the displaced alcohol, we can use the formula: weight = mass × acceleration due to gravity (\(g\)). In this case, we know the weight of the displaced alcohol (\(0.300 \mathrm{N}\)) and the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)). Mass of the displaced alcohol = \(\frac{Weight}{g}\) = \(\frac{0.300 \mathrm{N}}{9.81 \mathrm{m/s^2}}\) = \(0.0306 \mathrm{kg}\)
05

Calculate the density of the alcohol

To find the density of the alcohol, we will use the formula: density = \(\frac{mass}{volume}\). We are given the volume of the displaced alcohol as \(3.90 \times 10^{-5} \mathrm{m}^{3}\), and we have found the mass of the displaced alcohol (\(0.0306 \mathrm{kg}\)). Density of the alcohol = \(\frac{0.0306 \mathrm{kg}}{3.90 \times 10^{-5} \mathrm{m}^{3}}\) = \(784.62 \mathrm{kg/m^3}\)
06

Find the specific gravity of the alcohol

Specific gravity is a ratio of the density of a fluid to the density of a standard reference fluid (usually water). Specific gravity is a unitless quantity, and for water, specific gravity = 1. The density of water is approximately \(1000 \mathrm{kg/m^3}\). Specific gravity of the alcohol = \(\frac{Density \ of \ alcohol}{Density \ of \ water}\) = \(\frac{784.62 \ \mathrm{kg/m^3}}{1000 \ \mathrm{kg/m^3}}\) = \(0.7846\) The specific gravity of the alcohol is approximately 0.7846.

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